# RS Aggarwal (2015) Solutions for Class 10 Math Chapter 21 - Summative Assessment Ii

RS Aggarwal (2015) Solutions for Class 10 Math Chapter 21 Summative Assessment Ii are provided here with simple step-by-step explanations. These solutions for Summative Assessment Ii are extremely popular among class 10 students for Math Summative Assessment Ii Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal (2015) Book of class 10 Math Chapter 21 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal (2015) Solutions. All RS Aggarwal (2015) Solutions for class 10 Math are prepared by experts and are 100% accurate.

#### Page No 803:

#### Question 1:

If the sum of the roots of the equation $3{x}^{2}-(3k-2)x-(k-6)=0$ is equal to the product of its roots, then *k* = ?

(a) 1

(b) −1

(c) 0

(d) 2

#### Answer:

(d) 2

Let the roots of the equation be $\alpha \mathrm{and}\beta $.

Then,

$\alpha +\beta =\frac{-\left[-\left(3k-2\right)\right]}{3}=\frac{3k-2}{3}$ and $\alpha \beta =\frac{-\left(k-6\right)}{3}=\frac{-k+6}{3}$

Therefore,

$\frac{3k-2}{3}=\frac{-k+6}{3}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow 3k-2=-k+6$

$\Rightarrow 4k=8\Rightarrow k=2$

#### Page No 803:

#### Question 2:

The number of all-digit numbers divisible by 6 is

(a) 12

(b) 15

(c) 16

(d) 18

#### Answer:

(b) 15

All 2 − digit numbers divisible by 6, are

12, 18, 24, 30........96

This is an *AP*, in which *a* = 12, *d* = (18 − 12) = 6 and *l *= 96

Let the number of these terms be *n*.

Then, *T _{n}* = 96

⇒

*a*+ (

*n*− 1)

*d*= 96

⇒ 12 + (

*n*− 1) × 6 = 96

⇒ (

*n*− 1) × 6 = 84

⇒ (

*n*− 1) = 14 ⇒

*n*= 15

#### Page No 803:

#### Question 3:

A fair die is thrown once. The probability of getting a composite number is

(a) $\frac{1}{3}$

(b) $\frac{1}{6}$

(c) $\frac{2}{3}$

(d) 0

#### Answer:

(a) $\frac{1}{3}$

In a single throw of a die, the possible outcomes are

1 , 2, 3, 4, 5 and 6

The total number of possible outcomes = 6

Let *E* be the event of getting a composite number.

Then, the favourable outcomes are 4 and 6.

Number of favourable outcomes = 2

∴ P(getting a composite number) = *P*(*E*) = $\frac{2}{6}=\frac{1}{3}$

#### Page No 803:

#### Question 4:

Which of the following statements is true?

(a) The tangents drawn at the end points of a chord of a circle are parallel.

(b) From a point *P* on the exterior of a circle, only two secants can be drawn through *P* to the circle.

(c) From a point *P* in the plane of a circle, two tangents can be drawn to the circle.

(d) The perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.

#### Answer:

(d) The perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.

Let O be the centre of the given circle.

A tangent PR has been drawn touching the circle at point P.

Draw QP ⊥ RP at point P, such that point Q lies on the circle. $\angle $OPR = 90° (Since radius ⊥ tangent)

Also, $\angle $QPR = 90° (Given)

∴ $\angle $OPR = $\angle $QPR

Now, the above case is possible only when the centre O lies on the line QP.

Hence, perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.

#### Page No 804:

#### Question 5:

In the given figure, *PA* and *PB* are tangents to a circle, such that *PA* = 8 cm and ∠*APB* = 60*°. *The length of the chord *AB* is

Figure

(a) 8 cm

(b) 10 cm

(c) 12 cm

(d) 6 cm

#### Answer:

(a) 8 cm

Since the lengths of tangents drawn from a point to a circle are equal, *PA* = *PB*.

Therefore, $\angle $PAB = $\angle $PBA = *x*° (say)

Then, *x*° + *x*° + 60° = 180°

⇒2*x*° = 120°

⇒ *x*° = 60°

∴ Each angle of ΔPAB is 60°; therefore, it is an equilateral triangle.

∴ *AB* =* PA* = *PB* = 8 cm

#### Page No 804:

#### Question 6:

The angle of depression of an object from a 60-m-high tower is 30*°.* The distance of the object from the tower is

(a) $20\sqrt{3}$ m

(b) $60\sqrt{3}$ m

(c) $40\sqrt{3}$ m

(d) 120 m

#### Answer:

(b) $60\sqrt{3}$ m

Let the height of the tower be AB = 60 m.

Let the angle of depression of the point C be 30°.

$\angle ACB=\angle PAC=30\xb0\phantom{\rule{0ex}{0ex}}$

∴ In the right-angled ΔABC,

$\mathrm{tan}30\xb0=\frac{AB}{BC}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{60}{BC}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow BC=60\sqrt{3}\mathrm{m}$

Hence, the distance of the object from the tower is $60\sqrt{3}$ m.

#### Page No 804:

#### Question 7:

In what ratio does the point *P*(2, −5) divide the line segment joining *A*(−3, 5) and *B*(4, −9)?

(a) 3 : 2

(b) 2 : 1

(c) 5 : 2

(d) 5 : 3

#### Answer:

(c) 5 : 2

Let the required ratio be *k* : 1.

Then, by section formula, the co-ordinates of *P* are

$P\left(\frac{4k-3}{k+1},\frac{-9k+5}{k+1}\right)$

∴ $\frac{4k-3}{k+1}=2\mathrm{and}\frac{-9\mathrm{k}+5}{\mathrm{k}+1}=-5$ [ Since *P*(2, -5) is given]

⇒ 4*k* - 3 = 2*k* + 2 and -9*k* + 5 = - 5*k* - 5

⇒ 2*k* = 5 and 4*k* = 10

∴ $k=\frac{5}{2}$ in each case.

So, the required ratio is $\frac{5}{2}:1$, which is 5 : 2.

#### Page No 804:

#### Question 8:

Three solid spheres of radii 6 cm, 8 cm and 10 cm are melted to form a sphere. The radius of the sphere so formed is

(a) 24 cm

(b) 16 cm

(c) 14 cm

(d) 12 cm

#### Answer:

(d) 12 cm

Radius (*r*_{1}) of 1^{st} sphere = 6 cm

Radius (*r*_{2}) of 2^{nd} sphere = 8 cm

Radius (*r*_{3}) of 3^{rd} sphere = 10 cm

Let the radius of the resulting sphere be *r*.

We know that the object formed by recasting these spheres will be same in volume as the sum of the volumes of these spheres. Volume of 3 spheres = Volume of resulting sphere

$\frac{4}{3}\mathrm{\pi}\left[{{r}_{1}}^{3}+{{r}_{2}}^{3}+{{r}_{3}}^{3}\right]=\frac{4}{3}{\mathrm{\pi r}}^{3}$

$\Rightarrow \frac{4}{3}\mathrm{\pi}\left[{6}^{3}+{8}^{3}+{10}^{3}\right]=\frac{4}{3}{\mathrm{\pi r}}^{3}\phantom{\rule{0ex}{0ex}}$

⇒ *r*^{3} = (216 + 512 + 1000) = 1728

⇒ *r* = 12 cm

Hence, the radius of the sphere so formed is 12 cm.

#### Page No 804:

#### Question 9:

Find the value of *p* for which the quadratic equation ${x}^{2}-2px+1=0$ has no real roots.

#### Answer:

Given equation: x^{2} − 2px + 1 = 0

Here, a = 1, *b* = − 2*p* , *c* = 1

For no real roots, *b*^{2} − 4*ac* < 0

*b*^{2} − 4*ac* = (−2*p*)^{2} − (4 ×1 × 1)

⇒ 4*p*^{2} − 4 < 0

⇒ 4(*p*^{2} − 1) < 0

⇒ (*p*^{2} − 1) < 0 [ Since 4 < 0, which is not possible]

⇒ −1 < *p* < 1

#### Page No 804:

#### Question 10:

Find the 10th term from the end of the AP 4, 9, 14, ..., 254.

*Or,* which term of the AP 24, 21, 18, 15, ... is the first negative term?

#### Answer:

Here, *a* = 4, *d* = (9 - 4) = 5, *l* = 254 and *n* = 10

Now, *n*^{th} term from the end = {*l* - (*n* - 1)*d*}

∴ 10^{th} term from the end = {254 - (10 - 1) × 5}

= {254 - (9 × 5)} = (254 - 45) = 209

Hence, the 10^{th} term from the end is 209.

OR

Here, *a* = 24, *d* = (21 - 24) = - 3

Let the *n*^{th} term of the given AP be the first negative term.

Then, *T*_{n} < 0 ⇒ {*a*+ (*n* - 1)*d*} < 0

⇒ { 24 + (*n* - 1) × ( - 3) < 0

⇒ (27 - 3*n*) < 0

⇒ 27 < 3*n*

⇒ 3*n* > 27

⇒ *n* > 9

∴ *n* = 10

Hence, the 10^{th} term is the first negative term of the given AP.

#### Page No 804:

#### Question 11:

A circle is touching the side BC of a ∆*ABC* at *P* and is touching *AB* and *AC* when produced to *Q* and *R*, respectively.

Prove that $AQ=\frac{1}{2}$ (perimeter of ∆*ABC*).

Figure

#### Answer:

We know that the lengths of tangents drawn from an external point to a circle are equal.

∴

AQ = AR ..........(i) [tangent from A]

BP = BQ ...........(ii) [tangent from B]

CP = CR ...........(iii) [tangent from C]

Perimeter of ΔABC

= AB + BC + AC

= AB + BP + CP + AC

= AB + BQ + CR + AC [using (ii) and (iii)]

= AQ + AR

= 2AQ [ using (i)]

∴ $\mathrm{AQ}=\frac{1}{2}\left(\mathrm{Perimeter}\mathrm{of}\u25b3\mathrm{ABC}\right)$

#### Page No 804:

#### Question 12:

Two vertices of ∆*ABC* are *A*(6, 4) and *B*(−2, 2) and its centroid is *G*(3, 4). Find the coordinates of its third vertex *C*.

#### Answer:

Two vertices of ΔABC are *A*(6, 4) and *B*(− 2, 2). Let the third vertex be *C*(*a*, *b*).

Then, the coordinates of its centroid are

$G\left(\frac{6-2+a}{3},\frac{4+2+b}{3}\right)\phantom{\rule{0ex}{0ex}}$

$\Rightarrow G\left(\frac{4+a}{3},\frac{6+b}{3}\right)$

But it is given that the centroid is G(3,4).

