Rd Sharma 2021 Solutions for Class 10 Maths Chapter 13 Areas Related To Circles are provided here with simple step-by-step explanations. These solutions for Areas Related To Circles are extremely popular among Class 10 students for Maths Areas Related To Circles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2021 Book of Class 10 Maths Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2021 Solutions. All Rd Sharma 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

#### Page No 13.12:

#### Question 1:

Find the circumference and area of a circle of radius 4.2 cm.

#### Answer:

We know that the circumference *C* and area *A* of a circle of radius *r* are given by and respectively.

Here,

So substituting the value of *r* in above formulas,

Circumference of the circle

Area of the circle

#### Page No 13.12:

#### Question 2:

Find the circumference of a circle whose area is 301.84 cm^{2}.

#### Answer:

Let* r* cm be the radius of the circle. Then

Area of a circle is

We know that the Circumference of circle of radius *r* is

So substituting the value of r in above formula

#### Page No 13.12:

#### Question 3:

Find the area of a circle whose circumference is 44 cm.

#### Answer:

Let* r* be the radius of the circle. Then

Circumference of the circle

We know that the area of a circle of radius *r *is

Substituting the value of *r* in above formula

#### Page No 13.12:

#### Question 4:

The circumference of a circle exceeds the diameter by 16.8 cm. Find the circumference of the circle.

#### Answer:

Let the radius of a circle be *r* cm, then diameter of circle is 2*r* cm and Circumference iscm.

It is given that the circumference exceeds the diameter of circle by 16.8 cm.

So,

Now the circumference is

#### Page No 13.12:

#### Question 5:

A horse is tied to a pole with 28 m long string. Find the area where the horse can graze. (Take π = 22/7).

#### Answer:

We know that the horse is tied to a pole with 28 m long string. So the horse can graze the area of a circle of radius 28 m.

Area of circle is

Hence the horse can graze area.

#### Page No 13.12:

#### Question 6:

A steel wire when bent in the form of a square encloses an area of 121 cm^{2}. If the same wire is bent in the form of a circle, find the area of the circle.

#### Answer:

Let *a* cm be the side of square. Then area of square is

We have,

Let the radius of circle be *r* cm. Then,

Now, we will calculate area of circle.

#### Page No 13.12:

#### Question 7:

The circumference of two circles are in the ratio 2 : 3. Find the ratio of their areas.

#### Answer:

Let the radius of two circles beandrespectively. Then their circumferences are and respectively and their areas are and respectively.

It is given that,

Now we will calculate the ratio of their areas,

Substituting the value of ,

Hence the ratio of their Areas is.

#### Page No 13.12:

#### Question 8:

The sum of the radii of two circles is 140 cm and the difference of their circumferences is 88 cm. Find the diameters of the circles.

#### Answer:

Let the radius of two circles beandrespectively. Then their circumferences are and respectively and their areas are and respectively.

It is given that the sum of the radii of two circles is 140 cm and difference of their circumferences is 88 cm. So,

……(A)

……(B)

Now, solving (A) and (B)

Thus diameters of circles are,

#### Page No 13.12:

#### Question 9:

Find the radius of a circle whose circumference is equal to the sum of the circumference of two circles of radii 15 cm and 18 cm .132974

#### Answer:

Let the radius be r cm.

The radius of circle A, r_{A}= 15 cm

The radius of circle A, r_{B }= 18

Circumference of circle = circumference of circle A + circumference of circle B

$\Rightarrow 2\mathrm{\pi r}=2{\mathrm{\pi r}}_{\mathrm{A}}+2{\mathrm{\pi r}}_{\mathrm{B}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{r}={\mathrm{r}}_{\mathrm{A}}+{\mathrm{r}}_{\mathrm{B}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{r}=15+18=33\mathrm{cm}$

#### Page No 13.12:

#### Question 10:

The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles.

#### Answer:

Let the radius of circles be , andrespectively. Then their areas are, and respectively.

It is given that,

We have, and

Substituting the values of ,

Hence, the radius of circle is.

#### Page No 13.12:

#### Question 11:

The radii of two circles are 19 cm and 9 cm respectively. Find the radius and area of the circles which has it circumference equal to the sum of the circumferences of the two circles.

#### Answer:

Let the radius of circles be, andrespectively. Then their circumferences are, and respectively.

It is given that,

We have, and

Substituting the values of,

Hence the radius of the circle is.

We know that the area *A* of circle is

Substituting the value of *r*

Hence the area of the circle is.

#### Page No 13.12:

#### Question 12:

The area of circular playground is 22176 m^{2} . Find the cost of fencing this ground at the rate of ₹ 50 per metre.

#### Answer:

Area of circular playground = 22176 m^{2}

$\Rightarrow {\mathrm{\pi r}}^{2}=22176\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{22}{7}{\mathrm{r}}^{2}=22176\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{r}}^{2}=\frac{22176\times 7}{22}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{r}}^{2}=7056\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{r}=84\mathrm{cm}$

circumference = $2\mathrm{\pi r}=2\times \frac{22}{7}\times 84=528{\mathrm{cm}}^{2}$

Cost of fencing the ground = $528\times 50=Rs26400$

#### Page No 13.12:

#### Question 13:

The side of a square is 10 cm. Find the area of circumscribed and inscribed circles.

#### Answer:

It is given that the side of square is 10 cm.

So, the diameter of circle inscribed the square is 10 cm.

We know that the area *A* of circle inscribed the square is

Substituting the value of radius of inscribed circle,

Hence the area of circle inscribed the square is

Now we will find the diameter of circle circumscribed the square.

So,

We know that the area of circle inscribed the square is

Substituting the value of radius,

Hence the area of circle circumscribed the square is .

#### Page No 13.12:

#### Question 14:

If a square is inscribed in a circle, find the ratio of the areas of the circle and the square.

#### Answer:

Let ABCD be the square inscribed in a circle of radius *r*.

Here, OA = OB = *r*.

∴ OA^{2} + OB^{2} = AB^{2}

⇒ *r*^{2} + *r*^{2} = AB^{2}

⇒ 2*r*^{2} = AB^{2}

Now, area of square ABCD

Now we will find the ratio of area of the circle and the square.

Hence, the ratio of area of the circle and square is.

#### Page No 13.12:

#### Question 15:

The area of a circle inscribed in an equilateral triangle is 154 cm^{2}. Find the perimeter of the triangle. [Use π = 22/7 and $\sqrt{3}$ = 1.73]

#### Answer:

It is given that the area *A* of circle inscribed in an equilateral triangle is 154 cm^{2}.

We know that the area *A* of circle inscribed in an equilateral triangle is

Now, we will find the value of *r.*

Substituting the value of area*,*

Let the height of triangle be *h.* Then

If *a* is the side of triangle, then

Substituting the value of *h,*

Hence perimeter of triangle is .

#### Page No 13.12:

#### Question 16:

A field is in the form of a circle. A fence is to be erected around the field. The cost fencing would be Rs. 2640 at the rate of Rs. 12 per metre. The, the field is to be thoroughly ploughed at the cost of Re. 0.50 per m2. What is the amount required to plough the field? [Take π = 22/7].

#### Answer:

We know that the circumference *C* of a circle of radius *r* is

It is given that cost of fencing around the circular field would be Rs.2640 at the rate of Rs.12 per meter. So,

We know that the area *A* of circle of radius *r,*

Substituting the value of *r*

Since,

Hence, amount required to plough the field is.

#### Page No 13.12:

#### Question 17:

A park is in the form of a rectangle 120 m × 100 m. At the centre of the park there is a circular lawn, The area of park excluding lawn is 8700 m^{2}. Find the radius of the circular lawn. (Use π = 22/7).

#### Answer:

Let the radius of circular lawn be *r*. Then,

It is given that

Hence, radius of circular lawn is.

#### Page No 13.12:

#### Question 18:

A car travels 1 kilometre distance in which each wheel makes 450 complete revolutions. Find the radius of the its wheels.

#### Answer:

Let the radius of wheel be *r*. Thus, circumference *C* of the wheel

Since car travels 1 km distance in which wheel makes 450 complete revolutions. Then

We know that,

Hence the radius of wheel is.

#### Page No 13.12:

#### Question 19:

The area enclosed between the concentric circles is 770 cm^{2}. If the radius of the outer circle is 21 cm, find the radius of the inner circle.

#### Answer:

Let the radius of outer and inner two circles be *r*_{1} and *r*_{2} respectively.

Area enclosed between concentric circles = $\mathrm{\pi}{{r}_{1}}^{2}-\mathrm{\pi}{{r}_{2}}^{2}$

$\Rightarrow 770=\frac{22}{7}\left({21}^{2}-{{r}_{2}}^{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 245={21}^{2}-{{r}_{2}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {{r}_{2}}^{2}=441-245\phantom{\rule{0ex}{0ex}}\Rightarrow {{r}_{2}}^{2}=196\phantom{\rule{0ex}{0ex}}\Rightarrow {{r}_{2}}^{2}={14}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}_{2}=14\mathrm{cm}$

Hence, the radius of inner circle is 14 cm.

#### Page No 13.12:

#### Question 20:

An archery target has three regions formed by three concentric circles as shown in figure15.8. If the diameters of the concentric circles are in the ratios 1 : 2 : 3 , then find the ratio of the areas of three regions .

#### Answer:

Let the three regions be A , B and C.

The diameters are in the ratio 1 : 2 : 3.

Let the diameters be 1*x*, 2*x* and 3*x*

Then the radius will be $\frac{x}{2},\frac{2x}{2}and\frac{3x}{2}$

Area of region A = ${\mathrm{\pi r}}_{\mathrm{A}}^{2}=\mathrm{\pi}{\left(\frac{\mathrm{x}}{2}\right)}^{2}=\frac{{\mathrm{\pi x}}^{2}}{4}$

Area of region B = ${\mathrm{\pi r}}_{\mathrm{B}}^{2}-{\mathrm{\pi r}}_{A}^{2}=\mathrm{\pi}{\left(\mathrm{x}\right)}^{2}-\mathrm{\pi}{\left(\frac{\mathrm{x}}{2}\right)}^{2}=\frac{3\mathrm{\pi}{\left(\mathrm{x}\right)}^{2}}{4}$

Area of region C = ${\mathrm{\pi r}}_{C}^{2}-{\mathrm{\pi r}}_{\mathrm{B}}^{2}-{\mathrm{\pi r}}_{A}^{2}=\mathrm{\pi}{\left(\frac{3\mathrm{x}}{2}\right)}^{2}-\mathrm{\pi}{\left(\mathrm{x}\right)}^{2}-\mathrm{\pi}{\left(\frac{\mathrm{x}}{2}\right)}^{2}=\mathrm{\pi}{\left(\frac{3\mathrm{x}}{2}\right)}^{2}-\frac{3{\mathrm{\pi x}}^{2}}{4}=\frac{5{\mathrm{\pi x}}^{2}}{4}$

Thus, ratio of the areas of regions A, B and C will be

$\frac{{\mathrm{\pi x}}^{2}}{4}$:$\frac{3\mathrm{\pi}{\left(\mathrm{x}\right)}^{2}}{4}$:$\frac{5{\mathrm{\pi x}}^{2}}{4}$

⇒1:3:5

#### Page No 13.13:

#### Question 21:

The wheel of a motor cycle is of radius 35 cm . How many revolutions per minute must the wheel make so as to keep a speed of 66 km / hr ?

#### Answer:

Radius of the wheel,* r *= 35 cm

circumference = $2\mathrm{\pi r}=2\mathrm{\pi}\times 35=220\mathrm{cm}$

Distance covered by the wheel in one revolution = 220 cm

Let the number of revolutions required be *x*.

So, distance covered by the wheel in x revolutions = 220*x *

Distance covered by the wheel in one minute = 220*x*

The wheel covers $\frac{66\times 1000\times 100}{60}=110000\mathrm{cm}$ in one minute

$So,220x=110000\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{110000}{220}=500$

Hence, 500 revolutions per min are needed.

#### Page No 13.13:

#### Question 22:

A circular pond is 17.5 m in diameter. It is surrounded by a 2m wide path, Find the cost of constructing the path at the rate of ₹ 25 per m^{2 }.

#### Answer:

Diameter of the pond = 17.5 m

Radius of the pond = 8.75 m

Radius of the pond with the path = 8.75 + 2 = 10.75 m

Area of the path = Area of the pond along with the path − area of the pond

$\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{path}=\mathrm{\pi}\left[{\mathrm{R}}^{2}-{\mathrm{r}}^{2}\right]\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\left[{\left(10.75\right)}^{2}-{\left(8.75\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\left[\left(2\right)\left(19.5\right)\right]\phantom{\rule{0ex}{0ex}}=122.46{\mathrm{m}}^{2}$

Cost of constructing the path = $25\times 122.46=\mathrm{Rs}3061.5$

#### Page No 13.13:

#### Question 23:

A circular park is surroundeed by a rod 21 m wide. If the radius of the park is 105 m, find the area of the road .

#### Answer:

Radius of the park, *r* = 105 m

Radius of the park with the road,* R* = 105 + 21 = 126 m

Area of the road = Area of the park with the road − Area of the park

$=\mathrm{\pi}{R}^{2}-\mathrm{\pi}{r}^{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\left[{\left(126\right)}^{2}-{\left(105\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\left(21\right)\left(231\right)\phantom{\rule{0ex}{0ex}}=15246{\mathrm{m}}^{2}$

#### Page No 13.13:

#### Question 24:

A square of diagonal 8 cm is inscribed in a circle. Find the area of the region lying outside the circle and inside the square .

#### Answer:

DISCLAIMER: There is some error in the given question.

We have solved the question by taking a square inscribed in a circle. Then finding the area inside the circle and outside the square.

Diagonal of the square = 8 cm.

Let the side of the square be *a* cm.

In triangle BCD,

$B{C}^{2}+C{D}^{2}=B{D}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}+{a}^{2}={8}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=32.....\left(\mathrm{i}\right)$

Radius of the circle will be *R* = 4 cm

Now area between the circle and the square will be

$\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{circle}-\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{square}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}{\left(R\right)}^{2}-{a}^{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}{\left(4\right)}^{2}-32\phantom{\rule{0ex}{0ex}}=\left(16\mathrm{\pi}-32\right){\mathrm{cm}}^{2}$

#### Page No 13.13:

#### Question 25:

A path of 4 m width runs round a semi-circular grassy plot whose circumference is $163\frac{3}{7}\mathrm{m}$. Find:

(i) the area of the path

(ii) the cost of gravelling the path at the rate of ₹ 1.50 per square metre

(iii) the cost of turfing the plot at the rate of 45 paise per m^{2}.

#### Answer:

We have given AB = 4m and circumference of semicircle with radius OA as.

We are asked to find the area between the two semi-circles.

For that we will first find OA.

$\mathrm{\pi}r=163\frac{3}{7}$

Now we will substitute

$\frac{22}{7}\times r=163\frac{3}{7}\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{1144}{7}\times \frac{7}{22}\phantom{\rule{0ex}{0ex}}\Rightarrow r=52\phantom{\rule{0ex}{0ex}}\Rightarrow OA=52m$

Now we will find OB.

$\therefore OB=52+4\phantom{\rule{0ex}{0ex}}\therefore OB=56\mathrm{m}$

Now we will find the area between two semi-circles as given below,

$\therefore \mathrm{Area}=\frac{\mathrm{\pi}\times 56\times 56}{2}-\frac{\mathrm{\pi}\times 52\times 52}{2}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi}}{2}\left(1568-1352\right){\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi}}{2}\times 216\phantom{\rule{0ex}{0ex}}=\frac{22}{7\times 2}\times 216\phantom{\rule{0ex}{0ex}}=339.43{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}$

Therefore, area of the path is 339.43 m^{2}.

Now we will find the cost of gravelling the path.

$\mathrm{Cost}=339.43\times 1.50\phantom{\rule{0ex}{0ex}}=\mathrm{Rs}509.14$

Therefore, cost of gravelling the path is Rs 509.14.

Now we will find the cost of turfing the plot. For that we will find the area of the plot.

$\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{plot}=\frac{1}{2}{\mathrm{\pi r}}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times \frac{22}{7}{\left(52\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times \frac{22}{7}\times 52\times 52\phantom{\rule{0ex}{0ex}}=4249.14\phantom{\rule{0ex}{0ex}}\mathrm{Cost}\mathrm{of}\mathrm{turfing}\mathrm{the}\mathrm{plot}=4249.14\times 0.45\phantom{\rule{0ex}{0ex}}=\mathrm{Rs}1912.11$

Therefore, cost of the turfing the plot is Rs 1912.11.

**Disclaimer: Due to some error in the question, we get the different answers.**

#### Page No 13.13:

#### Question 26:

Find the area enclosed between two concentric circles of radii 3.5 cm and 7 cm. A third concentric circle is drawn outside the 7 cm circle , such that the area enclosed between it and the 7 cm circle is same as that between the two inner circles . Find the radius of the third circle correct to one decimal place.

#### Answer:

The area enclosed between the two circles of radii 3.5 cm and 7 cm

$=\mathrm{\pi}\left({7}^{2}-3.{5}^{2}\right)\phantom{\rule{0ex}{0ex}}=115.5{\mathrm{cm}}^{2}$

Let the radius of the outermost circle be *r* cm.

Area betweent the circles with radius* r* and 7 cm=Area between the circles with radius 7 cm and 3.5 cm

$\mathrm{\pi}\left({\mathrm{r}}^{2}-{7}^{2}\right)=115.5\phantom{\rule{0ex}{0ex}}\Rightarrow \left({\mathrm{r}}^{2}-{7}^{2}\right)=\frac{115.5}{\mathrm{\pi}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{r}}^{2}=36.75+49=85.75{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{r}=9.26\mathrm{cm}$

#### Page No 13.13:

#### Question 27:

A path of width 3.5 m runs around a semi-circular grassy plot whose perimeter is 72 m . Find the area of the path . $\left(\mathrm{Use}\mathrm{\pi}=\frac{22}{7}\right)$

#### Answer:

Let the radius of the semicircular plot be *r*.

