Math Ncert Exemplar 2019 Solutions for Class 7 Maths Chapter 9 Perimeter & Area are provided here with simple step-by-step explanations. These solutions for Perimeter & Area are extremely popular among class 7 students for Maths Perimeter & Area Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Math Ncert Exemplar 2019 Book of class 7 Maths Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Math Ncert Exemplar 2019 Solutions. All Math Ncert Exemplar 2019 Solutions for class 7 Maths are prepared by experts and are 100% accurate.

#### Page No 266:

#### Question 1:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

Observe the shapes 1, 2, 3 and 4 in the figures. Which of the following statements is not correct?

(a) Shapes 1, 3 and 4 have different areas and different perimeters.

(b) Shapes 1 and 4 have the same area as well as the same perimeter.

(c) Shapes 1, 2 and 4 have the same area.

(d) Shapes 1, 3 and 4 have the same perimeter.

#### Answer:

**Shape 1:**

Perimeter = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

= 22 units

Area = 18 × 1

= 18 square units**Shape 2:**

Perimeter = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

= 18 units

Area = 18 × 1

= 18 square units**Shape 3:**

Perimeter = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

= 22 units

Area = 16 × 1

= 16 square units**Shape 4:**

Perimeter = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

= 22 units

Area = 18 × 1

= 18 square units

Hence, the correct answer is option (a).

#### Page No 267:

#### Question 2:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

A rectangular piece of dimensions 3 cm × 2 cm was cut from a rectangular sheet of paper of dimensions 6 cm × 5 cm. Area of remaining sheet of paper is

(a) 30 cm^{2}

(b) 36 cm^{2}

(c) 24 cm^{2 }

(d) 22 cm^{2}

#### Answer:

Given,

Area of the bigger rectangle

Area of the smaller rectangle

Area of the remaining sheet of paper = Area of the bigger rectangle − Area of the smaller rectangle

#### Page No 268:

#### Question 3:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

36 unit squares are joined to form a rectangle with the least perimeter. Perimeter of the rectangle is

(a) 12 units

(b) 26 units

(c) 24 units

(d) 36 units

#### Answer:

Area of rectangle = 36 square units

Therefore, the sides of a rectangle are 4 cm and $9\phantom{\rule{0ex}{0ex}}\text{cm}$.

Perimeter × (4 + 9)

$=\; 26$ units

Hence, the correct answer is option (b).

#### Page No 268:

#### Question 4:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

A wire is bent to form a square of side 22 cm. If the wire is rebent to form a circle, its radius is

(a) 22 cm

(b) 14 cm

(c) 11 cm

(d) 7 cm

#### Answer:

Given: Side of a square =

Perimeter of the square and the circumference of the circle are equal.

Let the radius of the circle be r cm.

∴ Circumference of circle $=$ Perimeter of square

$\Rightarrow 2\mathrm{\pi}r=4\times \mathrm{side}\phantom{\rule{0ex}{0ex}}\Rightarrow 2\times \frac{22}{7}\times r=4\times 22\phantom{\rule{0ex}{0ex}}\Rightarrow r=4\times 22\times \frac{1}{2}\times \frac{7}{22}\phantom{\rule{0ex}{0ex}}\Rightarrow r=2\times 7\phantom{\rule{0ex}{0ex}}\Rightarrow r=14\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

Thus, the radius of the circle is 14 cm.

Hence, the correct answer is option (b).

#### Page No 268:

#### Question 5:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

Area of the circle obtained in Question 4 is

(a) 196 cm^{2}

(b) 212 cm^{2}

(c) 616 cm^{2}

(d) 644 cm^{2}

#### Answer:

The radius of circle (*r*) = 14 cm

$\begin{array}{rcl}\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{circle}& =& \mathrm{\pi}{r}^{2}\\ & =& \frac{22}{7}\times 14\times 14{\mathrm{cm}}^{2}\\ & =& 616{\mathrm{cm}}^{2}\end{array}$

Hence, the correct answer is option (c).