∴ $\frac{4+a}{3}=3$ and $\frac{6+b}{3}=4$

⇒ 4 + *a* = 9 and 6 + *b *= 12

⇒ *a* = 5 and *b* = 6

Hence, the third vertex of ΔABC is C(5, 6).

#### Page No 805:

#### Question 13:

A box contains 150 oranges. If one orange is taken out from the box at random and the probability of its being rotten is 0.06, then find the number of good oranges in the box.

#### Answer:

Total number of oranges = 150

Probability of it being good =1 - 0.06 = 0.94

Let the number of good oranges be *x.*

*P(*getting good oranges) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{good}\mathrm{oranges}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{oranges}}$

$0.94=\frac{x}{150}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow x=0.94\times 150=141$

Hence, the required number of good oranges in the box is 141.

#### Page No 805:

#### Question 14:

A toy is in the form of a cone mounted on a hemisphere of common base radius 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy.

#### Answer:

Given that the diameter of the common base of the cone and hemisphere is 7 cm.

∴ Radius of the cone (*r* ) = 7 cm

The total height of the toy is 31 cm.

∴

The height of the cone (*h*) = 31 - 7 = 24 cm

Surface area of the hemisphere = $2{\mathrm{\pi r}}^{2}$

= $\left(2\times \frac{22}{7}\times 7\times 7\right)$ = 308 cm^{2}

Slant height of the cone (*l*) = $\sqrt{{r}^{2}+{h}^{2}}=\sqrt{{\left(7\right)}^{2}+{\left(24\right)}^{2}}=\sqrt{625}$ = 25 cm

∴ Curved surface area of the cone = $\mathrm{\pi rl}=\left(\frac{22}{7}\times 7\times 25\right)$ = 550 cm^{2}

Hence,

Total surface area of the toy = Surface area of the hemisphere + Curved surface area of the cone

= (308 + 550) cm^{2}= 858 cm^{2}

#### Page No 805:

#### Question 15:

Solve: ${a}^{2}{b}^{2}{x}^{2}-(4{b}^{4}-3{a}^{4})x-12{a}^{2}{b}^{2}=0.$

#### Answer:

The given equation is:

*a*^{2}*b*^{2}*x*^{2} − (4*b*^{4} − 3*a*^{4})*x* − 12*a*^{2}*b*^{2} = 0

Comparing it with *Ax*^{2} + *Bx* + *C* = 0, we get:

*A* = *a*^{2}*b*^{2} , *B* = − (4*b*^{4} − 3*a*^{4}) and *C* = − 12*a*^{2}*b*^{2}

∴ *D* = (*B*^{2}^{ }− 4*AC*)

= (3*a*^{4} − 4*b*^{4}^{ })^{2} − 4(*a*^{2}*b*^{2})( − 12*a*^{2}*b*^{2} )

= 9*a*^{8} + 16*b*^{8} − 24*a*^{4}*b*^{4} + 48*a*^{4}*b*^{4}^{ }

= (3*a*^{4})^{2} + (4*b*^{4})^{2} + 2 × 3*a*^{4} × 4*b*^{4}

= (3*a*^{4}^{ }+ 4*b*^{4})^{2} ≥ 0

Hence, the given equation has real roots. These are

$\alpha =\frac{-B+\sqrt{D}}{2A}=\frac{\left(4{b}^{4}-3{a}^{4}\right)+\left(3{a}^{4}+4{b}^{4}\right)}{2{a}^{2}{b}^{2}}=\frac{8{b}^{4}}{2{a}^{2}{b}^{2}}=\frac{4{b}^{2}}{{a}^{2}}\phantom{\rule{0ex}{0ex}}$

$\beta =\frac{-B-\sqrt{D}}{2A}=\frac{\left(4{b}^{4}-3{a}^{4}\right)-\left(3{a}^{4}+4{b}^{4}\right)}{2{a}^{2}{b}^{2}}=\frac{-6{a}^{4}}{2{a}^{2}{b}^{2}}=\frac{-3{a}^{2}}{{b}^{2}}$

Hence, $\frac{4{b}^{2}}{{a}^{2}}\mathrm{and}\frac{-3{\mathrm{a}}^{2}}{{\mathrm{b}}^{2}}$ are the roots of the given equation.

#### Page No 805:

#### Question 16:

If the 8th term of an AP is 31 and its 15th term is 16 more than the 11 the term, find AP.

*Or*, find the sum of all two-digit odd positive numbers.

#### Answer:

Here,

*a*_{8} = 31 and *a*_{15} = 16 + *a*_{11}

⇒ 31 = *a *+ 7*d* .............(i)

and *a* + 14*d* = 16 + *a* + 10*d* .............(ii)

By solving equation (ii), we get:

4*d* = 16 ⇒ *d* = 4

On substituting *d* = 4 in (i), we get:

31 = *a* + 7 × 4 ⇒ *a* = 3

Since *a* = 3 and *d* = 4, required AP = 3, 7 , 11, 15 .....

**OR**

All two - digit odd positive numbers are 11, 13, 15, 17 ......99.

This is an AP in which *a* = 11, *d* = (13 - 11) = 2 and *l *= 99.

Let the number of terms be *n*. Then,

T_{n} = 99

⇒ *a* + (*n* - 1) × *d *= 99

⇒ 99 = 11 + (*n* - 1) × 2

⇒ 99 = 11 + 2*n* - 2

⇒ 90 = 2*n*

⇒ *n* = 45

Required sum = $\frac{n}{2}\left(a+l\right)$

= $\frac{45}{2}\left(11+99\right)=\left(\frac{45}{2}\times 110\right)$

= (45 × 55) = 2475

#### Page No 805:

#### Question 17:

In the adjoining figure, *PA* and *PB* are tangents drawn from an external point *P* to a circle with centre *O*.

Prove that ∠*APB* = 2∠*OAB*

Figure

*Or,* in the adjoining figure, quadrilateral *ABCD* is circumscribed. If the radius of the incircle with centre *O* is 10 cm and *AD* ⊥ *DC*, find the value of *x*.

Figure

#### Answer:

**Given:** PA and PB are the tangents to a circle with centre O from a point P outside it.

**To prove:** $\angle $APB = 2$\angle $OAB

**Proof : **Let $\angle $APB = *x*°

We know that the tangents to a circle from an external point are equal. So, PA = PB

Since, the angles opposite to the equal sides of a triangle are equal,

PA = PB ⇒ $\angle $PBA = $\angle $PAB

Also, the sum of the angles of a triangle is 180°.

∴ $\angle $APB + $\angle $PAB +$\angle $PBA = 180°

⇒ *x*° + 2$\angle $PAB = 180° [Since ∠PBA = ∠PAB]

$\Rightarrow \angle \mathrm{PAB}=\frac{1}{2}\left(180\xb0-\mathrm{x}\xb0\right)=\left(90\xb0-\frac{1}{2}\mathrm{x}\xb0\right)$

But PA is a tangent and OA is the radius of the given circle.

∴ $\angle $OAB + $\angle $PAB = 90°

$\angle \mathrm{OAB}+\left(90\xb0-\frac{1}{2}\mathrm{x}\xb0\right)=90\xb0\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \angle \mathrm{OAB}=\frac{1}{2}\mathrm{x}\xb0=\frac{1}{2}\angle \mathrm{APB}$

Hence, $\angle $APB = 2$\angle $OAB

**OR **

**Disclaimer : **In the figure side AB is given 27 cm, instead of AB , BP should be 27 cm.

Since the lengths of tangents from an external point to a circle are equal, we have:

BP = BQ = 27 cm

∴ QC = (BC - BQ) = (38 - 27) cm = 11 cm

⇒ CQ = CR = 11 cm

Join OR. Then, SORD is a rectangle.

∴ OS = DR = 10 cm

Hence, DC = (DR + RC) = (10 + 11) cm = 21 cm

#### Page No 805:

#### Question 18:

Draw a circle of radius 6 cm. From a point 10 cm away from its centre construct a pair of tangents to the circle. Measure the length of each of the tangent segments.

#### Answer:

Given that tangents PQ and PR are drawn from a point P to a circle with centre O.

Here, OQ = OR = 6 cm and OP = 10 cm.

Now,

PQ is tangent at Q and OQ is the radius through Q.

∴ OQ ⊥ QP

Similarly,

OR ⊥ PR

In the right-angled ΔOQP and ΔORP, we get:

OQ = OR [ radii of the same circle]

OP = OP [ common side]

∴ Δ OQP ≅ Δ ORP [ By RHS congruence]

Hence QP = PR

Again,

By Pythagoras' theorem, we get:

⇒ OP^{2} = OQ^{2} + QP^{2}

⇒ (10)^{2} = (6)^{2} + QP^{2}

⇒ QP^{2} = [100 - 36] = 64 ⇒ QP = 8 cm

Thus, QP = RP = 8 cm

#### Page No 805:

#### Question 19:

The three vertices of a parallelogram *ABCD,* taken in order, are *A*(1, −2), *B*(3, 6) and *C*(5, 10). Find the coordinates of the fourth vertex *D*.

#### Answer:

Let *A*(1, −2), *B*(3, 6) and *C*(5, 10) be the three vertices of a parallelogram *ABCD* and let its fourth vertex be D(*a*, *b*).

Join AC and BD.

Let AC and BD intersect at point O.

We know that the diagonals of a parallelogram bisect each other.

So, O is the mid-point of AC as well as of BD.

Mid-point of AC = $\left(\frac{1+5}{2},\frac{-2+10}{2}\right)$ ⇒ (3, 4)

Mid-point of BD = $\left(\frac{3+a}{2},\frac{6+b}{2}\right)$

∴ $\frac{3+a}{2}=3$ and $\frac{6+b}{2}=4$

⇒ 3 + *a* = 6 and 6 + *b* = 8

⇒ *a* = 3 and *b* = 2

Hence, the fourth vertex is D(3, 2).

#### Page No 805:

#### Question 20:

Find the third vertex a ∆*ABC* if two of its vertices are *B*(−3, 1) and *C*(0, −2) and its centroid is at the origin.

#### Answer:

Two vertices of ABC are B(- 3, 1) and C(0, -2). Let the third vertex be A(*a*, *b*).

Then, the coordinates of its centroid are

$G\left(\frac{-3+0+a}{3},\frac{1-2+b}{3}\right)\phantom{\rule{0ex}{0ex}}$

$=G\left(\frac{-3+a}{3},\frac{-1+b}{3}\right)$

But is is given that the centroid is G(0,0).

Therefore, $\frac{-3+a}{3}=0$ and $\frac{-1+b}{3}=0$

⇒ - 3+ *a* = 0 and - 1+* b* = 0

⇒ *a* = 3 and *b* = 1

Hence, the third vertex of ∆*ABC* is *A*(3 ,1).