Perimeter of the semi-circular grassy plot = $\mathrm{\pi r}+2\mathrm{r}=72$

$\Rightarrow r=14\mathrm{cm}$

Given that the width of the plot = 3.5 m

Thus, the outer radius = 3.5 + 14 = 17.5 m

Area of the path = $\frac{{\mathrm{\pi R}}^{2}}{2}-\frac{{\mathrm{\pi r}}^{2}}{2}$

$=\frac{\mathrm{\pi}}{2}\left({R}^{2}-{r}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi}}{2}\left({\left(17.5\right)}^{2}-{\left(14\right)}^{2}\right)\phantom{\rule{0ex}{0ex}}=173.25{\mathrm{m}}^{2}$

#### Page No 13.13:

#### Question 28:

A circular pond is of diameter 17.5 m . It is surrounded by a 2m wide path . Find the cost of constructing the path at the rate of ₹25 per square metre $\left(Use\mathrm{\pi}=\frac{22}{7}\right)$

#### Answer:

Diameter, *d* = 17.5 m

Radius, *r* = $\frac{17.5}{2}\mathrm{m}$

Radius of the pond with the 2 m wide path = $2+\frac{17.5}{2}$ m

Area of the circular path = Area of the pond with the path − area of the pond

$=\mathrm{\pi}{\left(2+\frac{17.5}{2}\right)}^{2}-\mathrm{\pi}{\left(\frac{17.5}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=122.57{\mathrm{m}}^{2}$

Cost of constructing the path = $25\times 122.57=\mathrm{Rs}3064.2$

#### Page No 13.13:

#### Question 29:

The outer circumference of a circular race-track is 528 m . The track is everywhere 14 m wide. Calculate the cost of levelling the track at the rate of 50 paise per square metre . $\left(\mathrm{Use}\mathrm{\pi}=\frac{22}{7}\right)$

#### Answer:

Let the radius of the inner circle and the race track be *R* m.

Outer circumference of the race track = 528 m

$\Rightarrow 2\mathrm{\pi}R=528\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{528}{2\mathrm{\pi}}=84\mathrm{m}$

Total radius of the outer circle = 84 − 14 = 70 m

Area of the circular track = Area of the outer circle − area of inner circle

$=\mathrm{\pi}{\left(84\right)}^{2}-\mathrm{\pi}{\left(70\right)}^{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\left[\left(84-70\right)\left(84+70\right)\right]\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\left(14\right)\left(154\right)\phantom{\rule{0ex}{0ex}}=6776{\mathrm{m}}^{2}$

Cost of levelling the track = $0.5\times 6776=\mathrm{Rs}3388$

#### Page No 13.13:

#### Question 30:

A road which is 7 m wide surrounds a circular park whose circumference is 352 m . Find the area of the road .

#### Answer:

Width of the road = 7 m

circumference of the circular park = 352 m

$\Rightarrow 2\mathrm{\pi r}=352\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{r}=\frac{352}{2\mathrm{\pi}}=56\mathrm{m}$

Area of the road = Area of the circular park including the path − area of the circular park

$=\mathrm{\pi}{\left(\mathrm{r}+7\right)}^{2}-{\mathrm{\pi r}}^{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\left[{\left(56+7\right)}^{2}-{\left(56\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\left[{63}^{2}-{56}^{2}\right]\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\left[7\times 119\right]\phantom{\rule{0ex}{0ex}}=2618{\mathrm{m}}^{2}$

#### Page No 13.13:

#### Question 31:

Prove that the area of a circular path of uniform width* h* surrounding a circular region of radius* r* is $\mathrm{\pi h}\left(2r+h\right)$.

#### Answer:

The width of the circular path = h

Let the inner circle be region A and the outer circle be region B

Radius of region A =* r*

Radius of region B =* r + h*

Area of the circular path = Area of region B − Area of region A

$=\mathrm{\pi}{\left(\mathrm{r}+\mathrm{h}\right)}^{2}-{\mathrm{\pi r}}^{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\left({\mathrm{r}}^{2}+{\mathrm{h}}^{2}+2\mathrm{rh}-{\mathrm{r}}^{2}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{\pi h}\left(\mathrm{h}+2\mathrm{r}\right)$

Hence Proved

#### Page No 13.24:

#### Question 1:

Find, in terms of π, the length of the arc that subtends an angle of 30° at the centre of a circle of radius 4 cm.

#### Answer:

The arc length *l* of a sector of an angle *θ* in a circle of radius r is given by

It is given that and. Substituting the value of* r* and *θ* in above equation,

#### Page No 13.24:

#### Question 2:

Find the angle subtended at the centre of a circle of radius 5 cm by an arc of length (5π/3) cm.

#### Answer:

We know that the arc length *l* of a sector of an angle *θ* in a circle of radius *r* is

It is given that and length . Substituting these value in above equation,

Hence, the angle subtended at the centre of circle is.

#### Page No 13.24:

#### Question 3:

An arc of length 20π cm subtends an angle of 144° at the centre of a circle. Find the radius of the circle.

#### Answer:

We know that the arc length *l* of a sector of an angle *θ* in a circle of radius *r* is

It is given and angle.

Now we substitute the value of *l* and *θ* in above formula to find the value of radius *r* of circle.

#### Page No 13.24:

#### Question 4:

An arc of length 15 cm subtends an angle of 45° at the centre of a circle. Find in terms of π, the radius of the circle.

#### Answer:

We know that the arc length *l* of a sector of an angle *θ* in a circle of radius r is

It is given that and angle.

Now we substitute the value of *l* and *θ* in above formula to find the value of radius *r* of circle.

#### Page No 13.24:

#### Question 5:

Find the angle subtended at the centre of a circle of radius 'a' by an arc of length (aπ/4) cm.

#### Answer:

We know that the arc length *l* of a sector of an angle *θ* in a circle of radius *r* is

It is given and radius.

Now we substitute the value of *l* and *r* in above formula to find the value of angle *θ* subtended at the centre of circle.

#### Page No 13.24:

#### Question 6:

A sector of a circle of radius 4 cm contains an angle of 30°. Find the area of the sector.

#### Answer:

We know that the area *A* of a sector of an angle *θ* in the circle of radius *r* is given by

It is given that and angle.

Now we substitute the value of *r* and *θ* in above formula,

#### Page No 13.24:

#### Question 7:

A sector of a circle of radius 8 cm contains an angle of 135°. Find the area of the sector.

#### Answer:

We know that the area *A* of a sector of an angle *θ* in the circle of radius *r* is given by

It is given thatand.

Now we substitute the value of *r* and *θ* in above formula,

#### Page No 13.25:

#### Question 8:

The area of a sector of a circle of radius 2 cm is π cm^{2}. Find the angle contained by the sector.

#### Answer:

We know that the area *A* of a sector of an angle *θ* in the circle of radius *r* is given by

It is given that and area.

Now we substitute the value of *r* and *A* in above formula to find the value of *θ*,

#### Page No 13.25:

#### Question 9:

The area of a sector of a circle of radius 5 cm is 5 π cm^{2} . Find the angle contained by the sector.

#### Answer:

We know that the area *A* of a sector of an angle *θ* in the circle of radius *r* is given by

It is given that radius and area.

Now we substitute the value of *r* and *A* in above formula to find the value of *θ*,

#### Page No 13.25:

#### Question 10:

Find the area of sector of the circle of radius 5 cm , if the corresponding arc length is 3.5 cm .

#### Answer:

Radius, *r* = 5 cm

Arc length, *l* = 3.5 cm

$\mathrm{Area}=\frac{1}{2}lr$

$\Rightarrow A=\frac{1}{2}\left(3.5\right)\left(5\right)=8.75{\mathrm{cm}}^{2}$

#### Page No 13.25:

#### Question 11:

In a circle of radius 35 cm, an arc subtends an angle of 72° at the centre. Find the length of the arc and area of the sector.

#### Answer:

We know that the arc length *l *and area *A* of a sector of an angle *θ* in the circle of radius *r* is given by and respectively.

It is given that, and.

We will calculate the arc length using the value of *r* and *θ*,

Now, we will find the value of area *A* of the sector

#### Page No 13.25:

#### Question 12:

The perimeter of a sector of a circle of radius 5.7 m is 27.2 m. Find the area of the sector.

#### Answer:

We know that the area A of a sector of circle of radius r and arc length l is given by

Let OAB is the given sector. Then,

arc AB = 15.8 m

So, *l* = 15.8 m

Now substituting the value of *r* and *l* in above formula,

$A=\frac{1}{2}\times 15.8\times 5.7\phantom{\rule{0ex}{0ex}}=45.03{m}^{2}$

#### Page No 13.25:

#### Question 13:

The perimeter of a certain sector of a circle of radius 5.6 m is 27.2 m. Find the area of the sector.

#### Answer:

We know that the area *A* of a sector of circle of radius *r* and arc length *l* is given by

Let OAB is the given sector. Then,

Perimeter of sector OAB = 27.2 m

OA + OB + arc AB = 27.2 m

5.6 + 5.6 + arc AB = 27.2 m

11.2 + arc AB = 27.2 m

arc AB = 16 m

So, *l* = 16 m

Now substituting the value of *r* and *l* in above formula,

$A=\frac{1}{2}\times 16\times 5.6\phantom{\rule{0ex}{0ex}}=44.8{m}^{2}$

#### Page No 13.25:

#### Question 14:

A sector is cut-off from a circle of radius 21 cm. The angle of the sector is 120°. Find the length of its arc and the area.

#### Answer:

We know that the arc length *l* and area *A* of a sector of circle at an angle *θ* of radius *r* is given by and angle.

Let OAB is the given sector.

It is given that and angle.

Now using the value of *r* and *θ,* we will find the value of* l* and *A*,

Arc length,

Area of sector,

#### Page No 13.25:

#### Question 15:

The minute hand of a clock is $\sqrt{21}$ cm long. Find the area described by the minute hand on the face of the clock between 7.00 AM and 7.05 AM.

#### Answer:

We know that the area *A* of a sector of circle at an angle *θ* of radius *r* is given by

.

We have,

Thus,

#### Page No 13.25:

#### Question 16:

The minute hand of a clock is 10 cm long. Find the area of the face of the clock described by the minute hand between 8 AM and 8.25 AM.

#### Answer:

We know that the area *A* of a sector of circle at an angle *θ* of radius *r* is given by

We have,

∴ Required area

= $\frac{150\xb0}{360\xb0}\times \frac{22}{7}\times {\left(10\right)}^{2}$

= 130.95 cm^{2}

#### Page No 13.25:

#### Question 17:

A sector of 56° cut out from a circle contains area 4.4 cm^{2}. Find the radius of the circle.

#### Answer:

We know that the area *A* of a sector of circle at an angle *θ* of radius *r* is given by

.

It is given that, and angle .

We can find the value of *r* by substituting these values in above formula,

#### Page No 13.25:

#### Question 18:

Area of sector of central angle 200^{0} of a circle is 770 cm^{2} . Find the length of the corresponding arc of this sector .

#### Answer:

Area of the sector = 770 cm^{2}

Central angle, $\theta ={200}^{\xb0}$

$\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{sector}=\frac{\theta}{360}\times \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{circle}\phantom{\rule{0ex}{0ex}}\Rightarrow 770=\frac{200}{360}\times \mathrm{\pi}{r}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow r\approx 21$

Now, $\mathrm{Area}\mathrm{of}\mathrm{sector}=\frac{1}{2}lr$

$\Rightarrow 770=\frac{1}{2}l\times 21\phantom{\rule{0ex}{0ex}}\Rightarrow l=\frac{220}{3}\mathrm{cm}$

#### Page No 13.25:

#### Question 19:

The length of minute hand of a clock is 5 cm . Find the area swept by the minute hand during the time period 6:05 am and 6:40 am.

#### Answer:

Minute hand of the clock describes a circle of radius equal to its length i.e 5 cm.

The minute hand rotates through 6^{o} in one minute.

So, the minute hand will sweep 210^{o }in one minute.

Area swept by the minute hand in one minute is the area of a sector of angle 6^{o }in a circle of radius 5 cm.

Area swept by the minute hand in 35 minutes is the area of a sector of angle 210^{o }in a circle of radius 5 cm

$\mathrm{Area}\mathrm{of}\mathrm{sector}=\frac{\theta}{360}\times \mathrm{\pi}{r}^{2}$

$=\frac{210}{360}\times \mathrm{\pi}\times {\left(5\right)}^{2}\phantom{\rule{0ex}{0ex}}=45\frac{5}{6}{\mathrm{cm}}^{2}$

#### Page No 13.25:

#### Question 20:

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

#### Answer:

Angle make by the minute hand in 1 minute = 6^{∘}

Angle make by the minute hand in 5 minute = 5 ⨯ 6^{∘} = 30^{∘}

Area of the sector having central angle is given by

$\frac{30\xb0}{360\xb0}\mathrm{\pi}{\left(14\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{12}\times \frac{22}{7}{\left(14\right)}^{2}\phantom{\rule{0ex}{0ex}}=51.33{\mathrm{cm}}^{2}$

Hence, the area swept by minute hand in 5 minutes is 51.33 cm^{2}

#### Page No 13.25:

#### Question 21:

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find (i) the length of the arc (ii) area of the sector formed by the arc. (Use π = 22/7)

#### Answer:

Here, we have θ = 60° and *r* = 21 cm

(i) The length of the arc is given by

$\frac{60\xb0}{360\xb0}\times 2\mathrm{\pi}\left(21\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{6}\times 2\times \frac{22}{7}\times 21\phantom{\rule{0ex}{0ex}}=22\mathrm{cm}$

(ii) Area of the sector formed by the arc is given by

$\frac{60\xb0}{360\xb0}\mathrm{\pi}{\left(21\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{6}\times \frac{22}{7}{\left(21\right)}^{2}\phantom{\rule{0ex}{0ex}}=231{\mathrm{cm}}^{2}$

#### Page No 13.25:

#### Question 22:

From a circular piece of cardboard of radius 3 cm two sectors of 90^{0} have been cutoff . Find the perimeter of the remaining portion nearest hundredth centimeters ( Take $\mathrm{\pi}$ = 22/ 7).

#### Answer:

Since the sectors are of central angle ${90}^{\xb0}$ so, the sector is in the form of a quadrant

Two such quadrants are cut off. So, a semicircle is left.

Perimeter of the semicircle = $\mathrm{\pi}r$

$=\frac{22}{7}\times \left(3\right)\phantom{\rule{0ex}{0ex}}=9.428\mathrm{cm}$

#### Page No 13.25:

#### Question 23:

The area of sector is one-twelfth that of the complete circle. Find the angle of the sector .

#### Answer:

We know, $A=\frac{\theta}{360}\times \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{circle}$ .....(i)

Let the area of the circle be *Ar. *

Thus area of the sector *= $\frac{1}{12}Ar$ .....*(ii)

From (i) and (ii) we have

$\frac{1}{12}Ar=\frac{\theta}{360}\times Ar\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{360}{12}=\theta \phantom{\rule{0ex}{0ex}}\Rightarrow \theta =30\xb0$

#### Page No 13.25:

#### Question 24:

*AB* is a chord of a circle with centre *O* and radius 4 cm. *AB* is of length 4 cm. Find the area of the sector of the circle formed by chord *AB*.

#### Answer:

We have to find the area of the sector AOB formed by the chord AB.

We have and. So,

Let. Then,

In, we have

Hence,

Now, using the value of and *r* we will find the area of sector AOB,

#### Page No 13.25:

#### Question 25:

In a circle of radius 6 cm, a chord of length 10 cm makes an angle of 110° at the centre of the circle. Find:

(i) the circumference of the circle

(ii) the area of the circle

(iii) the length of the arc *AB*,

(iv) the area of the sector *OAB*.

#### Answer:

It is given that the radius of circle, and angle at the centre of circle.

(i) We know that the Circumference C of circle of radius *r* is,

$\begin{array}{rcl}C& =& 2\mathrm{\pi r}\\ & =& 2\times \frac{22}{7}\times 6\\ & =& \frac{264}{7}\\ & =& 37.71\mathrm{cm}\end{array}$

(ii) We know that the Area A of circle of radius *r* is,

(iii) We know that the arc length *l* of a sector of an angle *θ* in a circle of radius r is

(iv) We know that the area *A* of a sector of an angle *θ* in the circle of radius *r* is given by

#### Page No 13.25:

#### Question 26:

In the following figure, shows a sector of a circle, centre O, containing an angle ~~0~~°. Prove that:

(i) Perimeter of the shaded region is $\left(\mathrm{tan}\theta +sec\theta +\frac{\mathrm{\pi \theta}}{180}-1\right)$

(ii) Area of the shaded region is $\frac{{r}^{2}}{2}\left(\mathrm{tan}\theta -\frac{\mathrm{\pi \theta}}{180}\right)$

#### Answer:

It is given that the radius of circle is *r* and the angle.

In,

It is given that.

(i) We know that the arc length *l* of a sector of an angle *θ* in a circle of radius r is

Now we substitute the value of *OC, OA* and *l* to find the perimeter of sector *AOC*,

Hence,

(ii) We know that area A of the sector at an angle θ in the circle of radius r is

.

Thus

Hence,

#### Page No 13.26:

#### Question 27:

Figure 15.18 shows a sector of a circle of radius r cm containing an angle $\theta \xb0$. The area of the sector is A cm^{2} and perimeter of the sector is 50 cm. Prove that

(i) $\theta =\frac{360}{\theta}\left(\frac{25}{r}-1\right)$

(ii) *A* = 25r − r^{2
}

#### Answer:

It is given that the radius of circle is *r* cm and angle.

(i) We know that the arc length *l* of a sector of an angle *θ* in a circle of radius r is

Now we substitute the value of *OB, OA* and *l* to find the perimeter of sector *AOB*,

(ii) We know that area A of the sector at an angle θ in the circle of radius r is

. Thus

Substituting the value of *θ,*

#### Page No 13.32:

#### Question 1:

*AB* is a chord of a circle with centre *O* and radius 4 cm. *AB* is of length 4 cm and divides the circle into two segments. Find the area of the minor segment.

#### Answer:

We know that the area of minor segment of angle *θ* in a circle of radius *r* is,

It is given that the chord *AB* divides the circle in two segments.

We have and. So,

Let. Then,

In, we have

Hence,

Now using the value of r and *θ*, we will find the area of minor segment

#### Page No 13.32:

#### Question 2:

A chord *PQ* of length 12 cm subtends an angle of 120° at the centre of a circle. Find the area of the minor segment cut off by the chord *PQ*.

#### Answer:

We know that the area of minor segment of angle *θ* in a circle of radius *r* is,

It is given that the chord *PQ* divides the circle in two segments.

We have and. So,

Since,

In, we have

Thus the radius of circle is .

Now using the value of radius *r* and angle *θ* we will find the area of minor segment

#### Page No 13.32:

#### Question 3:

A chord of a circle of radius 14 cm makes a right angle at the centre. Find the areas of the minor and major segments of the circle.

#### Answer:

We know that the area of minor segment of angle *θ* in a circle of radius *r* is,

It is given that the chord of the circle of radius makes right angle at the centre.

So,

Substituting the value of r and angle *θ* in above formula,

Area of minor segment

Hence, area of minor segment is

#### Page No 13.32:

#### Question 4:

A chord 10 cm long is drawn in a circle whose radius is 5$\sqrt{2}$ cm. Find area of both the segments.

#### Answer:

We know that the area of minor segment of angle *θ* in a circle of radius *r* is,

It is given that the chord AB divides the circle in two segments.