#### Page No 805:

#### Question 21:

Cards marked with all 2-digit numbers are placed in a box and are mixed thoroughly. One card is drawn at random. Find the probability that the number on the card is

(a) divisible by 10

(b) a perfect square

(c) a prime number less than 25

#### Answer:

Cards are marked with numbers 10, 11, 12, …, 99.

Total number of cards = 90

**(a)** The number of cards divisible by 10 are 10, 20, 30, 40, 50, 60, 70, 80, 90.

The number of favourable cases = 9

Probability of getting a card with a number divisible by 10 = $\frac{9}{90}=\frac{1}{10}$

**(b) **The numbers of cards which are perfect squares are 16, 25, 36, 49, 64, 81

The number of favourable cases = 6

Probability of getting a card that is a perfect square = $\frac{6}{90}=\frac{1}{15}$

**(c)** Prime numbers less than 25 are 2, 3, 5, 7, 11 , 13, 17, 19, 23.

The number of favourable cases = 9

Probability of getting a card with a prime number less than 25 = $\frac{9}{90}=\frac{1}{10}$

#### Page No 806:

#### Question 22:

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

#### Answer:

Let the radius of the park be *r *metres.

Then, its circumference = $2\mathrm{\pi r}$

Therefore,

$2\mathrm{\pi r}=352\Rightarrow 2\times \frac{22}{7}\times \mathrm{r}=352\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \mathrm{r}=\left(352\times \frac{7}{44}\right)=56$

Thus, the inner radius = 56 m

Outer radius = (56 + 7) m = 63 m

Area of the road = $\mathrm{\pi}\left[{\left(63\right)}^{2}-{\left(56\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}$ m^{2}

$=\frac{22}{7}\times \left(63+56\right)\left(63-56\right)\phantom{\rule{0ex}{0ex}}$ m^{2}

$=\left(\frac{22}{7}\times 119\times 7\right)$ m^{2} = 2618 m^{2}

#### Page No 806:

#### Question 23:

A round table cover shown in the adjoining figure has six equal designs. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs 0.50 per cm^{2}.

Figure

*Or*, in an equilateral triangle of side 12 cm, a circle is inscribed touching its sides. Find the area of the portion of the triangle not included in the circle.

#### Answer:

Given: radius = OA = OB = 28 cm

Area of the shaded region = (area of the circle - area of the hexagon)

= ${\mathrm{\pi r}}^{2}-6\times \mathrm{ar}\left(\u25b3\mathrm{OAB}\mathrm{with}\mathrm{each}\mathrm{side}r\mathrm{cm}\right)$

$\Rightarrow \left[\left(\frac{22}{7}\times 28\times 28\right)-\left(6\times \frac{\sqrt{3}}{4}\times 28\times 28\right)\right]\phantom{\rule{0ex}{0ex}}$cm^{2}

$\Rightarrow \left(28\times 28\right)\left[\frac{22}{7}-\frac{3}{2}\times 1.73\right]\phantom{\rule{0ex}{0ex}}$cm^{2}

$\Rightarrow 784\times \left(3.142-2.595\right)$ cm^{2} = (784 × 0.547) cm^{2} = 428.84 cm^{2}

Rate of design = Rs 0.50 per cm^{2}

Hence, total cost of designing = Rs (429.84 × 0.50) = Rs 214.42

**OR**

Let ΔABC be an equilateral triangle of side 12 cm.

Let AD ⊥ BC. Then D is the mid-point of BC.

∴ BD = DC = 6 cm and AB = 12 cm

∴ AD = $\sqrt{A{B}^{2}-B{D}^{2}}\phantom{\rule{0ex}{0ex}}$ cm

$=\sqrt{{\left(12\right)}^{2}-{\left(6\right)}^{2}}\phantom{\rule{0ex}{0ex}}$ cm

$=\sqrt{144-36}$ cm = $\sqrt{108}$ cm = $6\sqrt{3}$cm

Let O be the centre of the inscribed circle.

Then, O is the centroid of ΔABC.

∴ AO : OD = 2 : 1 and OD = $\frac{1}{3}AD=\frac{1}{3}\times 6\sqrt{3}=2\sqrt{3}$

∴ *r* = OD = $2\sqrt{3}$cm

Required area = (Area of ΔABC) - (Area of the incircle)

= $\left[\frac{\sqrt{3}}{4}\times {\left(12\right)}^{2}-3.14\times {\left(2\sqrt{3}\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}$ cm^{2}

$=\left(36\sqrt{3}-3.14\times 12\right)\phantom{\rule{0ex}{0ex}}$ cm^{2}

$=\left(62.28-37.68\right)$ cm^{2} = 24.6 cm^{2}

#### Page No 806:

#### Question 24:

If a sphere has the same surface area as the total surface area of a circular cone of height 40 cm and radius 30 cm, find the radius of the sphere.

#### Answer:

Given: radius of the cone (*r*) = 30 cm and height of the cone (*h*) = 40 cm

Slant height of the cone (*l*) = $\sqrt{{r}^{2}+{h}^{2}}=\sqrt{{\left(30\right)}^{2}+{\left(40\right)}^{2}}$

$\Rightarrow \sqrt{900+1600}\Rightarrow \sqrt{2500}=50$ cm

Total surface area of the circular cone = $\mathrm{\pi r}\left(\mathrm{l}+\mathrm{r}\right)$ sq. units

$=\mathrm{\pi}\times 30\left(50+30\right)$ cm^{2}

TSA of a sphere is same as TSA of a circular cone.

Therefore,

Total surface area of a sphere = Total surface area of circular cone

$4{\mathrm{\pi r}}^{2}=\mathrm{\pi}\times 30\times 80\phantom{\rule{0ex}{0ex}}$

$\Rightarrow {\mathrm{r}}^{2}=\frac{\left(30\times 80\right)}{4}=600\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \mathrm{r}=10\sqrt{6}$ cm

Hence, radius of the sphere will be $10\sqrt{6}$cm.

#### Page No 806:

#### Question 25:

A two-digit number is such that product of its digits is 35. If 18 is added to the number, the digits interchange their places. Find the number.

*Or*, two water taps together can fill a tank in $9\frac{3}{8}$ hours. The larger tap takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

#### Answer:

Let the tens digit be *x* and the units digit be *y*.

∴ Product of the digits = *xy *= 35............(i)

Original number is (10*x* + y).

If the digits interchange their position, the number becomes (10*y* + *x*).

Thus,

(10*x* + y) + 18 = (10*y* + *x*)

9*x* − 9*y* = − 18

*x* − *y* = − 2 .............(ii)

From (i), we get: $y=\frac{35}{x}$

On substituting the value of *y* in (ii), we get:

$x-\frac{35}{x}=-2\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow {x}^{2}-35=-2x\Rightarrow {x}^{2}+2x-35=0\phantom{\rule{0ex}{0ex}}$

$\Rightarrow {x}^{2}+7x-5x-35=0\phantom{\rule{0ex}{0ex}}$

$\Rightarrow x\left(x+7\right)-5\left(x+7\right)=0\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \left(x-5\right)\left(x+7\right)=0$

⇒ *x* = 5 or *x* = − 7

Since the digits cannot be negative, *x* = 5 and $y=\frac{35}{5}=7$

Hence, the required number = 10(5) + 7 = 57

**OR**

Let the smaller tap fill the tank in *x* hours.

Then, the larger tap will fill it in (*x* − 10) hours.

Time taken by both together to fill the tank = $9\frac{3}{8}=\frac{75}{8}$ hours

Part filled by smaller tap in 1 hour = $\frac{1}{x}$

Part filled by larger tap in 1 hour = $\frac{1}{\left(x-10\right)}$

Part filled by both the taps in 1 hour = $\frac{8}{75}$

∴ $\frac{1}{x}+\frac{1}{\left(x-10\right)}=\frac{8}{75}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \frac{\left(x-10\right)+x}{x\left(x-10\right)}=\frac{8}{75}\Rightarrow \frac{\left(2x-10\right)}{x\left(x-10\right)}=\frac{8}{75}$

75(2*x* −10) = 8*x*(*x* −10) [By cross−multiplication]

⇒ 150*x* − 750 = 8*x*^{2} − 80*x*

⇒ 8*x*^{2} − 230*x* + 750 = 0

⇒ 4*x*^{2} − 115*x* + 375 = 0

⇒ 4*x*^{2} − 100*x* − 15*x* + 375 = 0 ⇒ 4*x*(*x* − 25) − 15(*x* − 25) = 0

⇒ (4*x* − 15)(*x* − 25) = 0 ⇒ (4*x* − 15) = 0 or (*x* − 25) = 0

⇒ $x=25orx=\frac{15}{4}$

⇒* x* = 25 $\left[\mathrm{Since}\mathrm{x}=\frac{15}{4}\Rightarrow \left(\mathrm{x}-10\right)0\right]$

Hence, the time taken by the smaller tap to fill the tank = 25 hours.

And the time taken by the larger tap to fill the tank = (25 − 10) hours = 15 hours.

#### Page No 806:

#### Question 26:

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact to the centre.

#### Answer:

**Given: **

PA and PB are the tangents drawn from a point P to a circle with centre O. Also, the line segments OA and OB are drawn.

**To prove:**$\angle $APB + $\angle $AOB = 180°

**Proof:**

We know that the tangent to a circle is perpendicular to the radius through the point of contact.

∴ PA ⊥ OA ⇒ $\angle $OAP = 90°

PB ⊥ OB ⇒ $\angle $OBP = 90°

∴ $\angle $OAP + $\angle $OBP = (90° + 90°)= 180° ......................(i)

But we know that the sum of all the angles of a quadrilateral is 360°.

∴ $\angle $OAP + $\angle $OBP + $\angle $APB + $\angle $AOB = 360° .................(ii)

From (i) and (ii), we get:

$\angle $APB + $\angle $AOB = 180°

#### Page No 806:

#### Question 27:

From the top of a 7-m-high building, the angle of elevation of the top of a cable tower is 60*° *and the angle of depression of its foot is 45*°. *Find the height of the tower.

#### Answer:

Let AB be the building that is 7 metres high. AE ⊥ CD, where CD is the cable tower.