We have and. So,

Let. Then,

In, we have

Hence,

Now using the value of r and *θ*, we will find the area of minor segment

#### Page No 13.32:

#### Question 5:

A chord AB of a circle, of radius 14 cm makes an angle of 60° at the centre of the circle. Find the area of the minor segment of the circle. (Use π = 22/7)

#### Answer:

We know that the area of minor segment of angle *θ* in a circle of radius *r* is,

It is given that,

Substituting these values in above formula

#### Page No 13.32:

#### Question 6:

Find the area of minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60^{0} .

#### Answer:

Area of the minor segment of the circle = $\frac{\theta}{360}\times {\mathrm{\pi r}}^{2}-{\mathrm{r}}^{2}\mathrm{sin}\frac{\mathrm{\theta}}{2}\mathrm{cos}\frac{{\displaystyle \mathrm{\theta}}}{{\displaystyle 2}}$

$=\frac{60}{360}\times \mathrm{\pi}{\left(14\right)}^{2}-{\left(14\right)}^{2}\mathrm{sin}\frac{60}{2}\mathrm{cos}\frac{60}{2}\phantom{\rule{0ex}{0ex}}={\left(14\right)}^{2}\left[\frac{1}{6}\mathrm{\pi}-\frac{1}{2}\times \frac{\sqrt{3}}{2}\right]\phantom{\rule{0ex}{0ex}}=\left(\frac{308}{3}-49\sqrt{3}\right){\mathrm{cm}}^{2}$

#### Page No 13.32:

#### Question 7:

A chord of a circle of radius 20 cm sub tends an angle of 90^{0} at the centre . Find the area of the corresponding major segment of the circle

( Use $\mathrm{\pi}=3.14$)

#### Answer:

We know area of minor segment of the circle is $A=\left\{\frac{\mathrm{\pi \theta}}{360}-\mathrm{sin}\frac{\theta}{2}\mathrm{cos}\frac{\theta}{2}\right\}{r}^{2}$

$\Rightarrow A=\left\{\frac{\mathrm{\pi}\times 90\xb0}{360}-\mathrm{sin}\frac{90}{2}\mathrm{cos}\frac{90}{2}\right\}{\left(20\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow A=\left(\frac{\mathrm{\pi}}{4}-\frac{1}{2}\right)\left(400\right)$

Area of the major segment = Area of the circle − area of the minor segment

$=\mathrm{\pi}{\left(20\right)}^{2}-\left(400\right)\left[\frac{\mathrm{\pi}}{2}-\frac{1}{2}\right]\phantom{\rule{0ex}{0ex}}=\left(400\right)\left[\mathrm{\pi}-\frac{\mathrm{\pi}}{2}+\frac{1}{2}\right]\phantom{\rule{0ex}{0ex}}=1142{\mathrm{cm}}^{2}$

DISCLAIMER: The answer given in the book is incorrect. If we take the radius 10 instead of 20, then the answer will match.

#### Page No 13.32:

#### Question 8:

The radius of a circle with centre O is 5 cm in the given figure. Two radii OA and OB are drawn at right angles to each other. Fin the areas of the segments made by the chord AB (Take π = 3.14).

#### Answer:

We know area of minor segment of the circle is $A=\left\{\frac{\mathrm{\pi \theta}}{360}-\mathrm{sin}\frac{\theta}{2}\mathrm{cos}\frac{\theta}{2}\right\}{r}^{2}$

$A=\left\{\frac{\mathrm{\pi}90}{360}-\mathrm{sin}\frac{90}{2}\mathrm{cos}\frac{90}{2}\right\}{\left(5\right)}^{2}\phantom{\rule{0ex}{0ex}}A=\left\{\frac{\mathrm{\pi}}{4}-\mathrm{sin}45\times \mathrm{cos}45\right\}{\left(5\right)}^{2}\phantom{\rule{0ex}{0ex}}A=\left\{\frac{\mathrm{\pi}}{4}-\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}\right\}25\phantom{\rule{0ex}{0ex}}A=\left\{\frac{\mathrm{\pi}}{4}-\frac{1}{2}\right\}25=7.125{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}$

Area of the major segment = Area of the sector − area of minor segment

$=\mathrm{\pi}{\left(5\right)}^{2}-7.125\phantom{\rule{0ex}{0ex}}=71.375{\mathrm{cm}}^{2}$

#### Page No 13.33:

#### Question 9:

In the given figure, *AB* is the diameter of a circle, centre *O*, *C* is a point on the circumference such that ∠*COB* = $\theta $. The area of the minor segment cut off by *AC* is equal to twice the area of the sector *BOC*. Prove that

$\mathrm{sin}\frac{\theta}{2}\mathrm{cos}\frac{\theta}{2}=\mathrm{\pi}\left(\frac{1}{2}-\frac{\mathrm{\theta}}{120}\right)$

#### Answer:

We know that the area of minor segment of angle *θ* in a circle of radius *r* is,

.

It is given that,

So,

Area, *A* of minor segment cutoff by *AC* at angle

Now, since and

We know that the area of sector of a circle of radius *r* at an angle *θ* is

So, the area of sector *BOC, *

It is given that,

#### Page No 13.33:

#### Question 10:

A chord of a circle subtends an angle of $\theta $ at the centre of the circle. The area of the minor segment cut off by the chord is one eighth of the area of the circle. Prove that

$8\mathrm{sin}\frac{\theta}{2}\mathrm{cos}\frac{\theta}{2}+\mathrm{\pi}=\frac{\mathrm{\pi \theta}}{45}.$

#### Answer:

We know that the area of circle and area of minor segment of angle *θ* in a circle of radius *r* is given by, andrespectively.

It is given that,

#### Page No 13.56:

#### Question 1:

A plot is in the form of a rectangle *ABCD* having semi-circle on *BC* as shown in the following figure. If *AB* = 60 m and *BC* = 28 m, find the area of the plot.

#### Answer:

It is given that a plot is in form of rectangle ABCD having a semicircle on BC.

Since BC is the diameter of semicircle. Then, radius of semicircle is

=308 m^{2}

= 1680 m^{2}

Now,

Area of plot = Area of rectangle + Area of semicircle

= 1680 + 308 m^{2}

= 1988 m^{2}

#### Page No 13.56:

#### Question 2:

A play ground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m, find the area of the playground. (Take π = 22.7).

#### Answer:

It is given that a play ground has a shape of rectangle, with two semicircles on its smaller sides as diameter, added to its outside. So,

We have, sides of rectangle *l* = 36 m and *b* = 24.5 m.

Since, the diameter of semicircle is. Then,

$r=\frac{24.5}{2}\phantom{\rule{0ex}{0ex}}=12.25m$

= 235.81 m^{2}

= 882 m^{2}

Thus, the area of playground is

= 1353.62 m^{2}

#### Page No 13.56:

#### Question 3:

Find the area of the circle in which a square of area 64 cm^{2} is inscribed. [Use π = 3.14]

#### Answer:

We have given area of the square.

Now we will find the diameter of the square.

We know that diagonal of the square is same as the diameter of the circle.

Now we will find the area of the circle as shown below.

Therefore, area of the circle is.

#### Page No 13.56:

#### Question 4:

A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 m have been cut. Find the area of the remaining part.

#### Answer:

It is given that, the quadrants of radius *r* have been cut from the four corners of a rectangular piece is of length and width.

We have to find the area of remaining part.

We know that,

Now,

#### Page No 13.56:

#### Question 5:

In the following figure, PQRS is a square of side 4 cm. Find the area of the shaded square.

#### Answer:

Area of the shaded region is equal to the area of the square minus area of four sectors with the same radius minus area of the circle.

We know that area of the four sectors with radius is equal to area of one circle.

Therefore, area of the shaded region is

#### Page No 13.56:

#### Question 6:

Four cows are tethered at four corners of a square plot of side 50 m, so that they just cannot reach one another. What area will be left ungrazed? (Fig. 15.66)

#### Answer:

It is given that four cows are tethered at four corner of square ABCD. We have to find the area of plot that will left ungrazed.

Let the side of square is *a.*

a = 25 + 25 m = 50 m

Area of square = a^{2}

$=50\times 50\phantom{\rule{0ex}{0ex}}=2500{m}^{2}$

#### Page No 13.56:

#### Question 7:

A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions $20\mathrm{m}\times 16\mathrm{m}$, find the area of the field in which the cow can graze .

#### Answer:

Shaded portion shows the area that is grazed by the cow. It is in the form of a quadrant of a circle with radius 14 m.

$\mathrm{Area}\mathrm{grazed}\mathrm{by}\mathrm{the}\mathrm{cow}=\frac{1}{4}\mathrm{\pi}{r}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\mathrm{\pi}{\left(14\right)}^{2}\phantom{\rule{0ex}{0ex}}=154{\mathrm{m}}^{2}$

#### Page No 13.56:

#### Question 8:

A calf is tied with a rope of length 6 m at the corner of a square grassy lawn of side 20 m . If the length of the rope is increased by 5.5 m , find the increase in area of the grassy lawn in which the calf can graze .

#### Answer:

The area grazed by the calf is in the form of a quadrant of a circle with radius 6 m.

$\mathrm{Area}\mathrm{grazed}\mathrm{by}\mathrm{the}\mathrm{calf}\mathrm{with}\mathrm{rope}6\mathrm{m}=\frac{1}{4}\mathrm{\pi}{\left(6\right)}^{2}\phantom{\rule{0ex}{0ex}}=28.28{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}$

When the rope length is increased then total rope length = 6 + 5.5 m = 11.5 m

$\mathrm{Area}\mathrm{covered}\mathrm{by}\mathrm{the}\mathrm{calf}\mathrm{for}\mathrm{grazing}\mathrm{with}\mathrm{rope}11.5\mathrm{m}=\frac{1}{4}\mathrm{\pi}{\left(11.5\right)}^{2}\phantom{\rule{0ex}{0ex}}=103.91{\mathrm{m}}^{2}$

Hence, increase in area of grassy lawn that is grazed = 103.91 − 28.28 = 75.63 m^{2}

#### Page No 13.56:

#### Question 9:

A square water tank has its side equal to 40 m. There are four semi-circular grassy plots all round it. Find the cost of turfing the plot at Rs. 1.25 per square metre (Take π = 3.14).

#### Answer:

It is given that the side of square.

Since four semicircular grassy plots rounds a square water tank. Then, diameter of semicircular plot is.

So, the radius of semicircle

Now, the total area of plot is sum of area of four semicircular plots.

Since,

#### Page No 13.56:

#### Question 10:

A rectangular park is 100 m by 50 m. It is surrounding by semi-circular flower beds all round. Find the cost of levelling the semi-circular flower beds at 60 paise per square metre (use π = 3.14).

#### Answer:

Since four semicircular flower beds rounds the rectangular park. Then, diameters of semicircular plots are and

So, the radius of semicircle at larger side of rectangle

And the radius of semicircle at smaller side of rectangle

Now, the total area of semicircular plot is sum of area of four semicircular plots.

Since,

#### Page No 13.56:

#### Question 11:

The inside perimeter of a running track (shown in the following figure) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If the track is everywhere 14 m wide. find the area of the track. Also find the length of the outer running track.

#### Answer:

It is given that, and

We know that the circumference C of semicircle of radius be *r* is

$C=\mathrm{\pi}r$

The inside perimeter of running track is the sum of twice the length of straight portion and circumferences of semicircles. So,

inside perimeter of running track = 400 *m*

$2l+2\mathrm{\pi}r\mathit{=}400m$

$\Rightarrow 2\times 90+2\times \frac{22}{7}\times r=400m$

$\Rightarrow r=\frac{220\times 7}{2\times 22}=35m$

Thus, radius of inner semicircle is 35 m.

Now,

radius of outer semi circle *r*' = 35 + 14 = 49 *m*

Area of running track = $2\times \mathrm{Area}\mathrm{of}\mathrm{rectangle}+2\times \mathrm{Area}\mathrm{of}\mathrm{outer}\mathrm{semi}\mathrm{circle}-2\times \mathrm{Area}\mathrm{of}\mathrm{inner}\mathrm{semicircle}$

$=2\times 90\times 14+2\times \frac{\mathrm{\pi}(49{)}^{2}}{2}-2\times \frac{\mathrm{\pi}(35{)}^{2}}{2}$

$=2520+\mathrm{\pi}\times \left(49+35\right)\left(49-35\right)$

$=2520+\frac{22}{7}\times 84\times 14$

$=2520+3696=6216{m}^{2}$

Hence, the area of running track = 6216 *m ^{2}*

Now, length L of outer running track is

L = 2 × *l + *2$\mathrm{\pi}$*r**'
$=2\times 90+2\mathrm{\pi}\times 49$*

$=180+2\times \frac{22}{7}\times 49$

$=180+308=488m$

Hence, the length L of outer running track is 488

*m*.

#### Page No 13.57:

#### Question 12:

Find the area of the following figure, in square cm, correct to one place of decimal. (Take π = 22/7).

#### Answer:

Let the area of square ABCD be *A*.

It s given that,

So,

It is given that a semicircle is attached to one side of the square.

We know that the area of semicircle of radius r is

Substituting the value of *r*,

From the above figure it is seen that a right angle triangle is cutoff from one side of square.

Now, the area of above figure is,

Hence area of given figure is

#### Page No 13.57:

#### Question 13:

In the following figure, from a rectangular region ABCD with AB = 20 cm, a right triangle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside' the region. Find the area of the shaded region. [Use π =$\frac{22}{7}$] [CBSE 2014]

#### Answer:

AD

^{2}= AE

^{2}+ DE

^{2}

= (9)

^{2}+ (12)

^{2}

= 81 + 144

= 225

∴ AD

^{2}= 225

⇒ AD = 15 cm

We know that the opposite sides of a rectangle are equal

AD = BC = 15 cm

Area of the shaded region = Area of rectangle − Area of triangle AED + Area of semicircle

$=\mathrm{AB}\times \mathrm{BC}-\frac{1}{2}\times \mathrm{AE}\times \mathrm{DE}+\frac{1}{2}\mathrm{\pi}{\left(\frac{\mathrm{BC}}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=20\times 15-\frac{1}{2}\times 9\times 12+\frac{1}{2}\times \frac{22}{7}\times {\left(\frac{15}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=300-54+88.3928\phantom{\rule{0ex}{0ex}}=334.3928{\mathrm{cm}}^{2}$

Hence, the area of shaded region is 334.39 cm

^{2}

#### Page No 13.57:

#### Question 14:

From each of the two opposite corners of a square of side 8 cm, a quadrant of a circle of radius 1.4 cm is cut. Another circle of radius 4.2 cm is also cut from the centre as shown in the following figure. Find the area of the remaining (Shaded) portion of the square. (Use π = 22/7)

#### Answer:

It is given that a circle of radius 4.2 cm and two quadrants of radius 1.4 cm are cut from a square of side 8 cm.

Let the side of square be *a.* Then,

Since the radius of circle is 4.2 cm. So,

$Areaofcircle={\mathrm{\pi r}}^{2}\phantom{\rule{0ex}{0ex}}=\frac{22}{7}\times 4.2\times 4.2\phantom{\rule{0ex}{0ex}}=55.44{\mathrm{cm}}^{2}$

Now area of quadrant of circle of radius 1.4 cm is,

$\mathrm{Area}\mathrm{of}\mathrm{shaded}\mathrm{region}=\mathrm{Area}\mathrm{of}\mathrm{square}-\mathrm{Area}\mathrm{of}\mathrm{circle}-2\times \mathrm{Area}\mathrm{of}\mathrm{quadrant}\phantom{\rule{0ex}{0ex}}=64-55.44-2\times 1.54\phantom{\rule{0ex}{0ex}}=5.48{\mathrm{cm}}^{2}$

#### Page No 13.57:

#### Question 15:

In the following figure, *ABCD* is a rectangle with *AB* = 14 cm and *BC* = 7 cm. Taking *DC*, *BC* and *AD* as diameters, three semi-circles are drawn as shown in the figure. Find the area of the shaded region.

#### Answer:

Area of the shaded region can be calculated as shown below,

Area of the shaded region = Area of rectangle − area of the semi-circle with diameter DC triangle + 2 area of two semicircles with diameters AD and BC

Substituting we get,

Therefore, area of the shaded region is.

#### Page No 13.58:

#### Question 16:

In the following figure, *ABCD* is a rectangle, having *AB *= 20 cm and *BC* = 14 cm. Two sectors of 180° have been cut off. Calculate:

(i) the area of the shaded region.

(ii) the length of the boundary of the shaded region.

#### Answer:

(i) We have given two semi-circles and a rectangle.

Area of the shaded region = Area of the rectangle − Area of the two semicircles ……..(1)

Substituting we get,

Therefore, area of shaded region is.

(ii) Now we will find length of the boundary of the shaded region.

Therefore, length of the boundary of the shaded region is.

#### Page No 13.58:

#### Question 17:

In the following figure, the square *ABCD *is divided into five equal parts, all having same area. The central part is circular and the lines *AE*, *GC*, *BF* and *HD* lie along the diagonals *AC* and *BD* of the square. If *AB* = 22 cm, find:

(i) the circumference of the central part.

(ii) the perimeter of the part *ABEF*.

#### Answer:

We have a square ABCD.

We have,

(i)We have to find the perimeter of the triangle. We have a relation as,

So,

So perimeter of the circular region,

(ii)We have to find the perimeter of ABEF. Let O be the centre of the circular region.

Use Pythagoras theorem to get,

Similarly,

Now length of arc EF,

So, perimeter of ABFE,

#### Page No 13.58:

#### Question 18:

In the following figure find the area of the shaded region. (Use π = 3.14)

#### Answer:

Area of shaded region = Area of square − Area of 4 semicircle having diameter 4 cm − Area of square having side 4 cm

$={\left(\mathrm{Side}\right)}^{2}-4\times \frac{1}{2}\mathrm{\pi}{r}^{2}-{\left(\mathrm{side}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(14\right)}^{2}-2\times 3.14\times {\left(\frac{4}{2}\right)}^{2}-{4}^{2}\phantom{\rule{0ex}{0ex}}=196-25.12-16\phantom{\rule{0ex}{0ex}}=154.88{\mathrm{cm}}^{2}$

Hence, the area of shaded region is 154.88 cm

^{2}

#### Page No 13.58:

#### Question 19:

In the following figure, *OACB* is a quadrant of a circle with centre *O* and radius 3.5 cm. If *OD* = 2 cm, find the area of the (i) quadrant *OACB* (ii) shaded region.

#### Answer:

It is given that OACB is a quadrant of circle with centre at O and radius 3.5 cm.

(i) We know that the area of quadrant of circle of radius *r* is,

Substituting the value of radius,

Hence, the area of OACB is.

(ii) It is given that radius of quadrant of small circle is 2 cm.

Let the area of quadrant of small circle be.

It is clear from the above figure that area of shaded region is the difference of larger quadrant and the smaller one. Hence,

#### Page No 13.58:

#### Question 20:

In the following figure a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 21 cm, find the area of the shaded region.

#### Answer:

Construction: Join OB

In right triangle AOB

OB

^{2}= OA

^{2}+ AB

^{2}

= 21

^{2}+ 21

^{2}

= 441 + 441

= 882

∴ OB

^{2}= 882

Area of the shaded region = Area of quadrant OPBQ − Area of Square OABC

$=\frac{1}{4}\mathrm{\pi}{\left(\mathrm{OB}\right)}^{2}-{\left(\mathrm{OA}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\times \frac{22}{7}\times 882-441\phantom{\rule{0ex}{0ex}}=693-441\phantom{\rule{0ex}{0ex}}=252{\mathrm{cm}}^{2}$

Hence, the area of the shaded region is 252 cm

^{2}.

#### Page No 13.59:

#### Question 21:

In the following figure, OABC is a square of side 7 cm. If OAPC is a quadrant of a circle with centre O, then find the area of the shaded region. (Use π = 22/7)

#### Answer:

Area of shaded region = Area of square OABC − Area of quadrant OAPC

$={\left(\mathrm{Side}\right)}^{2}-\frac{1}{4}\mathrm{\pi}{r}^{2}\phantom{\rule{0ex}{0ex}}={\left(7\right)}^{2}-\frac{1}{4}\times \frac{22}{7}\times 7\times 7\phantom{\rule{0ex}{0ex}}=49-38.5\phantom{\rule{0ex}{0ex}}=10.5{\mathrm{cm}}^{2}$

Hence, the area of the shaded region is 10.5 cm^{2}

#### Page No 13.59:

#### Question 22:

In the following figure, *OE* = 20 cm. In sector *OSFT*, square *OEFG* is inscribed. Find the area of the shaded region.

#### Answer:

We have to find the area of the shaded portion. We have, and OEFG is a square.

Use Pythagoras theorem to find OF as,

So, radius of the circle,

Therefore area of the shaded region,

So,

#### Page No 13.59:

#### Question 23:

Find the area of the shaded region in the following figure, if AC = 24 cm, BC = 10 cm and O is the centre of the circle. (Use π = 3.14)

#### Answer:

It is given a triangle ABC is cut from a circle.

In,

, Since any angle inscribed in semicircle is always right angle.

By applying Pythagoras theorem,

We know that the area A of circle of radius *r* is

Substituting the value of radius *r*,

#### Page No 13.59:

#### Question 24:

A circle is inscribed in an equilateral triangle ABC is side 12 cm, touching its sides (the following figure). Find the radius of the inscribed circle and the area of the shaded part.

#### Answer:

We have to find the area of the shaded portion. We havewhich is an equilateral triangle and.

We have O as the incentre and OP, OQ and OR are equal.

So,

Thus,

So area of the shaded region,

#### Page No 13.59:

#### Question 25:

In the following figure, an equilateral triangle ABC of side 6 cm has been inscribed in a circle. Find the area of the shaded region. (Take π = 3.14).

#### Answer:

We have to find the area of the shaded portion. We havewhich is an equilateral triangle and. Let *r* be the radius of the circle.

We have O as the circumcentre.

So,

Thus,

So area of the shaded region,

#### Page No 13.59:

#### Question 26:

A circular field has a perimeter of 650 m. A square plot having its vertices on the circumference of the field is marked in the field. Calculate the area of the square plot.

#### Answer:

We have a circular field in which a square field is marked.

Let the radius of the circle be *r*. We have,

Use Pythagoras theorem to find the side of square as,

So area of the square plot,

#### Page No 13.59:

#### Question 27:

Find the area of a shaded region in the the following figure,where a circular arc of radius 7 cm has been drawn with vertex A of an equilateral triangle ABC of side 14 cm as centre. (Use π = 22/7 and $\sqrt{3}$ = 1.73)

#### Answer:

In equilateral traingle all the angles are of 60°

∴ ∠BAC = 60°

Area of the shaded region = (Area of triangle ABC − Area of sector having central angle 60°) + Area of sector having central angle (360° − 60°)

$=\frac{\sqrt{3}}{4}{\left(\mathrm{AB}\right)}^{2}-\frac{60\xb0}{360\xb0}\mathrm{\pi}{\left(7\right)}^{2}+\frac{300\xb0}{360\xb0}\mathrm{\pi}{\left(7\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}}{4}{\left(14\right)}^{2}-\frac{1}{6}\times \frac{22}{7}{\left(7\right)}^{2}+\frac{5}{6}\times \frac{22}{7}{\left(7\right)}^{2}\phantom{\rule{0ex}{0ex}}=84.77-25.67+128.35\phantom{\rule{0ex}{0ex}}=187.45{\mathrm{cm}}^{2}$

Hence, the area of shaded region is 187.45 cm^{2}

#### Page No 13.60:

#### Question 28:

A regular hexagon is inscribed in a circle. If the area of hexagon is $24\sqrt{3}$ cm^{2}, find the area of the circle. (Use π = 3.14)

#### Answer:

Let the radius of the circle be *r* and side of hexagon be *a*.

Area of hexagon = $\frac{3\sqrt{3}}{2}{a}^{2}$

$\Rightarrow 24\sqrt{3}=\frac{3\sqrt{3}}{2}{a}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}=16\phantom{\rule{0ex}{0ex}}\Rightarrow a=4\mathrm{cm}$

In an regular hexagon inscribed in a circle, its side is equal the radius.

∴ *r* = *a* = 4 cm

Now, Area of circle is given by

${\mathrm{\pi r}}^{2}\phantom{\rule{0ex}{0ex}}=3.14\times 4\times 4\phantom{\rule{0ex}{0ex}}=50.24{\mathrm{cm}}^{2}$

#### Page No 13.60:

#### Question 29:

*ABCDEF* is a regular hexagon with centre O (in the following figure). If the area of triangle *OAB* is 9 cm^{2}, find the area of : (i) the hexagon and (ii) the circle in which the haxagon is incribed.

#### Answer:

We know that a regular hexagon is made up of 6 equilateral triangles.

We have given area of the one of the triangles.

We know that if a regular hexagon is inscribed in the circle, then the radius of the circle is same as the side of the regular hexagon.

We also know that a regular hexagon is made up of 6 equilateral triangles and we have area of one of the equilateral triangle.

Substituting the value of the given equilateral triangle we get,

Now we will find the area of the circle.

Substituting the values we get,

Now we will substitute we get,

Therefore, area of the hexagon and area of the circle are and respectively.

#### Page No 13.60:

#### Question 30:

Four equal circles, each of radius 5 cm, touch each other as shown in the following figure. Find the area included between them (Take π = 3.14).

#### Answer:

It is given that four equal circle touches each other as shown in figure.

Let the side of square is *a.*

We know that

$=\frac{1}{4}\times 78.5c{m}^{2}$

= 21.5 cm^{2}

#### Page No 13.60:

#### Question 31:

Four equal circles, each of radius a, touch each other. Show that the area between them is $\frac{6}{7}{a}^{2}$ (Take π = 22/7).

#### Answer:

It is given that four equal circles of radius *a* touches each other.

So,

Since circles touches each other, the lines joining their centre make a square ABCD.

The side of square is 2a.

#### Page No 13.60:

#### Question 32:

A child makes a poster on a chart paper drawing a square ABCD of side 14 cm. She draws four circles with centre A, B, C and D in which she suggests different ways to save energy. The circles are drawn in such a way that each circle touches externally two of the three remaining circles (in the following figure). In the shaded region she write a message 'Save Energy'. Find the perimeter and area of the shaded region.

(Use π = 22/7)

#### Answer:

Perimeter of the shaded portion = 4 ⨯ Length of the arc having central angle 90^{∘}

$=4\times \frac{90\xb0}{360\xb0}\times 2\times \frac{22}{7}\times \frac{14}{2}\phantom{\rule{0ex}{0ex}}=4\times \frac{1}{4}\times 2\times \frac{22}{7}\times 7\phantom{\rule{0ex}{0ex}}=44\mathrm{cm}$

Area of shaded portion = Area of square ABCD − 4 ⨯ Area of the arc having central angle 90^{∘}

$={\left(14\right)}^{2}-4\times \frac{90\xb0}{360\xb0}\times \frac{22}{7}\times {\left(\frac{14}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=196-4\times \frac{1}{4}\times \frac{22}{7}\times {\left(7\right)}^{2}\phantom{\rule{0ex}{0ex}}=196-154\phantom{\rule{0ex}{0ex}}=42{\mathrm{cm}}^{2}$

#### Page No 13.61:

#### Question 33:

The diameter of a coin is 1 cm (in the following figure). If four such coins be placed on a table so that the rim of each touches that of the other two, find the area of the shaded region (Take π = 3.1416).

#### Answer:

Look at the figure carefully shaded region is bounded between four sectors of the circle with same radius and a square of side 1 cm.

Therefore, the area of the shaded region is nothing but the difference the area of the square and area of one circle.

Substituting we get,

Therefore, area of the shaded region is.

#### Page No 13.61:

#### Question 34:

Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular card board of dimensions 14 cm × 7 cm. Find the area of the remaining card board. (Use π = 22/7).

#### Answer:

We know that we can cut two circular pieces of equal radii and maximum area from the rectangular cardboard whose diameter is equal to the width of the rectangular cardboard.

∴ Radii of two circuar pieces = Half of the width of the rectangular cardboard = 3.5 cm

Now,

Area of remaining cardboard = Area of rectangular cardboard − 2 ⨯ Area of circular piece having radius 3.5 cm

$=14\times 7-2\left(\frac{22}{7}\times 3.5\times 3.5\right)\phantom{\rule{0ex}{0ex}}=98-77\phantom{\rule{0ex}{0ex}}=21{\mathrm{cm}}^{2}$

Hence, the area of the remaining cardboard is 21 cm^{2}

#### Page No 13.61:

#### Question 35:

In the following figure, *AB* and *CD* are two diameters of a circle perpendicular to each other and *OD* is the diameter of the smaller circle. If *OA* = 7 cm, find the area of the shaded region.

#### Answer:

It is given that AB and CD are two diameters of a circle perpendicular to each other and OD is the diameter of small circle.

It is given that,

So, radius *r* of small circle is

We know that the area *A* of circle of radius *r* is.

Substituting the value of *r* in above formula,

Now, let the area of large circle be.

Using the value radius OA,

Hence,

#### Page No 13.61:

#### Question 36:

In the the following figure, PSR, RTQ and PAQ are three semicircles of diameter 10 cm, 3 cm and 7 cm respectively. Find the perimeter of shaded region. [CBSE 2014]

#### Answer:

Perimeter of shaded region = Length of the arc PAQ + Length of the arc PSR + Length of the arc RTQ

$=\frac{1}{2}\times 2\mathrm{\pi}{r}_{1}+\frac{1}{2}\times 2\mathrm{\pi}{r}_{2}+\frac{1}{2}\times 2\mathrm{\pi}{r}_{3}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 2\mathrm{\pi}\left(\frac{7}{2}\right)+\frac{1}{2}\times 2\mathrm{\pi}\left(\frac{10}{2}\right)+\frac{1}{2}\times 2\mathrm{\pi}\left(\frac{3}{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{7}{2}\mathrm{\pi}+5\mathrm{\pi}+\frac{3}{2}\mathrm{\pi}\phantom{\rule{0ex}{0ex}}=10\mathrm{\pi}\phantom{\rule{0ex}{0ex}}=31.4\mathrm{cm}$

Hence, the perimeter of shaded region is 31.4 cm.

#### Page No 13.61:

#### Question 37:

In the following figure, two circles with centres *A* and *B* touch each other at the point *C*. If *AC* = 8 cm and *AB* = 3 cm, find the area of the shaded region.

#### Answer:

Area of the shaded region can be calculated as shown below,

Area of the shaded region = Area of circle with radius AC − area of circle with radius BC

We have given radius of the outer circle that is 8 cm but we don’t know the radius of the inner circle.

We can calculate the radius of the inner circle as shown below,

Substituting we get,

Therefore, area of the shaded region is.

#### Page No 13.62:

#### Question 38:

In the following figure, ABCD is a square of side 2a, Find the ratio between

(i) the circumferences

(ii) the areas of the in circle and the circum-circle of the square.

#### Answer:

We have a square ABCD having. From the given diagram we can observe that,

Radius of incircle

Radius of circumcircle

(i) We have to find the ratio of the circumferences of the two circles. So the required ratio is,

(ii) We have to find the ratio of the areas of the two circles. So the required ratio is,

#### Page No 13.62:

#### Question 39:

In the following figure, there are three semicircles, A, B and C having diameter 3 cm each, and another semicircle E having a circle D with diameter 4.5 cm are shown. Calculate:

(i) the area of the shaded region

(ii) the cost of painting the shaded region at the rate of 25 paise per cm^{2} , to the nearest rupee.

#### Answer:

(i) Area of the shaded region can be calculated as shown below,

Area of the shaded region = Area of the semi-circle with diameter of 9 cm − area of 2 semi-circles with radius 3cm − area of the circle with centre D + area of semi-circle with radius 3 cm

Substituting we get,

Therefore, area of the shaded region is.

Now we will find the cost of painting the shaded region at the rate of 25 paise per cm^{2}.

paise

Therefore, cost of painting the shaded region to the nearest rupee is.

#### Page No 13.62:

#### Question 40:

In the following figure, *ABC* is a right-angled triangle, ∠*B* = 90°, *AB* = 28 cm and *BC* = 21 cm. With *AC* as diameter a semicircle is drawn and with *BC* as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.

#### Answer:

We have given two semi-circles and one circle.

Area of the shaded region = area of semicircle with diameter AC + area of right angled triangle ABC − area of sector

First we will find the hypotenuse of right angled triangle ABC.

Substituting we get,

Therefore, area of shaded region is.

#### Page No 13.63:

#### Question 41:

In the following figure, *O* is the centre of a circular arc and *AOB* is a straight line. Find the perimeter and the area of the shaded region correct to one decimal place. (Take π = 3.142)

#### Answer:

(i) Let us find the perimeter of the shaded region.

Therefore, perimeter of the shaded region is .

Now we will find the area of the shaded region can be calculated as shown below,

Area of the shaded region = Area of the semi-circle − area of the right angled triangle

First, we will find the length of AB as shown below,

Substituting we get,

Therefore, area of the shaded region is.

#### Page No 13.63:

#### Question 42:

In the following figure, the boundary of the shaded region consists of four semi-circular arcs, the smallest two being equal. If the diameter of the largest is 14 cm and of the smallest is 3.5 cm, find

(i) the length of the boundary.

(ii) the area of the shaded region.

#### Answer:

(i) We will first find the length of the boundary.

Length of the boundary perimeter of semi-circle with diameter AB + boundary of semi-circle with diameter 7 cm

Therefore, length of the boundary is.

Now we will find the area of the shaded region as shown below,

Area of the shaded region = Area of the semi-circle with AB as a diameter − area of the semi-circle with radius AE − area of the semi-circle with radius BC + area of the semi-circle with diameter 7 cm.

Therefore, area of the shaded region is.

#### Page No 13.63:

#### Question 43:

In the following figure, *AB* = 36 cm and M is mid-point of *A*B. Semi-circles are drawn on *AB*, *AM* and *MB *as diameters. *A* circle with centre *C* touches all the three circles. Find the area of the shaded region.

#### Answer:

We have given two semi-circles and one circle.

Area of the shaded region = area of semicircle with diameter AB − area of two semicircles with diameters AM and MB - area of circle ……..(1)

Let us calculate the area of the semi-circle with AB as a diameter.

Now we will find the area of the semi-circle with AM as a diameter.

Area of the semi-circle with MB as a diameter is same as the area of the semi-circle with diameter with AM as a diameter.

Now we will find the area of the circle with centre C.

We know that radius of the circle is one sixth of AB.

Now we will substitute all these values in equation (1).

Therefore, area of shaded region is.

#### Page No 13.63:

#### Question 44:

In the following figure, *ABC* is a right angled triangle in which ∠*A* = 90°, *AB* = 21 cm and *AC* = 28 cm. Semi-circles are described on *AB*, *BC* and *AC* as diameters. Find the area of the shaded region.

#### Answer:

We have given three semi-circles and one right angled triangle.

……..(1)

Let us calculate the area of the semi-circle with AB as a diameter.

Now we will find the area of the semi-circle with AC as a diameter.

Now we will find the length of BC.

In right angled triangle ABC, we will use Pythagoras theorem,

Now we will calculate the area of the right angled triangle ABC.

Now we will find the area of the semi-circle with BC as a diameter.

Now we will substitute all these values in equation (1).

Therefore, area of shaded region is.

#### Page No 13.64:

#### Question 45:

In the following figure, shows the cross-section of railway tunnel. The radius *OA* of the circular part is 2 m. If ∠*AOB* = 90°, calculate:

(i) the height of the tunnel

(ii) the perimeter of the cross-section

(iii) the area of the cross-section.

figure

#### Answer:

We have a cross section of a railway tunnel. is a right angled isosceles triangle, right angled at O. let OM be perpendicular to AB.

(i) We have to find the height of the tunnel. We have,

Use Pythagoras theorem into get,

Let the height of the tunnel be h. So,

Thus,

Therefore,

$h=\left(2+\sqrt{2}\right)\mathrm{m}$

(ii)

Perimeter of cross-section is,

$=\left(3\mathrm{\pi}+2\sqrt{2}\right)m$

(iii)

$\left(3\mathrm{\pi}+2\right){m}^{2}$

#### Page No 13.64:

#### Question 46:

In the following figure, shows a kite in which *BCD* is the shape of a quadrant of a circle of radius 42 cm. *ABCD* is a square and Δ *CEF* is an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region.

#### Answer:

We will find the area of the shaded region as shown below,

Area of the shaded region = area of quadrant + area of isosceles triangle ……..(1)

Substituting we get,

Therefore, area of the shaded region is.

#### Page No 13.64:

#### Question 47:

In the following figure, ABCD is a trapezium of area 24.5 cm^{2} , If AD || BC, ∠DAB = 90°, AD = 10 cm, BC = 4 cm and ABE is quadrant of a circle, then find the area of the shaded region. [CBSE 2014]

#### Answer:

$\Rightarrow 24.5=\frac{1}{2}\left(10+4\right)\times \mathrm{AB}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AB}=3.5\mathrm{cm}$

Area of shaded region = Area of trapezium ABCD − Area of quadrant ABE

$=24.5-\frac{1}{4}\times \frac{22}{7}\times {\left(3.5\right)}^{2}\phantom{\rule{0ex}{0ex}}=24.5-9.625\phantom{\rule{0ex}{0ex}}=14.875{\mathrm{cm}}^{2}$

Hence, the area of shaded region is 14.875 cm

^{2}

#### Page No 13.64:

#### Question 48:

In the given figure, ABCD is a trapezium with AB || DC, AB = 18 cm DC = 32 cm and the distance between AB and DC is 14 cm. Circles of equal radii 7 cm with centres A, B, C and D have been drawn. Then find the area of the shaded region.