Let *AE = BD = x *m

In ΔAED,

$\angle $EAD = 45° = angle of depression

∴ $\frac{\mathrm{AE}}{\mathrm{ED}}=\mathrm{cot}45\xb0\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \frac{\mathrm{x}}{7}=1\Rightarrow \mathrm{x}=7$

In ΔACE,

$\angle $CAE = 60° = angle of elevation

$\angle $AEC = 90°

∴ $\frac{\mathrm{CE}}{\mathrm{AE}}=\mathrm{tan}60\xb0\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \frac{\mathrm{h}}{x}=\sqrt{3}$

$\Rightarrow h=x\sqrt{3}=\left(7\times 1.73\right)=12.11$

Height of the tower = CD = CE + ED = (12.11+ 7) m = 19.11 m

#### Page No 806:

#### Question 28:

Puja works in a bank and she gets a monthly salary of Rs 35000 with annual increment of Rs 1500. What would be her monthly salary in the 10th year of her employment?

#### Answer:

Here *a* = 35000 , *d* = 1500 and T_{10} = ?

We know that

T* _{n}* =

*a*+ (

*n*- 1)

*d*

T

_{10}= 35000 + (10 - 1) × 1500

= 35000 + (9 × 1500) = (35000 + 13500) = 48,500

Hence, Puja's monthly salary in the 10

^{th}year will be Rs. 48,500.

#### Page No 806:

#### Question 29:

In the given figure, *ABDCA* represents the quadrant of a circle of radius 7 cm with centre *A*. Calculate the area of the shaded region.

#### Answer:

We have:

Shaded area = Area of quadrant - Area of ΔABE

= $\left[\frac{1}{4}{\mathrm{\pi r}}^{2}-\frac{1}{2}\times \mathrm{h}\times \mathrm{b}\right]\phantom{\rule{0ex}{0ex}}$sq. units

$\Rightarrow \left[\left\{\frac{1}{4}\times \frac{22}{7}\times 7\times 7\right\}-\left\{\frac{1}{2}\times 7\times 2\right\}\right]\phantom{\rule{0ex}{0ex}}$ cm^{2}

$\Rightarrow \left(\frac{77}{2}-7\right)=\frac{77-14}{2}=\frac{63}{2}=31.5$ cm^{2}

#### Page No 807:

#### Question 30:

The radii of the circular ends of a solid frustum of a cone are 33cm and 27 cm and its slant height is 10 cm. Find its capacity and total surface area.

#### Answer:

*R* = 33 cm, *r* = 27 cm and *l *= 10 cm

*l*^{2} = *h*^{2} + (*R* - *r*)^{2}

*h*^{2} = *l*^{2} - (*R* - *r*)^{2}

= {(10)^{2} - (33 - 27)^{2}} = (100 - 36) = 64

⇒ *h *= 8 cm

$\mathrm{Capacity}=\frac{1}{3}\mathrm{\pi h}\left({\mathrm{R}}^{2}+{\mathrm{r}}^{2}+\mathrm{Rr}\right)\phantom{\rule{0ex}{0ex}}$ cm^{3}

$\left\{\frac{1}{3}\times \frac{22}{7}\times 8\left({33}^{2}+{27}^{2}+33\times 27\right)\right\}\phantom{\rule{0ex}{0ex}}$ cm^{3}

$\left\{\frac{1}{3}\times \frac{22}{7}\times 8\left(1089+729+891\right)\right\}=\left\{\frac{1}{3}\times \frac{22}{7}\times 8\times 2709\right\}=22704$ cm^{3}

Again,

$\mathrm{Total}\mathrm{surface}\mathrm{area}=\mathrm{\pi}\left[{\mathrm{R}}^{2}+{\mathrm{r}}^{2}+\mathrm{l}\left(\mathrm{R}+\mathrm{r}\right)\right]\phantom{\rule{0ex}{0ex}}$ sq. cm

$=\frac{22}{7}\left[{33}^{2}+{27}^{2}+10\left(33+27\right)\right]\phantom{\rule{0ex}{0ex}}$ cm^{2}

$=\frac{22}{7}\left[1089+729+10\left(60\right)\right]\phantom{\rule{0ex}{0ex}}$ cm^{2}

$=\frac{22}{7}\left[2418\right]=7599.43$ cm^{2}

Hence, capacity of the frustum is 22704 cm^{3} and total surface area of frustum = 7599.43 cm^{2}.

#### Page No 807:

#### Question 31:

From an external point *P*, tangents *PA* and *PB* are drawn to a circle with centre *O*. If *CD* is the tangent to the circle at the point *E* and *PA* = 14 cm, find the perimeter of ∆*PCD*.

Figure

#### Answer:

**Given :** PA and PB are tangents to the circle with centre O.

CD is a tangent to the circle at E, which intersects PA and PB at C and D, respectively, and PA = 14 cm.

We know that tangents drawn from an external point to a circle are equal.

∴ PA = PB = 14 cm

CA = CE and DB = DE

Now, perimeter of ΔPCD = PC + CD + PD

= PC + (CE + ED) + PD

= PC + (CA + DB) + PD

= (PC + CA) + (DB + PD)

= PA + PB

= 14 cm + 14 cm = 28 cm

#### Page No 807:

#### Question 32:

Construct ∆*ABC* in which *BC* = 5.4 cm, *AB* = 4.5 cm and ∠*ABC* = 60*°*. Construct a triangle similar to this triangle, whose sides are $\frac{3}{4}$ the corresponding sides of ∆*ABC*.

#### Answer:

**Steps of construction: **

**Step 1: **Draw a line segment BC = 5.4 cm.

**Step 2: **Draw an angle of 60° at B, so that $\angle $XBC = 60°.

**Step 3:** With centre B and radius 4.5 cm, draw an arc intersecting XB at A.

**Step 4:** Join AC

Thus, ΔABC is the required triangle.

**Step 5:** Draw a line BY below BC.

**Step 6: **Cut off 4 equal distances from B, such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}.

**Step 7: **Join CB_{4}

**Step 8: **Draw C'B_{3} parallel to CB_{4}.

**Step 9: **Draw C'A' parallel to CA through C' intersecting BA produced to A'.

Thus, ΔA'BC' is the required similar triangle.

#### Page No 807:

#### Question 33:

A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of a red ball, find the number of blue balls in the bag.

#### Answer:

Let the number of blue balls in the bag be *x*.

Then, total number of balls = ( 5 + *x*)

Given: *P*(a blue ball) = 3 × *P*(a red ball)

$\therefore \frac{x}{\left(5+x\right)}=3\times \frac{5}{\left(5+x\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(5+x\right)=15\left(5+x\right)\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(5+x\right)-15\left(5+x\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(5+x\right)\left(x-15\right)\phantom{\rule{0ex}{0ex}}\mathrm{Either}5+x=0\mathrm{or}x-15=0\phantom{\rule{0ex}{0ex}}\mathrm{So}\mathrm{either}x=-5\mathrm{or}x=15\phantom{\rule{0ex}{0ex}}\mathrm{But}\mathrm{number}\mathrm{of}\mathrm{ball}s\mathrm{cannot}\mathrm{be}\mathrm{negative}.\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},x=15$

Hence, the number of blue balls in the bag is 15.

#### Page No 807:

#### Question 34:

In what ratio is the line segment joining the points (−2, −3) and (3, 7) divided by the *y*-axis? Also, find the coordinates of the point of division.

#### Answer:

Let the *y* - axis cut the join A(-2, -3) and B(3, 7) at the point *P* in the ratio *k* : 1.

Then, by section formula, the co-ordinates of *P* are

$\left(\frac{3k-2}{k+1},\frac{7k-3}{k+1}\right)$

But *P* lies on the *y* - axis; so, its abscissa is 0.

∴ $\frac{3k-2}{k+1}=0\Rightarrow 3k-2=0\Rightarrow k=\frac{2}{3}$

So, the required ratio is $\frac{2}{3}:1$, which is 2 : 3.

$\mathrm{On}\mathrm{s}\mathrm{ubstituting}\mathrm{k}=\frac{2}{3}\mathrm{in}\left(\frac{7\mathrm{k}-3}{\mathrm{k}+1}\right),\mathrm{we}\mathrm{get}:$

$P\left(0,\frac{7\times {\displaystyle \frac{2}{3}}-3}{{\displaystyle \frac{2}{3}}+1}\right)\Rightarrow P\left(0,1\right)$

Hence, the point of intersection of *AB* and the *y* - axis is * P*(0, 1).

#### Page No 808:

#### Question 1:

The value of *k,* for which the equation $2{x}^{2}kx+3=0$ has two real equal roots, are

(a) $\pm 2\sqrt{3}$

(b) $\pm 3\sqrt{2}$

(c) $\pm 2\sqrt{6}$

(d) $\pm \sqrt{6}$

#### Answer:

(c) $\pm 2\sqrt{6}$

The given equation is 2*x*^{2} + *kx* + 3 = 0.

This is of the form *ax*^{2} + *bx* + *c* = 0, where, *a* = 2, *b* = *k* and *c* = 3.

Therefore,

*D* = (*b*^{2} - 4*ac*) = (*k*^{2} - 4 × 2 × 3) = (*k*^{2} - 24)

For real and equal roots, we must have:

*D* = 0 ⇒ *k*^{2} - 24 = 0 ⇒*k*^{2}^{ }= 24 ⇒ $k=\pm \sqrt{24}=\pm 2\sqrt{6}$

Hence, $2\sqrt{6}$ and $-2\sqrt{6}$ are the required values of *k*.

#### Page No 808:

#### Question 2:

How many terms are there in the AP 7, 11, 15, ..., 139?

(a) 31

(b) 32

(c) 33

(d) 34

#### Answer:

(d) 34

In the given *AP*, we have:

*a* = 7 and *d* = (11 - 7) = 4

Suppose there are *n* terms in the given *AP*. Then,

*T _{n}* = 139 ⇒

*a*+ (

*n*- 1)d = 139

⇒ 7 + (

*n*- 1) × 4 = 139

⇒ 4

*n*= 136

⇒

*n*= 34

Hence, there are 34 terms in the given

*AP*.

#### Page No 808:

#### Question 3:

One card is drawn from a well-shuffled deck of 52 cards. The probability of drawing a 10 of a black suit is

(a) $\frac{1}{13}$

(b) $\frac{1}{26}$

(c) $\frac{1}{52}$

(d) $\frac{3}{52}$

#### Answer:

(b) $\frac{1}{26}$

Number of possible outcomes = 52

Number of favourable outcomes = 2

Therefore,

P(drawing a 10 of a black suit) = $\frac{2}{52}=\frac{1}{26}$

#### Page No 808:

#### Question 4:

In a circle of radius 7 cm, tangent *PT* is drawn from a point *P,* such that *PT* = 24 cm. If *O* is the centre of the circle, then *OP* = ?

Figure

(a) 30 cm

(b) 28 cm

(c) 25 cm

(d) 31 cm

#### Answer:

(c) 25 cm

*OP*^{2} = *OT*^{2} + *PT*^{2} = (7)^{2} + (24)^{2}

⇒ *OP*^{2} = ( 49 + 576) = 625 ⇒ *OP* = $\sqrt{625}=25$ cm

#### Page No 809:

#### Question 5:

The ratio in which the line segment joining the points *A*(−3, 2) and *B*(6, 1) is divided by the *y*-axis is

(a) 3 : 1

(b) 1 : 3

(c) 2 : 1

(d) 1 : 2

#### Answer:

(d) 1 : 2

Let the *y* - axis cut the line segment joining A(- 3, 2) and B(6, 1) at the point P in the ratio *k* : 1.

Then, by section formula, the co-ordinates of P are

$\left(\frac{6k-3}{k+1},\frac{k+2}{k+1}\right)$

But P lies on the *y*-axis. So, its abscissa is 0.

∴ $\frac{6k-3}{k+1}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 6k-3=0\Rightarrow 6k=3\Rightarrow k=\frac{3}{6}=\frac{1}{2}$

So, the required ratio is $\frac{1}{2}:1$, which is 1 : 2.

#### Page No 809:

#### Question 6:

The distance of the point *P*(6, −6) from the origin is

(a) 6 units

(b) $\sqrt{6}$ units

(c) $3\sqrt{2}$ units

(d) $6\sqrt{2}$ units

#### Answer:

(d) $6\sqrt{2}$ units

Let *P*(6, - 6) be the given point and *O*(0, 0) be the origin.

Then, *OP* = $\sqrt{{\left(6-0\right)}^{2}+{\left(-6-0\right)}^{2}}=\sqrt{{6}^{2}+{\left(-6\right)}^{2}}$

= $\sqrt{36+36}=\sqrt{72}=\sqrt{36\times 2}=6\sqrt{2}$ units

#### Page No 809:

#### Question 7:

A kite is flowing at a height of 75 cm from the level ground, attached to a string inclined at 60*° *to the horizontal. The length of the string, with no slack in it, is

(a) $50\sqrt{2}\mathrm{m}$

(b) $25\sqrt{3}\mathrm{m}$

(c) $50\sqrt{3}\mathrm{m}$

(d) 37.5 m

#### Answer:

(c) $50\sqrt{3}\mathrm{m}$

Let *OB* be the length of the string from the level ground and *O *be the point of the observer.

Then, *AB* = 75 m and $\angle $OAB = 90° and $\angle $AOB = 60°.

Let *OB* be *l* metres.

From the right-angled Δ*OAB*, we have:

$\frac{OB}{AB}=\mathrm{cosec}60\xb0=\frac{2}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \frac{l}{75}=\frac{2}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow l=\left(75\times \frac{2}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\right)=\left(25\times 2\times \sqrt{3}\right)=50\sqrt{3}$ m

#### Page No 809:

#### Question 8:

A solid metal cone with radius of base 12 cm and height 24 cm is melted to form solid spherical balls of diameter 6 cm each. The number of balls formed is

(a) 16

(b) 32

(c) 24

(d) 28

#### Answer:

(b) 32

Radius of the cone = 12 cm and its height = 24 cm

Volume of cone = $\frac{1}{3}\mathrm{\pi}{r}^{2}\mathrm{h}=\left(\frac{1}{3}\times \mathrm{\pi}\times 12\times 12\times 24\right){\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}$

$=\left(48\times 24\right)\mathrm{\pi}{\mathrm{cm}}^{2}$

Volume of each ball = $\frac{4}{3}{\mathrm{\pi R}}^{3}=\frac{4}{3}\mathrm{\pi}\times 3\times 3\times 3$ [ Since diameter of each ball is 6 cm]

$=\left(36\mathrm{\pi}\right){\mathrm{cm}}^{3}$

Thus,

Number of balls formed = $\frac{\mathrm{Volume}\mathrm{of}\mathrm{solid}\mathrm{cone}}{\mathrm{Volume}\mathrm{of}\mathrm{each}\mathrm{ball}}\phantom{\rule{0ex}{0ex}}$

$=\frac{\left(48\times 24\mathrm{\pi}\right)}{36\mathrm{\pi}}=32$

#### Page No 809:

#### Question 9:

If the roots the equation $(a-b){x}^{2}+(b-c)x+(c-a)=0$ are equal, prove that *b* + *c* = 2*a*.

#### Answer:

Given equation is (*a* - *b*)*x*^{2} + (*b* - *c*)*x* + (*c* - *a*) = 0

It is of the form *Ax*^{2} + B*x* + *C* = 0

We know that *D* = *B*^{2} - 4*AC*

∴ *D* = (*b* - *c*)^{2} - 4(*a* - *b*)(*c* - *a*) = 0

For real and equal roots, *D* = 0

Now, *D* = 0

⇒ (*b* - *c*)^{2} - 4(*a* - *b*)(*c* - *a*) = 0

⇒*b*^{2} - 2bc + *c*^{2}^{ }- 4(*ac* - *a*^{2} - *bc* + *ab* ) = 0

⇒*b*^{2} - 2bc + *c*^{2}^{ }- 4*ac* + 4*a*^{2} + 4*bc* - 4*ab* = 0

⇒ 4*a*^{2} + *b*^{2} + *c*^{2}^{ }- 4*ab* + 2*bc* - 4*ac* = 0

⇒ (-2*a*)^{2} + *b*^{2} + *c*^{2} + 2(-2*a*)*b* + 2*bc** *+ 2*c*(- 2*a*) = 0

⇒ [ (- 2*a** *) + *b* + *c*]^{2} = 0

⇒ (- 2*a* + *b* + *c*)^{2} = 0

⇒ - 2*a* + *b* + *c* = 0

⇒ *b* + *c* = 2*a*

#### Page No 809:

#### Question 10:

Find the 10th term from the end of the AP 4, 9, 14, ..., 254.

*Or*, Which term of the AP 3, 15, 27, 34, ... will be 132 more than its 54th term?

#### Answer:

Here, *a* = 4, *d* = (9 − 4) = 5, *l* = 254 and *n* = 10

Now, *n*^{th} term from the end = {*l* − (*n* − 1)*d*}

∴ 10^{th} term from the end = {254 − (10 − 1) × 5}

= {254 − (9 × 5)} = (254 − 45) = 209

Hence, the 10^{th} term from the end is 209.

OR

Here, *a* = 3, *d* = (15 − 3) = 12

*T _{n}* = 132 +

*T*

_{54}

⇒

*a*+ (

*n*− 1)

*d*= 132 + {3 + (54 − 1) × 12}

⇒ 3 + (

*n*− 1) × 12 = 132 + 639

⇒(

*n*− 1) × 12 = (771 − 3) = 768

⇒ (

*n*− 1) = $\frac{768}{12}=64$

∴

*n*= (64 + 1) = 65

Hence, the required term is the 65

^{th}term.

#### Page No 809:

#### Question 11:

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

#### Answer:

**Given:** CD and EF are the tangents at the end-points A and B of the diameter AB of a circle with centre O.

**To prove**: $\mathrm{CD}\parallel \mathrm{EF}$

**Proof:** CD is the tangent to the circle at the point A.

∴ $\angle $BAD = 90°

EF is the tangent to the circle at the point B.

∴ $\angle $ABE = 90°

Thus,

$\angle $BAD = $\angle $ABE (each equal to 90°)

But these are alternate angles.

∴ $\mathrm{CD}\parallel \mathrm{EF}$

#### Page No 809:

#### Question 12:

From an external point *P,* tangents *PA* and *PB* are drawn to a circle with centre *O.* If *CD* is the tangent to the circle at the point *E* and *PA* = 14 cm, find the perimeter of ∆*PCD*.

Figure

#### Answer:

**Given :** PA and PB are tangents to the circle with centre O.

CD is a tangent to the circle at E, which intersects PA and PB at C and D, respectively and PA = 14 cm.

We know that tangents drawn from as external point to a circle are equal.

∴ PA = PB = 14 cm

CA = CE and DB = DE

Now, perimeter of ΔPCD = PC + CD + PD

= PC + (CE + ED) + PD

= PC + (CA + DB) + PD

= (PC + CA) + (DB + PD)

= PA + PB

= 14 cm + 14 cm = 28 cm

#### Page No 809:

#### Question 13:

The area of the circular base of a cone is 616 cm^{2} and its height is 48 cm. Find its whose surface area.

#### Answer:

Area of circular base = 616 cm^{2}

Therefore,

$\pi {r}^{2}=616\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{22}{7}{r}^{2}=616\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=\frac{616\times 7}{22}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=196\phantom{\rule{0ex}{0ex}}\Rightarrow r=14\mathrm{cm}$

And *h* = 48 cm

Then slant height,* l* = $\sqrt{{r}^{2}+{h}^{2}}=\sqrt{{14}^{2}+{48}^{2}}=\sqrt{196+2304}=\sqrt{2500}=50\mathrm{cm}$

Total surface are of cone = $\mathrm{\pi r}\left(\mathrm{r}+\mathrm{l}\right)=\frac{22}{7}\times 14\times \left(14+50\right)$

$=\frac{22}{7}\times 14\times 64\phantom{\rule{0ex}{0ex}}=2816{\mathrm{cm}}^{2}$

#### Page No 809:

#### Question 14:

In the adjoining figure, the area enclosed between two concentric circles is 770 cm^{2} and the radius of the outer circle is 21 cm. Find the radius of the inner circle.

Figure

#### Answer:

Let *R *and *r* be the outer and inner radii of a ring.

Then, area of the ring = $\mathrm{\pi}\left({\mathrm{R}}^{2}-{\mathrm{r}}^{2}\right)$

Given, R = 21 cm

Area of the ring = 770 cm^{2}

Thus,

$770=\frac{22}{7}\left\{{\left(21\right)}^{2}-{r}^{2}\right\}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \frac{770\times 7}{22}=441-{r}^{2}$

⇒ 245 - 441 = - *r*^{2}

⇒ - 196 = - *r*^{2}^{ }

⇒ 196 = *r*^{2}

⇒ *r* = 14 cm

Therefore, inner radius = 14 cm

#### Page No 810:

#### Question 15:

Solve for *x*: $12ab{x}^{2}-(9{a}^{2}-8{b}^{2})x-6ab=0.$

#### Answer:

2*abx*^{2} − (9*a*^{2} − 8*b*^{2})*x* − 6*ab* = 0

⇒ 12*abx*^{2} − 9*a*^{2}*x* + 8*b*^{2}*x* − 6*ab* = 0

⇒ 3*ax*(4*bx** *− 3*a*) + 2*b*(4*bx** *− 3*a*) = 0

⇒ (3*ax* + 2*b*)(4*bx** *− 3*a*) = 0

⇒ (3*ax* + 2*b*) = 0 or (4*bx** *− 3*a*) = 0

⇒ 3*ax* = − 2*b* or 4*bx** *= 3*a*

⇒ $x=\frac{-2b}{3a}$ or $x=\frac{3a}{4b}$

#### Page No 810:

#### Question 16:

If the 8th term of an *AP* is 31 and its 15th term is 16 more than the 11th term, find the *AP*.