(Use $\mathrm{\pi}=\frac{22}{7}$ [CBSE 2014]

#### Answer:

$=\frac{1}{2}\left(\mathrm{AB}+\mathrm{DC}\right)\times 14-\left(\frac{\angle \mathrm{A}}{360\xb0}\mathrm{\pi}{r}^{2}+\frac{\angle \mathrm{B}}{360\xb0}\mathrm{\pi}{r}^{2}+\frac{\angle \mathrm{C}}{360\xb0}\mathrm{\pi}{r}^{2}+\frac{\angle \mathrm{D}}{360\xb0}\mathrm{\pi}{r}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(\mathrm{AB}+\mathrm{DC}\right)\times 14-\left(\frac{\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}}{360\xb0}\right)\mathrm{\pi}{r}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(18+32\right)\times 14-\frac{22}{7}{\left(7\right)}^{2}\phantom{\rule{0ex}{0ex}}=350-154\phantom{\rule{0ex}{0ex}}=196{\mathrm{cm}}^{2}$

Hence, the area of shaded region is 196 cm

^{2}

#### Page No 13.64:

#### Question 49:

From a thin metallic piece, in the shape of a trapezium *ABCD*, in which *AB* || *CD* and ∠*BCD* = 90°, a quarter circle *BEFC* is removed (in the following figure). Given *AB* = *BC* = 3.5 cm and *DE* = 2 cm, calculate the area of the remaining piece of the metal sheet.

#### Answer:

We have given a trapezium. We are asked to find the area of the shaded region.

We can find the area of the remaining part that is area of the shaded region as shown below.

……..(1)

Now we find the value of CD.

………(Since, CE is radius of the sector, therefore, CE = 3.5)

Substituting the values of CD and in equation (1),

Therefore, area of the remaining part is.

#### Page No 13.65:

#### Question 50:

In the following figure, ABC is an equilateral triangle of side 8 cm. A, B and C are the centres of circular arcs of radius 4 cm. Find the area of the shaded region correct upto 2 decimal places. (Take π =3.142 and $\sqrt{3}$ = 1.732).

#### Answer:

Area of the shaded region can be calculated as shown below,

Area of the shaded region = Area of equilateral triangle − 3 area of circular arc

Substituting and we get,

Therefore, area of the shaded region is.

#### Page No 13.65:

#### Question 51:

Sides of a triangular field are 15 m , 16 m , 17 m . With three corners of the field a cow , a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field . Find the area of the field which cannot be grazed by three animals.

#### Answer:

The area grazed by the cow, buffalo and the horse are in the form of sectors of the circle with radius 7 m.

Let the angle formed in the three sectors be ${\theta}_{1},{\theta}_{2},{\theta}_{3}$

Now the area of these sectors will be

$\mathrm{Area}\mathrm{of}\mathrm{sector}\mathrm{with}\mathrm{angle}{\theta}_{1}=\frac{\mathrm{\pi}{\mathrm{\theta}}_{1}{\left(7\right)}^{2}}{360}\phantom{\rule{0ex}{0ex}}\mathrm{Area}\mathrm{of}\mathrm{sector}\mathrm{with}\mathrm{angle}{\theta}_{2}=\frac{\mathrm{\pi}{\mathrm{\theta}}_{2}{\left(7\right)}^{2}}{360}\phantom{\rule{0ex}{0ex}}\mathrm{Area}\mathrm{of}\mathrm{sector}\mathrm{with}\mathrm{angle}{\theta}_{3}=\frac{\mathrm{\pi}{\mathrm{\theta}}_{3}{\left(7\right)}^{2}}{360}\phantom{\rule{0ex}{0ex}}$

Area of the triangle ABC wil be

$s=\frac{a+b+c}{2}=\frac{15+16+17}{2}=\frac{48}{2}=24\phantom{\rule{0ex}{0ex}}Area=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{24\left(24-15\right)\left(24-16\right)\left(24-17\right)}\phantom{\rule{0ex}{0ex}}=24\sqrt{21}{\mathrm{m}}^{2}$

Area of the field not grazed by the animals = Area of the triangle ABC − Area of the three sectors

$=24\sqrt{21}-\left[\left(\frac{{\mathrm{\pi \theta}}_{1}{\left(7\right)}^{2}}{360}\right)+\left(\frac{{\mathrm{\pi \theta}}_{2}{\left(7\right)}^{2}}{360}\right)+\left(\frac{{\mathrm{\pi \theta}}_{3}{\left(7\right)}^{2}}{360}\right)\right]\phantom{\rule{0ex}{0ex}}=24\sqrt{21}-\left(\frac{\pi {\left(7\right)}^{2}}{360}\right)\left({\theta}_{1}+{\theta}_{2}+{\theta}_{3}\right)\phantom{\rule{0ex}{0ex}}=24\sqrt{21}-\left(\frac{\pi {\left(7\right)}^{2}}{360}\right)\left(180\right)\left(\mathrm{Angle}\mathrm{sum}\mathrm{property}\right)\phantom{\rule{0ex}{0ex}}=\left(24\sqrt{21}-77\right){\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}$

#### Page No 13.65:

#### Question 52:

In the given figure, the side of square is 28 cm and radius of each circle is half of the length of the side of the square where O and O' are centres of the circles. Find the area of shaded region.

#### Answer:

We have,

Side of square = 28 cm and radius of each circle = $\frac{28}{2}$ cm

Area of the shaded region

= Area of the square + Area of the two circles − Area of the two quadrants

$={\left(28\right)}^{2}+2\times \mathrm{\pi}\times {\left(\frac{28}{2}\right)}^{2}-2\times \frac{1}{4}\times \mathrm{\pi}\times {\left(\frac{28}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(28\right)}^{2}+\frac{3}{2}\times \mathrm{\pi}\times {\left(\frac{28}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(28\right)}^{2}\left(1+\frac{3}{2}\times \frac{22}{7}\times \frac{1}{2}\times \frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}={\left(28\right)}^{2}\left(1+\frac{33}{28}\right)\phantom{\rule{0ex}{0ex}}={\left(28\right)}^{2}\times \frac{61}{28}\phantom{\rule{0ex}{0ex}}=28\times 61\phantom{\rule{0ex}{0ex}}=1708{\mathrm{cm}}^{2}$

Therefore, the area of the shaded region is 1708 cm^{2}.

#### Page No 13.65:

#### Question 53:

In a hospital used water is collected in a cylindrical tank of diameter 2 m and height 5 m. After recycling, this water is used to irrigate a park of hospital whose length is 25 m and breadth is 20 m. If tank is filled completely then what will be the height of standing water used for irrigating the park.

#### Answer:

Diameter of cylinder (*d*) = 2 m

Radius of cylinder (*r*) = 1 m

Height of cylinder (*H*) = 5 m

Volume of cylinderical tank, V* _{c}* = $\mathrm{\pi}{r}^{2}H$ = $\mathrm{\pi}\times {\left(1\right)}^{2}\times 5=5\mathrm{\pi}\mathrm{m}$

Length of the park (

*l*) = 25 m

Breadth of park (

*b*) = 20 m

height of standing water in the park =

*h*

Volume of water in the park

*= lbh = $25\times 20\times h$*

Now water from the tank is used to irrigate the park. So,

Volume of cylinderical tank = Volume of water in the park

$\Rightarrow 5\mathrm{\pi}=25\times 20\times h\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{5\mathrm{\pi}}{25\times 20}=h\phantom{\rule{0ex}{0ex}}\Rightarrow h=\frac{\mathrm{\pi}}{100}\mathrm{m}\phantom{\rule{0ex}{0ex}}\Rightarrow h=0.0314\mathrm{m}$

#### Page No 13.66:

#### Question 54:

In Fig. 13.112, a square *OABC *is inscribed in a quadrant *OPBQ*. If *OA *= 15 cm, find the area of shaded region (use π = 3.14).

#### Answer:

OABC is a square.

Side of the square = OA = 15 cm

We know

Length of the diagonal of square = $\sqrt{2}$ × Side of the square

∴ OB = Radius of the quadrant of the circle = Length of the diagonal of square = $15\sqrt{2}\mathrm{cm}$

Now,

Area of shaded region

= Area of quadrant OPQO − Area of the square OABC

$=\frac{1}{4}\times \mathrm{\pi}{\left(\mathrm{OB}\right)}^{2}-{\left(\mathrm{OA}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\times 3.14\times {\left(15\sqrt{2}\right)}^{2}-{\left(15\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\times 3.14\times 450-225$

$=353.25-225\phantom{\rule{0ex}{0ex}}=128.25{\mathrm{cm}}^{2}$

Thus, the area of the shaded region is 128.25 cm^{2}.

#### Page No 13.66:

#### Question 55:

In Fig. 13.113, *ABCD *is a square with side $2\sqrt{2}$ cm and inscribed in a circle. Find the area of the shaded region. (Use π = 3.14)

#### Answer:

ABCD is a square.

Side of the square = AB = $2\sqrt{2}\mathrm{cm}$

We know

Length of the diagonal of square = $\sqrt{2}$ × Side of the square

∴ BD = Diameter of the circle = Length of the diagonal of square = $\sqrt{2}\times 2\sqrt{2}$ = 4 cm

⇒ Radius of the circle = $\frac{\mathrm{BD}}{2}$ = 2 cm

Now,

Area of shaded region

= Area of the circle − Area of the square ABCD

$=\mathrm{\pi}{\left(\frac{\mathrm{BD}}{2}\right)}^{2}-{\left(\mathrm{AB}\right)}^{2}\phantom{\rule{0ex}{0ex}}=3.14\times {\left(2\right)}^{2}-{\left(2\sqrt{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=3.14\times 4-8$

$=12.56-8\phantom{\rule{0ex}{0ex}}=4.56{\mathrm{cm}}^{2}$

Thus, the area of the shaded region is 4.56 cm^{2}.

#### Page No 13.68:

#### Question 1:

If the circumference and the area of a circle are numerically equal, then diameter of the circle is

(a) $\frac{\mathrm{\pi}}{2}$

(b) 2$\mathrm{\pi}$

(c) 2

(d) 4

#### Answer:

We have given that circumference and area of a circle are numerically equal.

Let it be *x*.

Let *r* be the radius of the circle, therefore, circumference of the circle is and area of the circle will be.

Therefore, from the given condition we have,

………(1)

………(2)

Therefore, from equation (1) get . Now we will substitute this value in equation (2) we get,

Simplifying further we get,

Cancelling *x* we get,

Now we will cancel

………(3)

Now we will multiply both sides of the equation (3) by we get,

We can rewrite this equation as given below,

Comparing equation (4) with equation (1) we get *r* = 2.

Therefore, radius of the circle is 2. We know that diameter of the circle is twice the radius of the circle.

Therefore, diameter of the circle is.

Hence, option (*d*) is correct.

#### Page No 13.68:

#### Question 2:

If the difference between the circumference and radius of a circle is 37 cm, then using π = $\frac{22}{7}$, the circumference (in cm) of the circle is

(a) 154

(b) 44

(c) 14

(d) 7

#### Answer:

We know that circumference; C of the circle with radius *r* is equal to.

We have given difference between circumference and radius of the circle that is 37 cm.

Substituting we get,

Dividing both sides of the equation by, we get,

Therefore, circumference of the circle will be

$2\mathrm{\pi r}=2\times \frac{22}{7}\times 7\phantom{\rule{0ex}{0ex}}=44{\mathrm{cm}}^{2}$

Hence, the correct choice is (*b*).

#### Page No 13.68:

#### Question 3:

A wire can be bent in the form of a circle of radius 56 cm. If it is bent in the form fo a square, then its area will be

(a) 3520 cm^{2}

(b) 6400 cm^{2}

(c) 7744 cm^{2}

(d) 8800 cm^{2}

#### Answer:

We have given that a wire is bent in the form of circle of radius 56 cm. If we bent the same wire in the form of square of side a cm, the perimeter of the wire will not change.

perimeter of the circle = perimeter of the square

We know that *r* = 56 cm.

Now we will substitute the value of *r* in the equation,

………(1)

Dividing both sides of the equation by 4 we get,

Now we obtained side of the square. Now we can calculate the area of the square as given below.

Hence, the area of the square is .

Therefore, the correct answer is (*c*).

#### Page No 13.68:

#### Question 4:

If a wire is bent into the shape of a square, then the area of the square is 81 cm^{2} . When wire is bent into a semi-circular shape, then the area of the semi-circle will be

(a) 22 cm^{2}

(b) 44 cm^{2}

(c) 77 cm^{2}

(d) 154 cm^{2}

#### Answer:

We have given that a wire is bent in the form of square of side a cm such that the area of the square is . If we bent the same wire in the form of a semicircle with radius *r* cm, the perimeter of the wire will not change.

perimeter of the square = perimeter of semi circle

………(1)

We know that area of the square = .

Now we will substitute the value of *a* in the equation (1),

Now we will substitute.

Multiplying both sides of the equation by 7 we get,

Now we will divide both sides of the equation by 36 we get, *r* = 7

Therefore, radius of the semi circle is 7cm.

Now we will find the area of the semicircle.

Area of the semicircle

Therefore, area of the semicircle is .

Hence the correct answer is option (*c*).

#### Page No 13.68:

#### Question 5:

A circular park has a path of uniform width around it. The difference between the outer and inner circumferences of the circular path is 132 m. Its width is

(a) 20 m

(b) 21 m

(c) 22 m

(d) 24 m

#### Answer:

Let OA = *r* be the radius of the inner circle and OB = *r*′ be the radius of the outer circle.

Therefore, circumference of the inner circle and circumference of the outer circle

Here we have to find the width of the circular park that is we have to find .

We have given the difference between the circumferences of outer circle and inner circle.

Substituting we get,

Now we will multiply both sides of the equation by .

Therefore, the width is.

Hence the correct answer is option (*b*).

#### Page No 13.68:

#### Question 6:

The radius of a wheel is 0.25 m. The number of revolutions it will make to travel a distance of 11 km will be

(a) 2800

(b) 4000

(c) 5500

(d) 7000

#### Answer:

We have given the radius of the wheel that is 0.25 cm.

We know that distance covered by the wheel in one revolution.

Distance covered by the wheel in one revolution is equal to the circumference of the wheel.

………(1)

Distance moved is given as 11 km so we will first convert it to m.

11 km = 11000 m

Now we will substitute the values in equation (1),

Now we will substitute.

Simplifying equation (1) we get,

Therefore, it will make revolutions to travel a distance of 11 km.

Hence, the correct answer is option (*d*).

#### Page No 13.68:

#### Question 7:

The ratio of the outer and inner perimeters of a circular path is 23 : 22. If the path is 5 metres wide, the diameter of the inner circle is

(a) 55 m

(b) 110 m

(c) 220 m

(d) 230 m

#### Answer:

Let OA = *r* be the radius of the inner circle and OB = *r*′ be the radius of the outer circle.

Therefore, circumference of the inner circle and circumference of the outer circle

Here we have to find the diameter of the inner circle.

We have given the ratio of outer and inner perimeters of a circular path.

Simplifying the above equation we get,

……..(1)

We have also given that the path is 5 meters wide, that is we have given

……..(2)

We are asked to find the diameter of the inner circle hence, we will eliminate *r*′ using equations (1) and (2) for that we will multiply equation (2) by 22.

……..(3)

Subtracting equation (1) from equation (3) we get,

Therefore, radius of the inner circle is 110 meters.

Therefore, diameter of the inner circle meters

Therefore, diameter of the inner circle is .

Hence, the correct answer is option (*c*).

#### Page No 13.68:

#### Question 8:

The circumference of a circle is 100 m. The side of a square inscribed in the circle is

(a) 50$\sqrt{2}$

(b) $\frac{50}{\mathrm{\pi}}$

(c) $\frac{50\sqrt{2}}{\mathrm{\pi}}$

(d) $\frac{100\sqrt{2}}{\mathrm{\pi}}$

#### Answer:

We have given the circumference of the circle that is 100 cm. If *d* is the diameter of the circle, then its circumference will be.

We obtained diameter of the circle. Look at the figure, diameter of the circle is also the diagonal of the square ABCD.

We know that if we have diagonal of the circle we can calculate the side of the square, using the formula given below,

Substituting the value of diagonal we get,

Therefore, side of the inscribed square is.

Hence, the correct answer is option (*d*).

#### Page No 13.68:

#### Question 9:

The area of the incircle of an equilateral triangle of side 42 cm is

(a) $22\sqrt{3}c{m}^{2}$

(b) 231 cm^{2}

(c) 462 cm^{2}

(d) 924 cm^{2}

#### Answer:

Let ABC be the equilateral triangle such that AB = BC = CA = 42 cm. Also, let O be the centre and *r* be the radius of its incircle.

AB, BC and CA are tangents to the circle at M, N and P.

∴ OM = ON = OP = *r*

Area of ΔABC = Area (ΔOAB) + Area (ΔOBC) + Area (ΔOCA)

Area of the circle =

Hence, the correct answer is option (*c*).

#### Page No 13.68:

#### Question 10:

The area of incircle of an equilateral triangle is 154 cm^{2} . The perimeter of the triangle is

(a) 71.5 cm

(b) 71.7 cm

(c) 72.3 cm

(d) 72.7 cm

#### Answer:

Area of incircle of equilateral triangle is 154

We have to find the perimeter of the triangle. So we will use area to get,

As triangle is equilateral so,

So,

So,

Therefore perimeter of the triangle is,

Therefore the answer is (*d*).

#### Page No 13.68:

#### Question 11:

The area of the largest triangle that can be inscribed in a semi-circle of radius r, is

(a) r^{2}

(b) 2r^{2}

(c) r^{3}

(d) 2r^{3}

#### Answer:

The triangle with the largest area will be symmetrical as shown in the figure.

Let the radius of the circle be *r*.

Hence,

Therefore the answer is (*a*).

#### Page No 13.68:

#### Question 12:

The perimeter of a triangle is 30 cm and the circumference of its incircle is 88 cm. The area of the triangle is

(a) 70 cm^{2}

(b) 140 cm^{2}

(c) 210 cm^{2}

(d) 420 cm^{2}

#### Answer:

We have to find the area of the given triangle.

Perimeter of triangle is 30 cm.

Let the radius of the circle be *r*.

We have,

Therefore,

Therefore the answer is (*c*).

#### Page No 13.68:

#### Question 13:

The area of a circle is 220 cm^{2}. The area of ta square inscribed in it is

(a) 49 cm^{2}

(b) 70 cm^{2}

(c) 140 cm^{2}

(d) 150 cm^{2}

#### Answer:

Let BD be the diameter and diagonal of the circle and the square respectively.

We know that area of the circle

Area of the circle

Multiplying both sides of the equation by 7 we get,

Dividing both sides of the equation by 22 we get,

As we know that diagonal of the square is the diameter of the square.

………(1)

Substituting in equation (1) we get,

Area of the square

Therefore, area of the square is.

Hence, the correct answer is option (c).

#### Page No 13.69:

#### Question 14:

If the circumference of a circle increases from 4π to 8π, then its area is

(a) halved

(b) doubled

(c) tripled

(d) quadrupled

#### Answer:

Let the circumference

Therefore, area of the circle when radius of the circle is 2 can be calculated as below,

……..(1)

Now when circumference is, then the radius of the circle is calculated as below,

Therefore, area of the circle when radius of the circle is 2 can be calculated as below,

……..(2)

Therefore, from equation (1) and (2) we can say that its area is.

Hence, the correct answer is option (*d*).

#### Page No 13.69:

#### Question 15:

If the radius of a circle is diminished by 10%, then its area is diminished by

(a) 10%

(b) 19%

(c) 20%

(d) 36%

#### Answer:

Let *x* be the initial radius of the circle.

Therefore, its area is ……..(1)

It is given that the radius is diminished by 10%, therefore, its new radius is calculated as shown below,

Now we will find the percentage decreased in the area.

Therefore, its area is diminished by.

Hence, the correct answer is option (*b*).

#### Page No 13.69:

#### Question 16:

If the area of a square is same as the area of a circle, then the ratio of their perimeters, in terms of π, is

(a) π : $\sqrt{3}$

(b) 2 : $\sqrt{\mathrm{\pi}}$

(c) 3 : $\mathrm{\pi}$

(d) $\mathrm{\pi}:\sqrt{2}$

#### Answer:

We have given that area of a circle of radius *r* is equal to the area of a square of side *a*.

We have to find the ratio of the perimeters of circle and square.

………(1)

Now we will substitute in equation (1).

Therefore, ratio of their perimeters is.

Hence, the correct answer is (*b*).

#### Page No 13.69:

#### Question 17:

The area of the largest triangle that can be inscribed in a semi-circle of radius r is

(a) 2*r*

(b) *r*^{2}

(c) *r*

(d) $\sqrt{r}$

#### Answer:

The triangle with the largest area will be symmetrical as shown in the figure.

Let the radius of the circle be *r*.

Hence,

Therefore the answer is (*b*).

#### Page No 13.69:

#### Question 18:

The ratio of the areas of a circle and an equilateral triangle whose diameter and a side are respectively equal, is

(a) $\mathrm{\pi}:\sqrt{2}$

(b) $\mathrm{\pi}:\sqrt{3}$

(c) $\sqrt{3}:\mathrm{\pi}$

(d) $\sqrt{2}:\mathrm{\pi}$

#### Answer:

We are given that diameter and side of an equilateral triangle are equal.

Let d and a are the diameter and side of circle and equilateral triangle respectively.

We know that area of the circle

Area of the equilateral triangle

Now we will find the ratio of the areas of circle and equilateral triangle.

We know that radius is half of the diameter of the circle.

Now we will substitute in the above equation,

Therefore, ratio of the areas of circle and equilateral triangle is.

Hence, the correct answer is option (*b*).

#### Page No 13.69:

#### Question 19:

If the sum of the areas of two circles with radii r_{1} and r_{2} is equal to the area of a circle of radius r, then ${r}_{1}^{2}+{r}_{2}^{2}$

(a) >r^{2}

(b) =r^{2}

(c) <r^{2}

(d) None of these

#### Answer:

We have given area of the circle of radius *r*_{1}_{ }plus area of the circle of radius *r*_{2} is equal to the area of the circle of radius *r*.

Therefore, we have,

Cancelling, we get

Therefore, .

Hence, the correct answer is option (*b*).

#### Page No 13.69:

#### Question 20:

If the perimeter of a semi-circular protractor is 36 cm, then its diameter is

(a) 10 cm

(b) 12 cm

(c) 14 cm

(d) 16 cm

#### Answer:

We know that perimeter of a semi-circle of radius *r* ………(1)

We have given the perimeter of the semi-circle and we are asked to find the diameter of the semi-circle.

Therefore, substituting the perimeter of the semi-circle in equation (1) we get,

Multiplying both sides of the equation by 2 we get,

Substituting we get,

Now we will multiply both sides of the equation by 7.

Adding like terms we get,

Dividing both sides of the equation 72 we get,

Therefore, radius of the semi circle is 7cm.

Now we will find the diameter.

Hence, diameter of the semi-circle is.

Therefore, the correct answer is (*c*).

#### Page No 13.69:

#### Question 21:

The perimeter of the sector *OAB *shown in the following figure, is

(a) $\frac{64}{3}cm$

(b) 26 cm

(c) $\frac{64}{5}cm$

(d) 19 cm

#### Answer:

We know that perimeter of a sector of radius *r* ………(1)

We have given sector angle and radius of the sector and we are asked to find perimeter of the sector OAB.

Therefore, substituting the corresponding values of the sector angle and radius in equation (1) we get,

………(2)

We will simplify equation (2) as shown below,

Substituting we get,

Now we will make the denominator same.

Therefore, perimeter of the sector is.

Hence, the correct answer is option (*a*).

#### Page No 13.69:

#### Question 22:

If the perimeter of a sector of a circle of radius 6.5 cm is 29 cm, then its area is

(a) 58 cm^{2}

(b) 52 cm^{2}

(c) 25 cm^{2}

(d) 56 cm^{2}

#### Answer:

We know that perimeter of a sector of radius *r* ………(1)

We have given perimeter of the sector and radius of the sector and we are asked to find the area of the sector. For that we have to find the sector angle.

Therefore, substituting the corresponding values of perimeter and radius in equation (1) we get,

………(2)

We will simplify equation (2) as shown below,

Dividing both sides of the equation by , we get,

Subtracting 1 from both sides of the equation we get,

………(3)

We know that area of the sector

From equation (3), we get

Area of the sector

Substituting we get,

Area of the sector

Therefore, area of the sector is .

Hence, the correct answer is option (*b*).

#### Page No 13.69:

#### Question 23:

If the area of a sector of a circle bounded by an arc of length 5π cm is equal to 20π cm^{2}, then its radius is

(a) 12 cm

(b)16 cm

(c) 8 cm

(d) 10 cm

#### Answer:

We have given length of the arc and area of the sector bounded by that arc and we are asked to find the radius of the circle.

We know that area of the sector .

Now we will substitute the values.

……..(1)

……..(2)

Now we will divide equation (1) by equation (2),

Now we will cancel the like terms.

Therefore, radius of the circle is.

Hence, the correct answer is option (*c*).

#### Page No 13.69:

#### Question 24:

The area of the circle that can be inscribed in a square of side 10 cm is

(a) 40 π cm^{2}

(b) 30 π cm^{2}

(c) 100 π cm^{2}

(d) 25 π cm^{2}

#### Answer:

We know that ABCD is a square of length 10 cm. A circle is inscribed in the square therefore, all the sides of the square are become tangents of the circle.

By, the tangent property, we have

If we join PR then it will be the diameter of the circle of 10 cm.

Therefore, radius of the circle = 5cm

Therefore, area of the circle is.

Hence, the correct answer is option (*d*).

#### Page No 13.69:

#### Question 25:

If the difference between the circumference and radius of a circle is 37 cm, then its area is

(a) 154 cm^{2}

(b) 160 cm^{2}

(c) 200 cm^{2}

(d) 150 cm^{2}

#### Answer:

We have given the difference between circumference and radius of the circle.

Let C be the circumference, *r* be the radius and A be the area of the circle.

Therefore, from the given condition we have

Now we will substitute.

Now we will substitute the value of *r* in.

Now we will substitute.

Therefore, area of the circle is.

Hence, the correct answer is option (*a*).

#### Page No 13.69:

#### Question 26:

The area of a circular path of uniform width h surrounding a circular region of radius r is

(a) $\mathrm{\pi}(2\mathrm{r}+\mathrm{h})\mathrm{r}$

(b) $\mathrm{\pi}(2\mathrm{r}+\mathrm{h})h$

(c) $\mathrm{\pi}(\mathrm{h}+\mathrm{r})\mathrm{r}$

(d) $\mathrm{\pi}(\mathrm{h}+\mathrm{r})h$

#### Answer:

We have

Therefore, radius of the outer circle will be.

Now we will find the area between the two circles.

Cancelling we get,

Therefore, area of the circle is.

Hence, the correct answer is option (*b*).

#### Page No 13.69:

#### Question 27:

If *AB* is a chord of length $5\sqrt{3}$ cm of a circle with centre *O* and radius 5 cm, then area of sector *OAB* is

(a) $\frac{3\mathrm{\pi}}{8}c{m}^{2}\phantom{\rule{0ex}{0ex}}$

(b) $\frac{8\mathrm{\pi}}{3}c{m}^{2}$

(c) $25{\mathrm{\pi cm}}^{2}$

(d) $\frac{25\mathrm{\pi}}{3}c{m}^{2}$

#### Answer:

We have to find the area of the sector OAB.

We have,

So,

Hence,

Therefore area of the sector,

So answer is (*d*)

#### Page No 13.70:

#### Question 28:

The area of a circle whose area and circumference are numerically equal, is

(a) 2$\mathrm{\pi}$ sq. units

(b) 4 $\mathrm{\pi}$ sq. units

(c) 6$\mathrm{\pi}$ sq. units

(d) 8$\mathrm{\pi}$ sq. units

#### Answer:

We have given that circumference and area of a circle are numerically equal.

Let it be *x*.

Let r be the radius of the circle, therefore, circumference of the circle is and area of the circle will be.

Therefore, from the given condition we have,

………(1)

………(2)

Therefore, from equation (1) get . Now we will substitute this value in equation (2) we get,

Simplifying further we get,

Cancelling *x* we get,

Now we will cancel

………(3)

Now we will multiply both sides of the equation (3) by we get,

Therefore, area of the circle is.

Hence, option (*b*) is correct.

#### Page No 13.70:

#### Question 29:

If diameter of a circle is increased by 40%, then its area increase by

(a) 96%

(b) 40%

(c) 80%

(d) 48%

#### Answer:

If *d* is the original diameter of the circle, then the original radius is.

If diameter of the circle increases by 40%, then new diameter of the circle is calculated as shown below,

That is new diameter

So, new area will be.

Now we will calculate the change in area.

Therefore, its area is increased by.

Hence, the correct answer is option (*a*).

#### Page No 13.70:

#### Question 30:

In the following figure, the shaded area is

(a) 50 (π−2) cm^{2}

(b) 25 (π−2) cm^{2}

(c) 25 (π+2) cm^{2}

(d) 5 (π−2) cm^{2}

#### Answer:

Area of the shaded region is-

So the answer is (*b*).

#### Page No 13.70:

#### Question 31:

In the following figure, the area of the segment *PAQ* is

(a) $\frac{{a}^{2}}{4}\left(\mathrm{\pi}+2\right)$

(b) $\frac{{a}^{2}}{4}\left(\mathrm{\pi}-2\right)$

(c) $\frac{{a}^{2}}{4}\left(\mathrm{\pi}-1\right)$

(d) $\frac{{a}^{2}}{4}\left(\mathrm{\pi}+1\right)$

#### Answer:

We have to find area of segment PAQ.

We know that.

Substituting the values we get,

Substituting and we get,

Now we will make the denominator same.

Therefore, area of the segment PAQ is.

Hence, the correct answer is option (*b*).

#### Page No 13.70:

#### Question 32:

In the following figure, the area of segment *ACB* is

(a) $\left(\frac{\mathrm{\pi}}{3}-\frac{\sqrt{3}}{2}\right){r}^{2}$

(b) $\left(\frac{\mathrm{\pi}}{3}+\frac{\sqrt{3}}{2}\right){r}^{2}$

(c) $\left(\frac{\mathrm{\pi}}{3}-\frac{\sqrt{2}}{3}\right){r}^{2}$

(d) None of these

Figure

#### Answer:

We have to find area of segment ACB.

We know that.

Substituting the values we get,

Substituting and we get,

Therefore, area of the segment ACB is.

Hence, the correct answer is option (*d*).

#### Page No 13.70:

#### Question 33:

If the area of a sector of a circle bounded by an arc of length 5π cm is equal to 20π cm^{2}, then the radius of the circle

(a) 12 cm

(b) 16 cm

(c) 8 cm

(d) 10 cm

#### Answer:

We have given length of the arc and area of the sector bounded by that arc and we are asked to find the radius of the circle.

If *l* is the length of the arc, A is the area of the arc and *r* is the radius of the circle, then we know the expression of the area of the sector in terms of the length of the arc and radius of the circle.

Now we will substitute the corresponding values of length of the arc and area of the sector.

Multiplying both sides of the equation by 2 we get,

Dividing both sides of the equation by we get,

Therefore, radius of the circle is .

Hence, the correct answer is option (*c*).

#### Page No 13.70:

#### Question 34:

In the following figure, the ratio of the areas of two sectors S_{1} and S_{2} is

(a) 5 : 2

(b) 3 : 5

(c) 5 : 3

(d) 4 : 5

figure

#### Answer:

Area of the sector,

Area of the sector,

Now we will take the ratio,

Now we will simplify the ratio as below,

Substituting the values we get,

Therefore, ratio of the areas of the two sectors is .

Hence, the correct answer is option (*d*).

#### Page No 13.71:

#### Question 35:

If the area of a sector of a circle is $\frac{5}{18}$ of the area of the circle, then the sector angle is equal to

(a) 60°

(b) 90°

(c) 100°

(d) 120°

#### Answer:

We have given that area of the sector is of the area of the circle.

Therefore, area of the sector area of the circle

Now we will simplify the equation as below,

Now we will multiply both sides of the equation by 360,

Therefore, sector angle is.

Hence, the correct answer is option (*c*).

#### Page No 13.71:

#### Question 36:

If he area of a sector of a circle is $\frac{7}{20}$ of the area of the circle, then the sector angle is equal to

(a) 110°

(b) 130°

(c) 100°

(d) 126°

#### Answer:

We have given that area of the sector is of the area of the circle.

Therefore, area of the sector area of the circle

Now we will simplify the equation as below,

Now we will multiply both sides of the equation by 360,

Therefore, sector angle is.

Hence, the correct answer is option (*d*).

#### Page No 13.71:

#### Question 37:

In the following figure, If ABC is an equilateral triangle, then shaded area is equal to

(a) $\left(\frac{\mathrm{\pi}}{3}-\frac{\sqrt{3}}{4}\right){r}^{2}$

(b) $\left(\frac{\mathrm{\pi}}{3}-\frac{\sqrt{3}}{2}\right){r}^{2}$

(c) $\left(\frac{\mathrm{\pi}}{3}+\frac{\sqrt{3}}{4}\right){r}^{2}$

(d) $\left(\frac{\mathrm{\pi}}{3}+\sqrt{3}\right){r}^{2}$

Figure

#### Answer:

We have given that ABC is an equilateral triangle.

As we know that,.

Area of the shaded region = area of the segment BC.

Let

Substituting the values we get,

Substituting and we get,

Therefore, area of the shaded region is . Hence, the correct answer is option (*a*).

#### Page No 13.71:

#### Question 38:

In the following figure, the area of the shaded region is

(a) 3π cm^{2}

(b) 6π cm^{2}

(c) 9π cm^{2}

(d) 7π cm^{2}

#### Answer:

In the figure,

and,

Therefore, area of the shaded region is.

Hence, the correct answer is option (*a*).

#### Page No 13.71:

#### Question 39:

If the perimeter of a circle is equal to that of a square, then the ratio of their areas is

(a) 13 : 22

(b) 14 : 11

(c) 22 : 13

(d) 11 : 14

#### Answer:

We have given that perimeter of circle of radius *r* is equal to square of side *a*.

perimeter of the circle = perimeter of the square

Now we will substitute value of *r* in the following equation

Substituting we get,

Hence, ratio of the areas of the circle and square is.

Therefore, the correct answer is (*b*).

#### Page No 13.71:

#### Question 40:

The radius of a circle is 20 cm. It is divided into four parts of equal area by drawing three concentric circles inside it. Then, the radius of the largest of three concentric circles drawn is

(a) 10$\sqrt{5}$ cm

(b) 10$\sqrt{3}$ cm

(c) 10$\sqrt{5}$ cm

(d) 10$\sqrt{2}$ cm

#### Answer:

The circle can be divided into four parts of equal area by drawing three concentric circles inside it as,

It is given that OB = 20 cm. Let OA = *x*.

Since the circle is divided into four parts of equal area by the three concentric circles, we have,

Area of the fourth region = Area of the given circle

Therefore, the correct answer is (*b*).

#### Page No 13.71:

#### Question 41:

The area of a sector whose perimeter is four times its radius r units, is

(a) $\frac{{r}^{2}}{4}$ sq. units

(b) 2r^{2} sq. units

(c) r^{2} sq.units

(d) $\frac{{r}^{2}}{2}$sq. units

#### Answer:

We know that perimeter of the sector .

We have given that perimeter of the sector is four times the radius.

Subtracting 2*r* from both sides of the equation,

Dividing both sides of the equation 2*r* we get,

……….(1)

Let us find the area of the sector.

Substituting we get,

Hence, area of the sector is.

Therefore, the correct answer is (*c*).

#### Page No 13.71:

#### Question 42:

If a chord of a circle of radius 28 cm makes an angle of 90 ° at the centre, then the area of the major segment is

(a) 392 cm^{2}

(b) 1456 cm^{2}

(c) 1848 cm^{2}

(d) 2240 cm^{2}

#### Answer:

Area of major segment,

So the answer is (*d*)

#### Page No 13.71:

#### Question 43:

If area of a circle inscribed in an equilateral triangle is 48π square units, then perimeter of the triangle is

(a) 17$\sqrt{3}$ units

(b) 36 units

(c) 72 units

(d) 48$\sqrt{3}$ units

#### Answer:

Let the circle of radius *r *be inscribed in an equilateral triangle of side *a*.

Area of the circle is given as 48π.

⇒ π*r*^{2} = 48π

⇒ *r*^{2} = 48

Now, it is clear that ON⊥BC. So, ON is the height of ΔOBC corresponding to BC.

Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB = 3 × Area of ΔOBC

Thus, perimeter of the equilateral triangle = 3 × 24 units = 72 units

So the answer is (*c*).

#### Page No 13.71:

#### Question 44:

The hour hand of a clock is 6 cm long. The area swept by it between 11.20 am and 11.55 am is

(a) 2.75 cm^{2}

(b) 5.5 cm^{2}

(c) 11 cm^{2}

(d) 10 cm^{2}

#### Answer:

Hour hand movesin one minute.