*Or*, find the sum of all odd integers between 2 and 100 that are divisible by 3.

#### Answer:

Here,

*a*_{8} = 31 and *a*_{15} = 16 + *a*_{11}

⇒ 31 = *a *+ 7*d* .............(i)

And *a*_{15} = 16 + *a*_{11}

⇒ *a* + 14*d* = 16 + *a* + 10*d* .............(ii)

By solving equation (ii), we get:

4*d* = 16

⇒ *d* = 4

On substituting *d* = 4 in (i), we get:

31 = *a* + 7 × 4

⇒ *a* = 3

Since *a* = 3 and *d* = 4, required AP = 3, 7 , 11, 15 .....

**OR**

All odd integers between 2 and 100 divisible by 3 are

3, 9 , 15 , 21 ................ 99

Here *a* = 3 ,* d* = 6 and *n*^{th} term = 99

Let the number of terms be *n*. Then,

T* _{n}* = 99

⇒

*a*+ (

*n*- 1)×

*d*

⇒ 99 = 3+ (

*n*- 1) × 6

⇒ 99 = 3+ 6

*n*- 6

⇒ 102 = 6

*n*

⇒

*n*= 17

Required sum = $\frac{n}{2}\left(a+l\right)$

= $\frac{17}{2}\times \left(3+99\right)=\left(17\times 51\right)=867$

#### Page No 810:

#### Question 17:

Prove that the parallelogram circumscribing a circle is a rhombus.

*Or*, ∆*ABC* is drawn to circumscribe a circle of radius 4 cm, such that the segments *BD* and *DC* into which *BC* is divided by the point of contact *D,* are of lengths 8 cm and 6 cm respectively. Find *AB* and *AC*.

Figure

#### Answer:

**Given:** A parallelogram ABCD circumscribes a circle with centre O.

Thus, AB = CD and AD = BC

**To Prove :** AB = BC = CD = AD

**Proof :**

We know that the lengths of tangents drawn from an exterior point to a circle are equal.

∴ AP = AS .......(i) [tangents from A]

BP = BQ .......(ii) [tangents from B]

CR = CQ .......(iii) [tangents from C]

DR = DS .......(iv) [tangents from D]

∴ AB + CD = AP + BP + CR + DR

= AS + BQ + CQ + DS [From (i), (ii), (iii) and (iv)]

= (AS + DS) + (BQ + CQ)

= AD + BC

Thus, AB + CD = AD + BC

⇒ AB + AB = AD+ AD

⇒ 2AB = 2AD

⇒ AB = AD

∴ CD = AB = AD = BC

Hence, ABCD is a rhombus.

OR

Let the given circle touch the sides AB and AC of the triangle at point F and E, respectively, and let the length of the line segment AF be *x*.

In ΔABC, BF = BD = 6 cm (Tangents on the circle from point C)

CE = CD = 8 cm (Tangents on the circle from point B)

AE = AF = *x* cm (Tangents on the circle from point A)

AB = AF + FB = (*x* + 6) cm

BC = BD + DC = (8 + 6) = 14 cm

CA = CE + EA = (8 + *x* ) cm

∴ Perimeter of triangle (2*s*) = AB + BC + CA = (*x* + 8 + 14 + 6 + *x*) = (28 + 2*x*) cm

Semi-perimeter (*s*) = (14 + *x*) cm

Now,

$\mathrm{Area}\mathrm{of}\u25b3\mathrm{ABC}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\phantom{\rule{0ex}{0ex}}$

$=\sqrt{\left\{14+x\right\}\left\{\left(14+x\right)-14\right\}\left\{\left(14+x\right)-\left(6+x\right)\right\}\left(14+x\right)-\left(8+x\right)}\phantom{\rule{0ex}{0ex}}$

$=\sqrt{\left(14+x\right)\left(x\right)\left(8\right)\left(6\right)}=4\sqrt{3\left(14x+{x}^{2}\right)}$

Area of ΔOBC = $\frac{1}{2}\times \mathrm{OD}\times \mathrm{BC}=\frac{1}{2}\times 4\times 14=28\phantom{\rule{0ex}{0ex}}$

Area of ΔOCA = $\frac{1}{2}\times OE\times \mathrm{AC}=\frac{1}{2}\times 4\times \left(8+\mathrm{x}\right)=16+2\mathrm{x}\phantom{\rule{0ex}{0ex}}$

Area of ΔOAB = $\frac{1}{2}\times OF\times \mathrm{AB}=\frac{1}{2}\times 4\times \left(6+\mathrm{x}\right)=12+2\mathrm{x}$

Again,

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

$4\sqrt{3\left(14x+{x}^{2}\right)}=28+16+2x+12+2x\phantom{\rule{0ex}{0ex}}$

$\Rightarrow 4\sqrt{3\left(14x+{x}^{2}\right)}=56+4x\phantom{\rule{0ex}{0ex}}$

$\Rightarrow 4\sqrt{3\left(14x+{x}^{2}\right)}=4(14+x)\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \sqrt{3\left(14x+{x}^{2}\right)}=14+x$

Squaring both the sides, we get:

3(14*x** *+ *x*^{2}) = (14 + *x*)^{2}

⇒ 42*x* + 3*x*^{2} = 196 + 28*x* + *x*^{2}

⇒ 2*x*^{2} +14*x* -196 = 0

⇒ *x*^{2} + 7*x* - 98 = 0

⇒ *x*^{2} + 14*x* - 7*x* - 98 = 0

⇒ *x *(*x* + 14) - 7( *x* + 14) = 0

⇒ (*x* - 7)(*x* + 14) = 0

⇒ (*x *+14 ) = 0 or (*x* − 7) = 0

Therefore, *x* = − 14 or *x* = 7

However, *x* = − 14 is not possible as the length of the sides cannot be negative.

Therefore, *x* = 7

Hence, AB = *x* + 6 = 7 + 6 = 13 cm

AC = 8 + *x* = 8 + 7 = 15 cm

#### Page No 810:

#### Question 18:

Draw a circle of diameter 12 cm. From a point 10 cm away from its centre, construct a pair of tangents to the circle. Measure the length of each tangent segment.

#### Answer:

Given that tangents PQ and PR are drawn from a point P to a circle with centre O.

Here, OQ = OR = 6 cm and OP = 10 cm.

Now,

PQ is a tangent at Q and OQ is the radius through Q.

∴ OQ ⊥ QP

Similarly,

OR ⊥ PR

In the right-angled ΔOQP and ΔORP, we get:

OQ = OR [ radii of the same circle]

OP = OP [ common side]

∴ ΔOQP ≅ ΔORP [ By RHS congruence]

Hence QP = PR

Again,

By Pythagoras' theorem, we get:

⇒ OP^{2} = OQ^{2} + QP^{2}

⇒ (10)^{2} = (6)^{2} + QP^{2}

⇒ QP^{2} = [100 - 36] = 64 ⇒ QP = 8 cm

Thus, QP = RP = 8 cm

#### Page No 810:

#### Question 19:

Show that the points *A*(*a*, *a*), *B*(−*a*, −*a*) and $\mathrm{C}(-a\sqrt{3},a\sqrt{3})$ are the vertices of an equilateral triangle.

#### Answer:

The given points are $\mathrm{A}\left(\mathrm{a},\mathrm{a}\right),\mathrm{B}\left(-\mathrm{a},-\mathrm{a}\right)\mathrm{and}\mathrm{C}\left(-\mathrm{a}\sqrt{3},\mathrm{a}\sqrt{3}\right)$.

$\mathrm{AB}=\sqrt{{\left[a-\left(-a\right)\right]}^{2}+{\left[a-\left(-a\right)\right]}^{2}}=\sqrt{{\left(2a\right)}^{2}+{\left(2a\right)}^{2}}=\sqrt{4{a}^{2}+4{a}^{2}}=\sqrt{8{a}^{2}}=2\sqrt{2}a$ units

$\mathrm{BC}=\sqrt{{\left[-a-\left(-a\sqrt{3}\right)\right]}^{2}+{\left[-a-\left(a\sqrt{3}\right)\right]}^{2}}\phantom{\rule{0ex}{0ex}}$

$=\sqrt{{\left[a\left(\sqrt{3}-1\right)\right]}^{2}+{\left[a\left(\sqrt{3}+1\right)\right]}^{2}}\phantom{\rule{0ex}{0ex}}$

$=\sqrt{{a}^{2}\left(3+1-2\sqrt{3}\right)+{a}^{2}\left(3+1+2\sqrt{3}\right)}\phantom{\rule{0ex}{0ex}}$

$=\sqrt{{a}^{2}\left(4-2\sqrt{3}\right)+{a}^{2}\left(4+2\sqrt{3}\right)}=\sqrt{{a}^{2}\left(4-2\sqrt{3}+4+2\sqrt{3}\right)}=\sqrt{8{a}^{2}}=2\sqrt{2}a$ units

$CA=\sqrt{{\left[a-\left(-a\sqrt{3}\right)\right]}^{2}+{\left[a-a\sqrt{3}\right]}^{2}}\phantom{\rule{0ex}{0ex}}$

$=\sqrt{{\left[a\left(1+\sqrt{3}\right)\right]}^{2}+{\left[a\left(1-\sqrt{3}\right)\right]}^{2}}\phantom{\rule{0ex}{0ex}}$

$=\sqrt{{a}^{2}\left(1+3+2\sqrt{3}\right)+{a}^{2}\left(1+3-2\sqrt{3}\right)}\phantom{\rule{0ex}{0ex}}$

$=\sqrt{{a}^{2}\left(4+2\sqrt{3}\right)+{a}^{2}\left(4-2\sqrt{3}\right)}=\sqrt{{a}^{2}\left(4+2\sqrt{3}+4-2\sqrt{3}\right)}=\sqrt{8{a}^{2}}=2\sqrt{2}a$ units

∴ AB = BC = CA = $2\sqrt{2}a$ units

Hence, ΔABC is an equilateral triangle and each of its sides is $2\sqrt{2}a$ units.

#### Page No 810:

#### Question 20:

Find the area of a rhombus if its vertices are *A*(3, 0) *B*(4, 5), *C*(−1, 4) and *D*(−2, −1).

#### Answer:

Let *A *(3, 0), *B *(4, 5), *C *(−1, 4) and *D *(−2, −1) be the vertices of a rhombus ABCD.