So, area,

So the answer is (*b*)

#### Page No 13.72:

#### Question 45:

*ABCD* is a square of side 4 cm. If *E* is a point in the interior of the square such that Δ*CED* is equilateral, then area of Δ *ACE* is

(a) $2\sqrt{3}-1c{m}^{2}$

(b) $4\sqrt{3}-1c{m}^{2}$

(c) $6\sqrt{3}-1c{m}^{2}$

(d) $8\sqrt{3}-1c{m}^{2}$

#### Answer:

We have the following diagram.

Since is equilateral,

Therefore,

Now,

Since AC is diagonal of sqr.ABCD

Therefore,

Therefore we get,

Now, in , draw a perpendicular EM to the base AC.

So in ,

Therefore,

Now in ,

So the answer is (*b*)

#### Page No 13.72:

#### Question 46:

If the area of a circle is equal to the sum of the areas of two circles of diameters 10 cm and 24 cm, then diameter of the large circle (in cm) is

(a) 34

(b) 26

(c) 17

(d) 14

#### Answer:

Let the diameter of the larger circle be *d*

Now, Area of larger circle = Area of circle having diameter 10 cm + Area of circle having diameter 24 cm

$\Rightarrow \mathrm{\pi}{\left(\frac{d}{2}\right)}^{2}=\mathrm{\pi}{\left(\frac{10}{2}\right)}^{2}+\mathrm{\pi}{\left(\frac{24}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{d}{2}\right)}^{2}={\left(5\right)}^{2}+{\left(12\right)}^{2}$

$\Rightarrow {\left(\frac{d}{2}\right)}^{2}=25+144\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{d}{2}\right)}^{2}={13}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{d}{2}=13\phantom{\rule{0ex}{0ex}}\Rightarrow d=26\mathrm{cm}$

Hence, the correct answer is option (b).

#### Page No 13.72:

#### Question 47:

If π is taken as $\frac{22}{7}$, the distance (in metres) covered by a wheel of diameter 35 cm, in one revolution, is

(a) 2.2

(b) 1.1

(c) 9.625

(d) 96.25

#### Answer:

The disstance covered by the wheel in one revolution is equal to the circumference of the wheel.

$\mathrm{Circumference}=\mathrm{\pi}{r}^{2}\phantom{\rule{0ex}{0ex}}=\frac{22}{7}{\left(\frac{35}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{22}{7}{\left(\frac{35}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=962.5\mathrm{cm}$

The distance covered by the wheel in one revolution (in m) is given by

$\frac{962.5}{100}=9.625\mathrm{m}$

Hence, the correct answer is option (c).

#### Page No 13.72:

#### Question 48:

ABCD is a rectangle whose three vertices are B (4,0), C (4,3) and D (0, 3). The length of one of its diagonals is [CBSE 2014]

(a) 5

(b) 4

(c) 3

(d) 25

#### Answer:

BD = $\sqrt{{\left(0-4\right)}^{2}+{\left(3-0\right)}^{2}}$

$=\sqrt{{4}^{2}+{3}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{25}\phantom{\rule{0ex}{0ex}}=5\mathrm{units}$

Hence, the correct answer is option (a).

#### Page No 13.72:

#### Question 49:

Area of the largest triangle that can be inscribed in a semi-circle of radius *r* units is

(a)* r ^{2 }*sq. units (b) $\frac{1}{2}$

*r*sq. units (c) 2

^{2 }*r*sq. units (d) $\sqrt{2}$

^{2 }*r*sq. units

^{2 }#### Answer:

Area of a triangle = $\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$

$\Rightarrow Area=\frac{1}{2}\times \left(2r\right)\times r={r}^{2}$

So, the area of the largest triangle that can be inscribed in a semi-circle of radius *r* is *r*^{2}

Hence, the correct answer is option (a).

#### Page No 13.72:

#### Question 50:

If the sum of the areas of two circles with radii ${r}_{1}$ and ${r}_{2}$ is equal to the area of circle of radius $r$, then

(a) $r={r}_{1}+{r}_{2}$ (b) ${{r}_{1}}^{2}+{{r}_{2}}^{2}={r}^{2}$ (c) ${r}_{1}+{r}_{2}r$ (d) ${{r}_{1}}^{2}+{{r}_{2}}^{2}{r}^{2}$

#### Answer:

The radius of the two circles are ${r}_{1}\mathrm{and}{r}_{2}$.

Now, according to the given condition

Area of circle with radius r = Area of circle with radius ${r}_{1}$+ Area of circle with radius ${r}_{2}$

${\mathrm{\pi r}}^{2}={\mathrm{\pi r}}_{1}^{2}+{\mathrm{\pi r}}_{2}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{r}}^{2}={\mathrm{r}}_{1}^{2}+{\mathrm{r}}_{2}^{2}$

Hence, the correct answer is option (b).

#### Page No 13.72:

#### Question 51:

If the sum of the circumferences of the two circles with radii ${r}_{1}$ and ${r}_{2}$ is equal to the circumference of a circle of radius $r$, then

(a) $r={r}_{1}+{r}_{2}$ (b) ${r}_{1}+{r}_{2}r$ (c) ${r}_{1}+{r}_{2}r$ (d) None of these

#### Answer:

The circumference of the circle with radius r = circumference of the circle with radius ${r}_{1}$+ circumference of the circle with radius ${r}_{2}$

$2\mathrm{\pi r}=2{\mathrm{\pi r}}_{1}+2{\mathrm{\pi r}}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{r}={\mathrm{r}}_{1}+{\mathrm{r}}_{2}$

Hence, the correct answer is option (a).

#### Page No 13.72:

#### Question 52:

If the circumference of a circle and the perimeter of a square are equal , then

(a) Area of the circle = Area of the square (b) Area of the circle < Area of the square

(c) Area of the circle > Area of the square (d) nothing definite can be said

#### Answer:

Circumference of the circle = Perimeter of the square

$2\mathrm{\pi}r=4a\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}r\mathit{=}\frac{\mathit{2}\mathit{a}}{\mathit{\pi}}\phantom{\rule{0ex}{0ex}}$

Now area of the circle = $\mathrm{\pi}{r}^{2}=\mathrm{\pi}{\left(\frac{2\mathrm{a}}{\mathrm{\pi}}\right)}^{2}=\frac{4{\mathrm{a}}^{2}}{\mathrm{\pi}}$

Area of the square = *a*^{2}

So, the area of the circle > area of the square

Hence, the correct answer is option (c).

#### Page No 13.72:

#### Question 53:

If the perimeter of a circle is equal to that of a square , then the ratio of their areas is

(a) 22 : 7 (b) 14 : 11 (c) 7 : 22 (d) 11 : 14

#### Answer:

Perimeter of a circle = Perimeter of a square

$\Rightarrow 2\mathrm{\pi}r=4a\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{2a}{\mathrm{\pi}}$

Now area of the circle = $\mathrm{\pi}{r}^{2}=\mathrm{\pi}{\left(\frac{2\mathrm{a}}{\mathrm{\pi}}\right)}^{2}=\frac{4{\mathrm{a}}^{2}}{\mathrm{\pi}}$

Area of the square = *a*^{2
$\frac{\mathrm{Area}\mathrm{of}\mathrm{circle}}{\mathrm{Area}\mathrm{of}\mathrm{square}}=\frac{{\displaystyle \frac{4{a}^{2}}{\mathrm{\pi}}}}{{a}^{2}}=\frac{4}{{\displaystyle \frac{22}{7}}}=\frac{14}{11}$}

Hence, the required ratio is 14 : 11.

The correct answer is option (b).

#### Page No 13.73:

#### Question 1:

If the area of a circle is 154 cm^{2}, then its perimeter is __________.

#### Answer:

Let the radius of the circle be *r* cm.

Area of the circle = 154 cm^{2}

$\Rightarrow \mathrm{\pi}{r}^{2}=154{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{22}{7}\times {r}^{2}=154\phantom{\rule{0ex}{0ex}}\Rightarrow r=\sqrt{\frac{154\times 7}{22}}=7\mathrm{cm}$

∴ Perimeter of the circle

= Circumference of the circle

$=2\mathrm{\pi}r\phantom{\rule{0ex}{0ex}}=2\times \frac{22}{7}\times 7\phantom{\rule{0ex}{0ex}}=44\mathrm{cm}$

If the area of a circle is 154 cm^{2}, then its perimeter is _____44 cm_____.

#### Page No 13.73:

#### Question 2:

If the perimeter of a circle is equal to that of a square, then the ratio of their areas is ___________.

#### Answer:

Ans

#### Page No 13.73:

#### Question 3:

The area of the circle that can be inscribed in a square of side 6 cm is __________.

#### Answer:

If a circle is inscribed in a square, then the diameter of the circle is equal to the side of the square.

Let the radius of the circle be *r* cm.

Now,

Diameter of the circle = Side of the square

⇒ 2*r* = 6 cm

⇒ *r* = 3 cm

∴ Area of the inscribed circle $=\mathrm{\pi}{r}^{2}=\mathrm{\pi}\times {\left(3\right)}^{2}=9\mathrm{\pi}{\mathrm{cm}}^{2}$

The area of the circle that can be inscribed in a square of side 6 cm is $\overline{)9\mathrm{\pi}{\mathrm{cm}}^{2}}$.

#### Page No 13.73:

#### Question 4:

The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is __________.

#### Answer:

Let the radius of the circle be *r* cm.

Radius of first circle, *r*_{1} = 24 cm

Radius of second circle, *r*_{2} = 7 cm

Now,

Area of the circle = Area of first circle + Area of second circle

$\therefore \mathrm{\pi}{r}^{2}=\mathrm{\pi}{r}_{1}^{2}+\mathrm{\pi}{r}_{2}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}={r}_{1}^{2}+{r}_{2}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}={\left(24\right)}^{2}+{\left(7\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=576+49=625$

$\Rightarrow r=\sqrt{625}=25\mathrm{cm}$

∴ Diameter of circle = 2*r* = 2 × 25 = 50 cm

The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is ____50 cm____.

#### Page No 13.73:

#### Question 5:

The area of the square that can be inscribed in a circle of radius 8 cm is ___________.

#### Answer:

If a square is inscribed in a circle, then the length of diagonal of square is equal to the diameter of the circle.

Let the side of the square be *a* cm.

∴ Length of diagonal of square = $\sqrt{2}$ × Side of square = $\sqrt{2}a$

Radius of the circle, *r* = 8 cm

∴ Diameter of the circle = 2*r* = 2 × 8 = 16 cm

Now,

Length of diagonal of square = Diameter of the circle

$\Rightarrow \sqrt{2}a=16\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{16}{\sqrt{2}}=8\sqrt{2}\mathrm{cm}$

∴ Area of the square $={a}^{2}={\left(8\sqrt{2}\mathrm{cm}\right)}^{2}=128{\mathrm{cm}}^{2}$

The area of the square that can be inscribed in a circle of radius 8 cm is $\overline{)128{\mathrm{cm}}^{2}}$.

#### Page No 13.73:

#### Question 6:

It is proposed to build a single park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be ___________.

#### Answer:

Let the radius of the new circular park be *r* m.

Suppose the radii of two circular parks be *r*_{1} m and *r*_{2} m.

Diameter of the first circular park = 16 m

⇒ 2*r*_{1} = 16

⇒ *r*_{1} = 8 m

Diameter of the second circular park = 12 m

⇒ 2*r*_{2} = 12

⇒ *r*_{2} = 6 m

Now,

Area of the new circular park = Area of first circular park + Area of second circular park

$\mathrm{\pi}{r}^{2}=\mathrm{\pi}{r}_{1}^{2}+\mathrm{\pi}{r}_{2}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}={r}_{1}^{2}+{r}_{2}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{\mathit{2}}={\left(8\right)}^{2}+{\left(6\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{\mathit{2}}=64+36=100$

$\Rightarrow r=\sqrt{100}$ = 10 m

Thus, the radius of the new park is 10 m.

It is proposed to build a single park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be ____10 m____.

#### Page No 13.73:

#### Question 7:

The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm, is __________.

#### Answer:

Let the radius of the circle be *r* cm.

Suppose the radii of two given circles be *r*_{1} cm and *r*_{2} cm.

Diameter of the first circle = 36 cm

⇒ 2*r*_{1} = 36 .....(1)

Diameter of the second circle = 20 cm

⇒ 2*r*_{2} = 20 .....(2)

Now,

Circumference of the circle = Circumference of the first circle + Circumference of the second circle

$\Rightarrow 2\mathrm{\pi}r=2\mathrm{\pi}{r}_{1}+2\mathrm{\pi}{r}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 2r=2{r}_{1}+2{r}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 2r=36+20=56\left[\mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{56}{2}=28\mathrm{cm}$

Thus, the radius of the circle is 28 cm.

The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm, is _____28 cm_____.

#### Page No 13.73:

#### Question 8:

If the difference between outer and inner radii of a circular ring is 14 cm, then the difference between outer and inner circumferences, is __________.

#### Answer:

Let *r*_{1} and *r*_{2} be the radii of the inner and outer circular rings, respectively.

Given: *r*_{2 }− *r*_{1} = 14 cm .....(1)

∴ Difference between their outer and inner circumferences

$=2\mathrm{\pi}{r}_{2}-2\mathrm{\pi}{r}_{1}\phantom{\rule{0ex}{0ex}}=2\mathrm{\pi}\left({r}_{2}-{r}_{1}\right)\phantom{\rule{0ex}{0ex}}=2\times \frac{22}{7}\times 14\left[\mathrm{Using}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}=88\mathrm{cm}$

If the difference between outer and inner radii of a circular ring is 14 cm, then the difference between outer and inner circumferences, is _____88 cm_____.

#### Page No 13.73:

#### Question 9:

The area of a sector whose perimeter is four times its radius *r*, is ___________.

#### Answer:

Let *l* be the length of the arc of the corresponding sector.

Radius of the sector* *= *r *units

We know

Perimeter of the sector = *l* + 2*r*

∴ *l* + 2*r* = 4*r* (Given)

⇒ *l* = 4*r* − 2*r* = 2*r* .....(1)

∴ Area of the sector $=\frac{1}{2}lr=\frac{1}{2}\times 2r\times r={r}^{2}\mathrm{units}$ [Using (1)]

The area of a sector whose perimeter is four times its radius *r*, is _____ r^{2} units_____.

#### Page No 13.73:

#### Question 10:

Two circles touch each other externally. The sum of their areas is 490 π cm^{2}. Their centres are separated by 28 cm. The difference of their perimeters is ___________.

#### Answer:

Let *R* and *r* be the radii of the two circles. Suppose *R *> *r*.

We know that if two circles touch each other externally, then the distance between their centres is equal to the sum of the radii of the two circles.

∴ *R*_{ }+ *r* = 28 cm .....(1)

Also,

Sum of the areas of the two circles = $490\mathrm{\pi}$ cm^{2}

$\Rightarrow \mathrm{\pi}{R}^{2}+\mathrm{\pi}{r}^{2}=490\mathrm{\pi}\phantom{\rule{0ex}{0ex}}\Rightarrow {R}^{2}+{r}^{2}=490.....\left(2\right)$

Now,

${\left(R+r\right)}^{2}+{\left(R-r\right)}^{2}=2\left({R}^{2}+{r}^{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(28\right)}^{2}+{\left(R-r\right)}^{2}=2\times 490\left[\mathrm{Using}\left(1\right)\mathrm{and}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(R-r\right)}^{2}=980-784=196\phantom{\rule{0ex}{0ex}}\Rightarrow R-r=\sqrt{196}=14\mathrm{cm}.....\left(3\right)$

∴ Difference between their perimeters

$=2\mathrm{\pi}R-2\mathrm{\pi}r\phantom{\rule{0ex}{0ex}}=2\mathrm{\pi}\left(R-r\right)\phantom{\rule{0ex}{0ex}}=2\times \frac{22}{7}\times 14\left[\mathrm{Using}\left(3\right)\right]\phantom{\rule{0ex}{0ex}}=88\mathrm{cm}$

Two circles touch each other externally. The sum of their areas is 490 π cm^{2}. Their centres are separated by 28 cm. The difference of their perimeters is _______88 cm________.

#### Page No 13.73:

#### Question 11:

If *l *is the arc length of a sector of a circle of radius *r *and *A* is the area of the sector, then *A *: *l = __________.*

#### Answer:

Arc length of the sector* *= *l* units

Radius of the sector* *= *r *units

We know

Area of the sector, *A* $=\frac{1}{2}lr$

$\therefore \frac{A}{l}=\frac{r}{2}$

⇒ *A* : *l* = *r* : 2

If *l *is the arc length of a sector of a circle of radius *r *and *A* is the area of the sector, then *A *: *l *= _____ r : 2_____.

#### Page No 13.73:

#### Question 12:

If the circumferences of two circles are in the ratio *c*_{1} : *c*_{2}, then the ratio of their areas is __________.

#### Answer:

Let *R* and *r* be the radii of the two circles.

Now,

Circumference of first circle : Circumference of second circle = *c*_{1} : *c*_{2} (Given)

$\therefore \frac{2\mathrm{\pi}R}{2\mathrm{\pi}r}=\frac{{c}_{1}}{{c}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{R}{r}=\frac{{c}_{1}}{{c}_{2}}$

Squaring on both sides, we get

$\frac{{R}^{2}}{{r}^{2}}=\frac{{c}_{1}^{2}}{{c}_{2}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{\pi}{R}^{2}}{\mathrm{\pi}{r}^{2}}=\frac{{c}_{1}^{2}}{{c}_{2}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{Area}\mathrm{of}\mathrm{first}\mathrm{circle}}{\mathrm{Area}\mathrm{of}\mathrm{second}\mathrm{circle}}=\frac{{c}_{1}^{2}}{{c}_{2}^{2}}$

∴ Area of first circle : Area of second circle = ${c}_{1}^{2}:{c}_{2}^{2}$

If the circumferences of two circles are in the ratio *c*_{1} : *c*_{2}, then the ratio of their areas is $\overline{){c}_{1}^{2}:{c}_{2}^{2}}$.

#### Page No 13.73:

#### Question 13:

The area of a sector of angle θ° of a circle of radius *r *is _________.

#### Answer:

Radius of the sector = *r*

Angle of the sector = *θ*

∴ Area of the sector = $\frac{\theta}{360\xb0}\times \mathrm{\pi}{r}^{2}=\frac{\mathrm{\pi}{r}^{2}\theta}{360\xb0}$

The area of a sector of angle θ° of a circle of radius *r *is $\overline{)\frac{\mathrm{\pi}{r}^{2}\theta}{360\xb0}}$.

#### Page No 13.73:

#### Question 14:

The length of the arc of a sector of angle θ° of a circle of radius *r *is ___________.

#### Answer:

Radius of the sector = *r*

Angle of the sector = *θ*

∴ Length of arc of the sector = $\frac{\theta}{360\xb0}\times 2\mathrm{\pi}r=\frac{\mathrm{\pi}r\theta}{180\xb0}$

The length of the arc of a sector of angle θ° of a circle of radius *r *is $\overline{)\frac{\mathrm{\pi}r\theta}{180\xb0}}$.

#### Page No 13.73:

#### Question 15:

The area of the minor segment of angle θ° of a circle of radius *r *is ___________.

#### Answer:

Radius of the sector = *r*

Angle of the minor segment = *θ*

∴ Area of the minor segment = $\frac{\theta}{360\xb0}\times \mathrm{\pi}{r}^{2}-{r}^{2}\mathrm{sin}\frac{\theta}{2}\mathrm{cos}\frac{\theta}{2}=\left(\frac{\theta \mathrm{\pi}}{360\xb0}-\mathrm{sin}\frac{\theta}{2}\mathrm{cos}\frac{\theta}{2}\right){r}^{2}$

The area of the minor segment of angle θ° of a circle of radius *r *is $\overline{)\left(\frac{\theta \mathrm{\pi}}{360\xb0}-\mathrm{sin}\frac{\theta}{2}\mathrm{cos}\frac{\theta}{2}\right){r}^{2}}$.

#### Page No 13.73:

#### Question 16:

The difference in the areas of the regular hexagon circumscribing a circle of radius 10 cm and the regular hexagon inscribed in the circle is ___________.

#### Answer:

In the figure, O is the centre of circle of radius 10 cm. ABCDEF is a regular hexagon circumscribing the circle and PQRSTU is a regular hexagon inscribed in the circle.

Here, ∆OPQ is an equilateral triangle.

Length of each side of equilateral ∆OPQ = Radius of the circle = 10 cm

Now,

Area of the regular hexagon PQRSTU

= 6 × Area of equilateral ∆OPQ

$=6\times \frac{\sqrt{3}}{4}\times {\left(10\right)}^{2}$ $\left[\mathrm{Area}\mathrm{of}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}=\frac{\sqrt{3}}{4}{\left(\mathrm{Side}\right)}^{2}\right]$

$=150\sqrt{3}{\mathrm{cm}}^{2}.....\left(1\right)$

Also,

∆OAB is an equilateral triangle.

Length of an altitude of equilateral ∆OAB = Radius of the circle = 10 cm

We know

Length of an altitude of an equilateral triangle = $\frac{\sqrt{3}}{2}\left(\mathrm{Side}\right)$

$\therefore \frac{\sqrt{3}}{2}\times \mathrm{AB}=10\mathrm{cm}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AB}=\frac{20}{\sqrt{3}}\mathrm{cm}$

Area of the regular hexagon ABCDEF

= 6 × Area of equilateral ∆OAB

$=6\times \frac{\sqrt{3}}{4}\times {\left(\frac{20}{\sqrt{3}}\right)}^{2}$ $\left[\mathrm{Area}\mathrm{of}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}=\frac{\sqrt{3}}{4}{\left(\mathrm{Side}\right)}^{2}\right]$

$=200\sqrt{3}{\mathrm{cm}}^{2}.....\left(2\right)$

∴ Required difference in areas

= Area of the regular hexagon ABCDEF − Area of the regular hexagon PQRSTU

$=200\sqrt{3}-150\sqrt{3}\left[\mathrm{Using}\left(1\right)\mathrm{and}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}=50\sqrt{3}{\mathrm{cm}}^{2}$

The difference in the areas of the regular hexagon circumscribing a circle of radius 10 cm and the regular hexagon inscribed in the circle is $\overline{)50\sqrt{3}{\mathrm{cm}}^{2}}$.

#### Page No 13.73:

#### Question 17:

In Fig. 13.121, *ABCD *is a square of side 10 cm and a circle is inscribed in it. The area of the shaded part is __________.

#### Answer:

Let O be the centre of the circle. Suppose the circle touches the sides BC and CD of the square at E and F, respectively.

OE = OF * *(Radius of the circle)

Now, ∠OEC = ∠OFC = 90º (Radius is perpendicular to the tangent at the point of contact)

Also, OE = OF = CF = CE

∴ OECF is a square.

We know that if a circle is inscribed in a square, then the diameter of the circle is equal to the side of the square.

$\therefore \mathrm{OE}=\frac{\mathrm{BC}}{2}=\frac{10}{2}=5\mathrm{cm}$

⇒ Length of each side of the square OECF = 5 cm

Now,

Area of the shaded portion

= $\frac{1}{2}$(Area of the square OECF − Area of the quadrant OEFO)

$=\frac{1}{2}\left[{\left(\mathrm{OE}\right)}^{2}-\frac{1}{4}\times \mathrm{\pi}{\left(\mathrm{OE}\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[{\left(5\right)}^{2}-\frac{1}{4}\times \mathrm{\pi}{\left(5\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{25}{2}\left(1-\frac{\mathrm{\pi}}{4}\right)\phantom{\rule{0ex}{0ex}}=\frac{25}{2}\left(\frac{4-\mathrm{\pi}}{4}\right)$

$=\frac{100-25\mathrm{\pi}}{8}{\mathrm{cm}}^{2}$

In Fig. 13.121, ABCD is a square of side 10 cm and a circle is inscribed in it. The area of the shaded part is $\overline{)\frac{100-25\mathrm{\pi}}{8}{\mathrm{cm}}^{2}}$.

#### Page No 13.73:

#### Question 18:

In Fig. 13.122, two circles of radii 7 cm each are shown. *ABCD *is a rectangle and *AD *and *BC* are the radii. The area of the shaded region is __________.

#### Answer:

In the given figure,

BC = CE = DE = 7 cm (Radius of each circle)

Also, the two circles touch each other externally.

We know that if two circles touch each other externally, then the distance between their centres is equal sum of their radii.

CD = CE + ED = 7 + 7 = 14 cm

Now,

Area of the shaded region

= Area of the rectangle ABCD − (Area of quadrant BCEB + Area of quadrant ADEA)

$=\mathrm{CD}\times \mathrm{BC}-\left[\frac{1}{4}\times \mathrm{\pi}{\left(\mathrm{CE}\right)}^{2}+\frac{1}{4}\times \mathrm{\pi}{\left(\mathrm{DE}\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=14\times 7-\left[\frac{1}{4}\times \frac{22}{7}\times {\left(7\right)}^{2}+\frac{1}{4}\times \frac{22}{7}\times {\left(7\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=98-77\phantom{\rule{0ex}{0ex}}=21{\mathrm{cm}}^{2}$

In Fig. 13.122, two circles of radii 7 cm each are shown. ABCD is a rectangle and AD and BC are the radii. The area of the shaded region is _____21 cm ^{2}_____.

#### Page No 13.73:

#### Question 19:

The area of the largest circle that can be drawn inside a rectangle of length *a* cm and breadth *b* cm (*a > b*) is __________.

#### Answer:

The largest circle that can be drawn inside a rectangle of length *a* cm and breadth *b* cm (*a > b*) will have its diameter equal to the breadth of the rectangle.

Let the radius of the largest circle that can be drawn inside a rectangle of length *a* cm and breadth *b* cm (*a > b*) be *r* cm.

∴ Diameter of the circle = *b*

$\Rightarrow 2r=b\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{b}{2}$

∴ Area of the required circle $=\mathrm{\pi}{r}^{2}=\mathrm{\pi}{\left(\frac{b}{2}\right)}^{2}=\frac{\mathrm{\pi}{b}^{2}}{4}{\mathrm{cm}}^{2}$

The area of the largest circle that can be drawn inside a rectangle of length *a* cm and breadth *b* cm (*a > b*) is $\overline{)\frac{\mathrm{\pi}{b}^{2}}{4}{\mathrm{cm}}^{2}}$.

#### Page No 13.73:

#### Question 20:

The number of revolutions made by a circle of radius *r* to cover a distance *s* is ___________.

#### Answer:

In one revolution, the distance covered by a circle of radius *r* is equal to the circumference of the circle.

Circumference of the circle of radius *r* = $2\mathrm{\pi}r$

Number of revolutions made by a circle of radius *r* to cover a distance $2\mathrm{\pi}r$ = 1

∴ Number of revolutions made by a circle of radius *r* to cover a distance *s* = $\frac{s}{2\mathrm{\pi}r}$

The number of revolutions made by a circle of radius *r* to cover a distance *s* is $\overline{)\frac{s}{2\mathrm{\pi}r}}$.

#### Page No 13.74:

#### Question 1:

What is the ratio of the areas of a circle and an equilateral triangle whose diameter and a side are respectively equal?

#### Answer:

We are given that diameter and side of an equilateral triangle are equal.

Let *d* and *a* are the diameter and side of circle and equilateral triangle respectively.

Therefore *d* = *a*

We know that area of the circle

Area of the equilateral triangle

Now we will find the ratio of the areas of circle and equilateral triangle.

We know that radius is half of the diameter of the circle.

Now we will substitute in the above equation,

Therefore, ratio of the areas of circle and equilateral triangle is.

#### Page No 13.74:

#### Question 2:

If the circumference of two circles are in the ratio 2 : 3, what is the ratio of their areas?

#### Answer:

We are given ratio of circumferences of two circles. If and are circumferences of two circles such that

…… (1)

Simplifying equation (1) we get,

Let and are the areas of the respective circles and we are asked to find their ratio.

…… (2)

We know that substituting this value in equation (2) we get,

Therefore, ratio of their areas is.

#### Page No 13.74:

#### Question 3:

Write the area of the sector of a circle whose radius is* r* and length of the arc is l.

#### Answer:

We know that area of the sector of the circle of radius *r *

Length of the arc

But we have given that length of the arc

…… (1)

Area of the sector

Now we will adjust 2 in the following way,

Area of the sector

Area of the sector

From equation (1) we will substitute

Area of the sector

Area of the sector

Therefore, area of the sector =.

#### Page No 13.74:

#### Question 4:

What is the length (in terms of π) of the arc that subtends an angle of 36° at the centre of a circle of radius 5 cm?

#### Answer:

We have

We have to find the length of the arc.

Substituting the values we get,

……….(1)

Now we will simplify the equation (1) as below,

Therefore, length of the arc is

#### Page No 13.74:

#### Question 5:

What is the angle subtended at the centre of a circle of radius 6 cm by an arc of length 3 π cm?

#### Answer:

We have

We will find the angle subtended at the centre of a circle.

Substituting the values we get,

…… (1)

Now we will simplify the equation (1) as below,

Therefore, angle subtended at the centre of the circle is.

#### Page No 13.74:

#### Question 6:

What is the area of a sector of a circle of radius 5 cm formed by an arc of length 3.5 cm?

#### Answer:

We have

Now we will find the area of the sector.

Substituting the values we get,

…… (1)

Now we will simplify the equation (1) as below,

Therefore, area of the sector is.

#### Page No 13.74:

#### Question 7:

In a circle of radius 10 cm, an arc subtends an angle of 108° at the centre. what is the area of the sector in terms of π?

#### Answer:

We have given the radius of the circle and angle subtended at the centre of the circle.

Now we will find the area of the sector.

Substituting the values we get,

…… (1)

Now we will simplify the equation (1) as below,

Therefore, area of the sector is.

#### Page No 13.74:

#### Question 8:

If a square is inscribed in a circle, what is the ratio of the areas of the circle and the square?

#### Answer:

We have the following situation

Let *BD* be the diameter and diagonal of the circle and the square respectively.

We know that area of the circle

Area of the square

As we know that diagonal of the square is the diameter of the square.

…… (1)

Substituting in equation (1) we get,

Now we will find the ratio of the areas of circle and square.

Now we will simplify the above equation as below,

Therefore, ratio of areas of circle and square is.

#### Page No 13.74:

#### Question 9:

Write the formula for the area of a sector of angle $\theta $ (in degrees) of a circle of radius r.

#### Answer:

Let *r* be the radius of the circle and angle *θ* subtended at the centre of the circle.

Area of the sector of the circle

Therefore, area of the sector is.

#### Page No 13.74:

#### Question 10:

Write the formula for the area of a segment in a circle of radius r given that the sector angle is $\theta $ (in degrees).

#### Answer:

In this figure, centre of the circle is *O*, radius *OA* = *r* and

We are going to find the area of the segment *AXB*.

…… (1)

We know that

We also know that

Substituting these values in equation (1) we get,

Therefore, area of the segment is.

#### Page No 13.74:

#### Question 11:

If the adjoining figure is a sector of a circle of radius 10.5 cm, what is the perimeter of the sector? (Take $\mathrm{\pi}=22/7$)

#### Answer:

Given figure is a quadrant of a circle. We have given radius of sector that is 10.5 cm. Arc AB subtended an angle of 60° at the centre of the circle.

Perimeter of the sector

Substituting the values we get,

Perimeter of the sector …… (1)

Now we will simplify equation (1) as shown below,

Now we will substitute.

Therefore, perimeter of the given sector is.

#### Page No 13.74:

#### Question 12:

If the diameter of a semi-circular protractor is 14 cm, then find its perimeter.

#### Answer:

Let *AB* be the diameter of the semi-circular protractor.

We know that perimeter of the semicircle ……(1)

We have given the diameter of the protractor.

Therefore, radius of the protractor

So, radius of the protractor

Substituting the value of *r* in equation (1) we get,

Substituting we get,

Therefore, perimeter of the semi-circular protractor is.

#### Page No 13.74:

#### Question 13:

An arc subtends an angle of 90° at the centre of the circle of the radius 14 cm. Write the area of minor sector thus formed in terms of π.

#### Answer:

We have given an angle subtended by an arc at the centre of the circle and radius of the circle.

Now we will find the area of the minor sector.

Area of the minor sector =

Substituting the values we get,

Area of the minor sector = …… (1)

Now we will simplify the equation (1) as below,

Area of the minor sector =

Area of the minor sector =

Area of the minor sector = π × 7 × 7

Area of the minor sector = 49π

Therefore, area of the minor sector is.

#### Page No 13.75:

#### Question 14:

Find the area of the largest triangle that can be inscribed in a semi-circle of radius* r* units. [CBSE 2015]

#### Answer:

Radius of the semi circle *r* units

When a largest triangle inscribed in a semicircle, then base = *r* + *r* = 2*r* units

Thus, Area of triangle is given by

$\frac{1}{2}\times \mathrm{base}\times \mathrm{height}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times \left(2r\right)\times r\phantom{\rule{0ex}{0ex}}={r}^{2}\mathrm{square}\mathrm{units}$

#### Page No 13.75:

#### Question 15:

Find the area of sector of circle of radius 21 cm and central angle 120^{0}.

#### Answer:

Area of a sector of a circle = $\frac{\theta}{360}\times {\mathrm{\pi r}}^{2}$

$=\frac{120}{360}\times \pi {\left(21\right)}^{2}\phantom{\rule{0ex}{0ex}}=462{\mathrm{cm}}^{2}$

#### Page No 13.75:

#### Question 16:

What is the area of a square inscribed in a circle of diameter *p *cm ?

#### Answer:

The diameter of the circle = Diagonal of the square = *p*

We know that diagonal of a square = $\sqrt{2}a$

$\Rightarrow \sqrt{2}a=p\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{p}{\sqrt{2}}$

Thus, the area of the square = ${a}^{2}={\left(\frac{p}{\sqrt{2}}\right)}^{2}=\frac{{p}^{2}}{2}{\mathrm{cm}}^{2}$

#### Page No 13.75:

#### Question 17:

Is it true to say that area of segment of a circle is less than the area of its corresponding sector ? Why ?

#### Answer:

A circle has two segments, a major segment and a minor segment.

The area of the minor segment is less than the area of the corresponding sector.

But same is not true for the major segment.

#### Page No 13.75:

#### Question 18:

If the numerical value of the area of a circle is equal to the numerical value of its circumference , find its radius.

#### Answer:

Given that area of the circle = circumference

${\mathrm{\pi r}}^{2}=2\mathrm{\pi r}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{\pi r}}^{2}-2\mathrm{\pi r}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\pi r}\left(\mathrm{r}-2\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{r}=0\mathrm{or}\mathrm{r}=2\phantom{\rule{0ex}{0ex}}$

Since radius cannot be equal to 0 so, *r* = 2

#### Page No 13.75:

#### Question 19:

How many revolutions a circular wheel of radius *r *metres makes in covering a distance of *s* metres?

#### Answer:

In one revolution, the wheel covers a distance of $2\mathrm{\pi r}$

So, to cover a distance of s meters, the wheel has to make $\frac{s}{2\mathrm{\pi r}}$ revolutions.

#### Page No 13.75:

#### Question 20:

Find the ratio of the area of the circle circumscribing a square to the area of the circle inscribed in the square .

#### Answer:

Let the side of the square inscribed in a square be *a* units.

Diameter of the circle outside the square = Diagonal of the square = $\sqrt{2}a$

Radius = $\frac{\sqrt{2}a}{2}=\frac{a}{\sqrt{2}}$

So, the area of the circle circumscribing the square = $\mathrm{\pi}{\left(\frac{\mathrm{a}}{\sqrt{2}}\right)}^{2}$ .....(i)

Now, the radius of the circle inscribed in a square = $\frac{a}{2}$

Hence, area of the circle inscribed in a square = $\mathrm{\pi}{\left(\frac{\mathrm{a}}{2}\right)}^{2}$ .....(ii)

From (i) and (ii)

$\frac{\mathrm{Area}\mathrm{of}\mathrm{circle}\mathrm{circumscribing}\mathrm{a}\mathrm{square}}{\mathrm{Area}\mathrm{of}\mathrm{circle}\mathrm{inscribed}\mathrm{in}\mathrm{a}\mathrm{square}}=\frac{\mathrm{\pi}{\left({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}\right)}^{2}}{\mathrm{\pi}{\left({\displaystyle \frac{\mathrm{a}}{2}}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{4}}}\phantom{\rule{0ex}{0ex}}=\frac{2}{1}$

Hence, the required ratio is 2 : 1.

#### Page No 13.75:

#### Question 21:

A piece of wire 22 cm long is bent into the form of an arc of a circle subtending angle of 60° at its centre. Find the radius of the circle $\left[\mathrm{Use}\frac{\mathrm{\pi}}{27}\right]$

#### Answer:

Length of the wire = 22 cm

Angle subtended at the centre = $\theta $ = 60º

Length of the arc = $\frac{\theta}{360}\times 2\mathrm{\pi}r$

$\Rightarrow \frac{60}{360}\times 2\times \frac{22}{7}\times r=22\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{22}{7}r=22\times 3\phantom{\rule{0ex}{0ex}}\Rightarrow r=21\mathrm{cm}$

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