$\mathrm{Length}\mathrm{of}\mathrm{diagonal}\mathrm{AC}=\sqrt{{\left[3-\left(-1\right)\right]}^{2}+{\left(0-4\right)}^{2}}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$ units

$\mathrm{Length}\mathrm{of}\mathrm{diagonal}\mathrm{BD}=\sqrt{{\left[4-\left(-2\right)\right]}^{2}+{\left[5-\left(-1\right)\right]}^{2}}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2}$ units

Hence, area of rhombus = $\frac{1}{2}\times 4\sqrt{2}\times 6\sqrt{2}$ = 24 sq. units

#### Page No 810:

#### Question 21:

Cards marked with numbers 13, 14, 15, ..., 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card drawn is

(a) divisible by 5

(b) a perfect square

#### Answer:

Total numbers of cards = 60 - 13 + 1 = 48

**(a)** Cards with numbers divisible by 5 are 15, 20, 25, 30, 35, 40, 45, 50, 55, 60

Number of such cards = 10

∴ *P*(getting a number divisible by 5) = $\frac{10}{48}=\frac{5}{24}$

**(b)** Perfect squares are 16, 25, 36, 49.

Number of such cards = 4

∴ *P*(getting a perfect square) = $\frac{4}{48}=\frac{1}{12}$

#### Page No 810:

#### Question 22:

A window in a building is at a height of 10 m from the ground. The angle of depression of a point *P* on the ground from the window is 30*°. *The angle of elevation of the top of the building from the point *P* is 60°. Find the height of the building.

*Or*, in a violent storm, a tree is bent by the wind. The top of the tree meets the ground at an angle of 30°, at a distance of 30 metres from the root. At what height from the bottom did the tree get bent? What was the original height of the tree?

#### Answer:

Given, height of the window from the ground DB = 10 m

Let H be the height of the building.

Let O be the point of observation. Then,

DB = 10 m, $\angle $DPB = 30° and $\angle $APB = 60°

BP = *x* m

$\mathrm{tan}30\xb0=\frac{10}{x}$

$\frac{1}{\sqrt{3}}=\frac{10}{x}\Rightarrow x=10\sqrt{3}\mathrm{m}$ ...............(i)

Again,

$\mathrm{tan}60\xb0=\frac{H}{x}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \sqrt{3}=\frac{H}{x}$

$\Rightarrow H=\sqrt{3}x\phantom{\rule{0ex}{0ex}}$

$\Rightarrow H=\left(\sqrt{3}\times 10\sqrt{3}\right)\mathrm{m}=30\mathrm{m}$

Hence, required height of the building is 30 m.

OR

Let AB be the original height of the tree, which gets bent at a point C. After being bent, let the part CB take the position CD, meeting the ground at D. Then AD = 30 m, ∠ADC = 30°, ∠DAC = 90° and CD = CB

Let AC = *x* metres and CD = CB = *y* metres

From right-angled ΔDAC, we have:

$\frac{AC}{AD}=\mathrm{tan}30\xb0=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \frac{x}{30}=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow x=\left(30\times \frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\right)=10\sqrt{3}\mathrm{m}$

Also, from right-angled ΔDAC, we have:

$\frac{DC}{AD}=\mathrm{sec}30\xb0=\frac{2}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \frac{y}{30}=\frac{2}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow y=\left(30\times \frac{2}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\right)=20\sqrt{3}\mathrm{m}$

Thus AB = (*x* + *y*) metres = $\left(10\sqrt{3}+20\sqrt{3}\right)\mathrm{m}=30\sqrt{3}\mathrm{m}$

Hence, the tree got bent at a height of $10\sqrt{3}\mathrm{m}$ and the original height of the tree was $30\sqrt{3}\mathrm{m}$.

#### Page No 810:

#### Question 23:

A wire bent in the form of a circle of radius 42 cm is cut and again bent in the form of a square. Find the ratio of the areas of the regions enclosed by the circle and the square.

#### Answer:

Length of the wire = Circumference of the circle

= $\left(2\times \frac{22}{7}\times 42\right)=264$ cm

Therefore,

Perimeter of the square = Length of the wire = 264 cm

Hence, the side of the square = $\left(\frac{264}{4}\right)\mathrm{cm}=66\mathrm{cm}$

Thus,

Area of the circle = ${\mathrm{\pi r}}^{2}=\left(\frac{22}{7}\times 42\times 42\right)$ cm^{2} = 5544 cm^{2}

Area of the square = (Side)^{2} = (66 cm)^{2} = 4356 cm^{2}

Therefore,

$\frac{\mathrm{Area}\mathrm{of}\mathrm{circle}}{\mathrm{Area}\mathrm{of}\mathrm{square}}=\frac{5544}{4356}=\frac{504}{396}=\frac{14}{11}$

Hence,

Area of circle : Area of square = 14 : 11

#### Page No 810:

#### Question 24:

A metallic sphere of radius 10.5 cm is melted and then recast into smaller cones, each of radius 3.5 cm and height 3 cm. How many cones are obtained?

#### Answer:

Radius of the sphere = 10.5 cm

Volume of the sphere = $\frac{4}{3}{\mathrm{\pi r}}^{3}=\left(\frac{4}{3}\times \mathrm{\pi}\times 10.5\times 10.5\times 10.5\right)$ cm^{3} =$\left(\frac{4}{3}\mathrm{\pi}\times \frac{21}{2}\times \frac{21}{2}\times \frac{21}{2}\right)$ cm^{3} = $\left(\frac{3087\mathrm{\pi}}{2}\right)$ cm^{3}

Radius of each cone = 3.5 cm and height = 3 cm

Volume of each cone = $\left(\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}\right)$ = $\left(\frac{1}{3}\mathrm{\pi}\times 3.5\times 3.5\times 3\right)$ cm^{3} =$\left(\frac{1}{3}\mathrm{\pi}\times \frac{7}{2}\times \frac{7}{2}\times 3\right)$ cm^{3}^{ }= $\left(\frac{49\mathrm{\pi}}{4}\right)$ cm^{3}

Required number of cones = $\frac{\mathrm{Volume}\mathrm{of}\mathrm{sphere}}{\mathrm{Volume}\mathrm{of}\mathrm{each}\mathrm{cone}}=\left(\frac{3087\mathrm{\pi}\times 4}{2\times 49\mathrm{\pi}}\right)=126$

Hence, required number of cones is 126.

#### Page No 811:

#### Question 25:

In the given figure, ∆*ABC* is right-angled at *A*. Semi-circles are drawn on *AB*, *AC* and *BC* as diameters. It is given that *AB* = 3 cm and *AC* = 4 cm. Find the area of the shaded region.

Figure

#### Answer:

Given: *AB* = 3 cm and *AC* = 4

Thus, BC = 5 cm (By Pythagoras' theorem)

Area of shaded region = {ar (ΔABC + ar(semi-circle APB) + ar(semi-circle AQC)} - ar(semi-circle BAC)

= $\left[\left(\frac{1}{2}\times 3\times 4\right)+\left(\frac{1}{2}\mathrm{\pi}\times \frac{3}{2}\times \frac{3}{2}\right)+\left(\frac{1}{2}\mathrm{\pi}\times 2\times 2\right)-\left(\frac{1}{2}\mathrm{\pi}\times \frac{5}{2}\times \frac{5}{2}\right)\right]$ cm^{2}

= $\left\{6+\frac{1}{2}\mathrm{\pi}\left(\frac{9}{4}+4-\frac{25}{4}\right)\right\}$ cm^{2}

= (6 + 0) cm^{2} = 6 cm^{2}

#### Page No 811:

#### Question 26:

Rs 250 is divided equally among a certain number of children. If there were 25 more children, each would have received 50 paise less. Find the number of children.

*Or*, the hypotenuse of a right-angled triangle is 6 cm more than twice the shortest side. If the third side is 2 cm less than the hypotenuse, find the sides of the triangle.

#### Answer:

Let the required number of children be *x*.

And total amount = Rs 250 = 25000 paise

Then, share of each child = $\frac{25000}{x}$ paise

If there were (*x* + 25) children, share of each child = $\frac{25000}{\left(x+25\right)}$paise

∴ $\frac{25000}{x}-\frac{25000}{\left(x+25\right)}=\frac{50}{1}$

$\Rightarrow 25000\left(\frac{1}{x}-\frac{1}{x+25}\right)=\frac{50}{1}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow 25000\left[\frac{x+25-x}{x\left(x+25\right)}\right]=\frac{50}{1}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \frac{25000\times 25}{50}={x}^{2}+25x\phantom{\rule{0ex}{0ex}}$

$\Rightarrow 12500={x}^{2}+25x\phantom{\rule{0ex}{0ex}}$

$\Rightarrow {x}^{2}+25x-12500=0\phantom{\rule{0ex}{0ex}}$

$\Rightarrow {x}^{2}+125x-100x-12500=0\phantom{\rule{0ex}{0ex}}$

$\Rightarrow x\left(x+125\right)-100\left(x+125\right)=0\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \left(x-100\right)\left(x+125\right)=0\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \left(x-100\right)=0\mathrm{or}\left(\mathrm{x}+125\right)=0\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \mathrm{x}=100\mathrm{or}\mathrm{x}=-125$

Since the number of children cannot be negative, *x* = 100

OR

Let the length of the shortest side be *x* cm.

Then, hypotenuse = (2*x* + 6) cm

And the third side = (2*x* + 6 - 2) cm = (2*x* + 4) cm

By Pythagoras' theorem, we have:

(2*x*+ 6)^{2} = *x*^{2} + (2*x* + 4)^{2}

$\Rightarrow {\left(2x\right)}^{2}+2\times 2x\times 6+{6}^{2}={x}^{2}+{\left(2x\right)}^{2}+2\times 2x\times 4+{4}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 4{x}^{2}+24x+36={x}^{2}+4{x}^{2}+16x+16$

⇒ *x*^{2} - 8*x* - 20 = 0

⇒ *x*^{2} - 10*x* + 2*x* - 20 = 0

⇒ *x*(*x* - 10) + 2(*x* - 10) = 0

⇒ (*x* + 2)(*x* - 10) = 0

⇒ *x* = - 2 or *x* = 10

Since, the side of a triangle is never negative, the length of the shortest side = 10 cm

Length of the hypotenuse = (2*x* + 6) cm = 26 cm

Length of the third side = ( 26 - 2) = 24 cm

Hence, the sides of the triangle are 10 cm, 26 cm and 24 cm.

#### Page No 811:

#### Question 27:

If the sum of first *n*, 2*n* and 3*n* terms of an AP be S_{1}, S_{2} and S_{3} respectively, then prove that S_{3} = 3 (S_{2}_{ }− S_{1}).

#### Answer:

Let *a* be the first term and *d* the common difference of the given AP. Then,

S_{1} = sum of first *n* terms of the given AP

S_{2} = sum of first 2*n* terms of the given AP

S_{3}_{ }= sum of first 3*n* terms of the given AP

Therefore,

${S}_{1}=\frac{n}{2}\left\{2a+\left(n-1\right)d\right\}\phantom{\rule{0ex}{0ex}}$

${S}_{2}=\frac{2n}{2}\left\{2a+\left(2n-1\right)d\right\}\phantom{\rule{0ex}{0ex}}$

${S}_{3}=\frac{3n}{2}\left\{2a+\left(3n-1\right)d\right\}$

Thus,

$3\left({S}_{2}-{S}_{1}\right)=3\left[\left\{2na+n\left(2n-1\right)d\right\}-\left\{na+\frac{1}{2}n\left(n-1\right)d\right\}\right]\phantom{\rule{0ex}{0ex}}$

$=3\left[na+\frac{3}{2}{n}^{2}d-\frac{1}{2}nd\right]=\frac{3n}{2}\left[2a+3nd-d\right]\phantom{\rule{0ex}{0ex}}$

$=\frac{3n}{2}\left[2a+\left(3n-1\right)d\right]={S}_{3}$

Hence, S_{3} = 3 (S_{2}_{ }− S_{1})

#### Page No 811:

#### Question 28:

The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of $1500\sqrt{3}\mathrm{m}$, find its speed.

#### Answer:

Let *A* and *B* be the two positions of the jet plane and *O* be the point of observation. Let *OX* be the horizontal ground.

Then, $\angle $XOA = 60° and $\angle $XOB = 30°

Draw AL ⊥ OX and BM ⊥ OX.

Let AL = BM = *h* metres = $1500\sqrt{3}\mathrm{m}$

From right ΔOLA, we have:

$\frac{\mathrm{OL}}{\mathrm{AL}}=\mathrm{cot}60\xb0=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \frac{\mathrm{OL}}{1500\sqrt{3}}=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \mathrm{OL}=1500$

From right ΔOMB, we have:

$\frac{\mathrm{OM}}{\mathrm{MB}}=\mathrm{cot}30\xb0=\sqrt{3}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \frac{\mathrm{OM}}{1500\sqrt{3}}=\sqrt{3}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \mathrm{OM}=4500$

Therefore,

LM = (OM - OL) = (4500 - 1500) = 3000 m

Thus, the jet plane covers 3000 m in 15 seconds.

Hence, the speed of the jet plane = $\left(\frac{3000}{15}\times \frac{60\times 60}{1000}\right)\mathrm{kmph}=720\mathrm{kmph}$

#### Page No 811:

#### Question 29:

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

#### Answer:

**Given: ** A circle with centre O and a tangent AB at point P of the circle.

**To prove: **OP ⊥ AB

**Construction: **Take a point Q, other than P, on the tangent AB. Join OQ.

**Proof:**

Q is a point on the tangent AB other than the point of contact P.

∴ Q lies outside the circle.

Let OQ intersect the circle at R.

Then,

OR < OQ [ a part is less than the whole] ................(i)

But OP = OR [radii of the same circle] .................(ii)

∴ OP < OQ [ from (i) and (ii)]

Thus, OP is shorter than any other line segment joining O to any point of AB, other than P.

In other words, OP is the shortest distance between the point O and the line AB.

But the shortest distance between a point and a line is the perpendicular distance.

∴ OP ⊥ AB.

#### Page No 811:

#### Question 30:

A quadrilateral *ABCD* is drawn to circumscribe a circle, as shown in the given figure.

Prove that *AB* + *CD* = *AD* + *BC.*

Figure

#### Answer:

We know that the lengths of tangents drawn from an exterior point to a circle are equal.

∴ AP = AS .......(i) [tangents from A]

BP = BQ .......(ii) [tangents from B]

CR = CQ .......(iii) [tangents from C]

DR = DS .......(iv) [tangents from D]

∴ AB + CD = AP + BP + CR + DR

= AS + BQ + CQ + DS [From (i), (ii), (iii) and (iv)]

= (AS + DS) + (BQ + CQ)

= AD + BC

Thus, AB + CD = AD + BC

#### Page No 811:

#### Question 31:

A solid is made up of a cube and a hemisphere is attached on its top, as shown in the figure. Each edge of the cube measures 5 cm and the hemisphere has a diameter of 4.2 cm. Find the total area to be painted.

*Or*, the diameters of the lower and upper ends of a bucket, in the form of a frustum of a cone, are 10 cm and 30 cm respectively. If its heights is 24 cm, find

(i) the capacity of the bucket

(ii) the area of the metal sheet used to make the bucket

#### Answer:

Total surface area of the cube = 6 × (edge)^{2}

= (6 × 5 × 5) cm^{2}

Area to be painted = (Total surface area of the cube) - (Base area of the hemisphere) + (CSA of the hemisphere)

= $\left(150-{\mathrm{\pi r}}^{2}+2{\mathrm{\pi r}}^{2}\right)\phantom{\rule{0ex}{0ex}}$ cm^{2}

$=\left(150+{\mathrm{\pi r}}^{2}\right)\phantom{\rule{0ex}{0ex}}$cm^{2}

$=\left(150+\frac{22}{7}\times 2.1\times 2.1\right)\phantom{\rule{0ex}{0ex}}$ cm^{2}

$=\left(150+\frac{693}{50}\right)$ cm^{2}

= (150 + 13.86) cm^{2}^{ }= 163.86 cm^{2}

Hence, required area to be painted is 163.86 cm^{2}.

**OR**

We have:

*R* = 15 cm, *r* = 5 cm and *h *= 24 cm

*l*^{2} = *h*^{2} + (*R* - *r*)^{2}

= ((24)^{2} + (15 - 5)^{2}} = (576 + 100) = 676

⇒ *l *= 26 cm

(i)

$\mathrm{Capacity}\mathrm{of}\mathrm{the}\mathrm{bucket}=\frac{1}{3}\mathrm{\pi h}\left({\mathrm{R}}^{2}+{\mathrm{r}}^{2}+\mathrm{Rr}\right){\mathrm{cm}}^{3}\phantom{\rule{0ex}{0ex}}$

$=\frac{1}{3}\left\{3.14\times 24\left({15}^{2}+{5}^{2}+15\times 5\right)\right\}{\mathrm{cm}}^{3}\phantom{\rule{0ex}{0ex}}$

$=\frac{1}{3}\left\{3.14\times 24\left(225+25+75\right)\right\}{\mathrm{cm}}^{3}=\frac{1}{3}\left\{3.14\times 24\times 325\right\}{\mathrm{cm}}^{3}$

= 8164 cm^{3}

(ii)

$\mathrm{Required}\mathrm{area}\mathrm{of}\mathrm{t}\mathrm{he}\mathrm{metal}\mathrm{sheet}=\mathrm{\pi}\left[{\mathrm{r}}^{2}+\mathrm{l}\left(\mathrm{R}+\mathrm{r}\right)\right]\mathrm{sq}.\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

$=3.14\left[{5}^{2}+26\left(15+5\right)\right]{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}$

$=3.14\times \left(25+520\right){\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}$

$=3.14\times 545{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}$

$=1711.3{\mathrm{cm}}^{2}$

#### Page No 811:

#### Question 32:

Find the value of *k* for which the points *A*(−1, 3), *B*(2, *k*) and *C*(5, −1) are collinear.

#### Answer:

Let *A*(−1, 3), *B*(2, *k*) and *C*(5, − 1) be the given points. Then,

(*x*_{1} = − 1, *y*_{1} = 3), (*x*_{2} = 2 , *y*_{2} = *k*) and (*x*_{3} = 5, *y*_{3} = − 1)

Since the given points are collinear, the area of the triangle formed by them must be 0.

⇒$\frac{1}{2}$[*x*_{1}(*y*_{2} − *y*_{3}) + *x*_{2}(*y*_{3} − *y*_{1}) + *x*_{3}(*y*_{1} − *y*_{2})] = 0

⇒ − 1(*k* + 1) + 2( − 1 − 3) + 5(3 − *k*) = 0

⇒ − 1(*k* + 1) + 2( − 4) + 5(3 − *k*) = 0

⇒ − *k* − 1 − 8 + 15 − 5*k* = 0

⇒ − 6*k* = − 6

⇒ 6*k* = 6 ⇒ *k* = 1

Hence, the required value of *k* is 1.

#### Page No 812:

#### Question 33:

Two dice are thrown at the same time. Find the probability that the sum of two numbers appearing on the top of the dice is more than 9.

#### Answer:

When two dice are thrown simultaneously, the possible outcomes are :

(1,1); (1,2); (1,3); (1,4); (1,5); (1,6)

(2,1); (2,2); (2,3); (2,4); (2,5); (2,6)

(3,1); (3,2); (3,3); (3,4); (3,5); (3,6)

(4,1); (4,2); (4,3); (4,4); (4,5); (4,6)

(5,1); (5,2); (5,3); (5,4); (5,5); (5,6)

(6,1); (6,2); (6,3); (6,4); (6,5); (6,6)

Number of possible outcomes = 36

Let *E* be the event that the sum of the numbers appearing on the top of the two dice is more than 9.

The favourable outcomes are:

(4, 6), (5, 5), (5, 6), (6, 4), (6, 5) ,(6, 6)

Number of favourable outcomes = 6

∴ *P*(getting a sum more than* 9*) = *P*(*E*) = $\frac{6}{36}=\frac{1}{6}$

#### Page No 812:

#### Question 34:

A circus tent is cylindrical to a height of 3 m and conical above it. If its base radius is 52.5 m and the slant height of the conical portion is 53 m, find the area of canvas needed to make the tent.

#### Answer:

**For cylinder:**

Height of cylinder (*h*) = 3 m

Base radius of cylinder (*r*) = 52.5 m

**For cone:**

Slant height of cone (*l*) = 53 m

Base radius of cone = Base radius of cylinder (*r*) = 52.5 m

∴ Area of canvas required = $\left(\mathrm{\pi rl}+2\mathrm{\pi rh}\right)\mathrm{sq}.\mathrm{units}$

= $\mathrm{\pi r}\left(\mathrm{l}+2\mathrm{h}\right)$ sq. units

= $\frac{22}{7}\times 52.5\left(53+2\times 3\right)$ m^{2}

= $\frac{22}{7}\times 52.5\times 59$ m^{2} = 9,735 m^{2}

Hence, area of canvas needed to make the tent = 9,735 m^{2}

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