#### Page No 268:

#### Question 6:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

Area of a rectangle and the area of a circle are equal. If the dimensions of the rectangle are 14cm × 11 cm, then radius of the circle is

(a) 21 cm

(b) 10.5 cm

(c) 14 cm

(d) 7 cm

#### Answer:

Length of rectangle (

Breadth of rectangle

Area of circle

$\Rightarrow \mathrm{\pi}{r}^{2}=l\times b\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{22}{7}\times {r}^{2}=14\times 11\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=14\times 11\times \frac{7}{22}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=7\times 7\phantom{\rule{0ex}{0ex}}\Rightarrow r=7\mathrm{cm}$

Thus, the radius of the circle is 7 cm.

Hence, the correct answer is option (d).

#### Page No 268:

#### Question 7:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

Area of shaded portion in the given figure is

(a) 25 cm^{2}

(b) 15 cm^{2}

(c) 14 cm^{2}

(d) 10 cm^{2}

#### Answer:

Length of the rectangle (

Breadth of the rectangle ($4\phantom{\rule{0ex}{0ex}}\text{cm}$

Area of the shaded portion $\times $ area of the rectangle

$=\frac{1}{2}\times 5\times 4{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}=10{\mathrm{cm}}^{2}$

Thus, the area of shaded portion in the given figure is 10 cm^{2}.

Hence, the correct answer is option (d).

#### Page No 268:

#### Question 8:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

Area of parallelogram ABCD is not equal to

(a) DE × DC

(b) BE × AD

(c) BF × DC

(d) BE × BC

#### Answer:

Area of parallelogram = Base × Height

Area of parallelogram ABCD = AD × BE

= BC × BE (âˆµAD = BC)

Area of parallelogram ABCD = DC × BF

= AB × BF (âˆµAB = DC)

So, area of parallelogram ABCD ≠ DE × DC

Hence, the correct answer is option (a).

#### Page No 269:

#### Question 9:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

Area of triangle MNO in the given figure is

(a) $\frac{1}{2}\mathrm{MN}\times \mathrm{NO}$

(b) $\frac{1}{2}\mathrm{NO}\times \mathrm{MO}$

(c) $\frac{1}{2}\mathrm{MN}\times \mathrm{OQ}$

(d) $\frac{1}{2}\mathrm{NO}\times \mathrm{OQ}$

#### Answer:

Area of triangle = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$

Area of triangle MNO = $\frac{1}{2}\times \mathrm{NO}\times \mathrm{OQ}$

$=\frac{1}{2}\times \mathrm{MP}\times \mathrm{OQ}$

Hence, the correct answer is option (c).

#### Page No 269:

#### Question 10:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

Ratio of area of âˆ†MNO to the area of parallelogram MNOP in the same given figure is

(a) 2 : 3

(b) 1 : 1

(c) 1 : 2

(d) 2 : 1

#### Answer:

Area of triangle MNO: Area of parallelogram MNOP

$=\frac{{\displaystyle \frac{1}{2}}\times \mathrm{Base}\times \mathrm{Height}}{\mathrm{Base}\times \mathrm{Height}}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{1}{2}}\times \mathrm{NO}\times \mathrm{QO}}{\mathrm{MP}\times \mathrm{OQ}}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{1}{2}\times \mathrm{MP}\times \mathrm{QO}}}{\mathrm{MP}\times \mathrm{OQ}}\left(\because \mathrm{MP}=\mathrm{NO}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}=1:2$

Hence, the correct answer is option (c).

#### Page No 269:

#### Question 11:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

Ratio of areas of âˆ†MNO, âˆ†MOP and âˆ†MPQ in the given figure is

(a) 2 : 1 : 3

(b) 1 : 3 : 2

(c) 2 : 3 : 1

(d) 1 : 2 : 3

#### Answer:

$\begin{array}{rcl}\mathrm{Area}\mathrm{of}\u25b3\mathrm{MNO}& =& \frac{1}{2}\times \mathrm{NO}\times \mathrm{MO}\\ & =& \frac{1}{2}\times 4\times 5{\mathrm{cm}}^{2}\\ & =& 10{\mathrm{cm}}^{2}\end{array}$

$\begin{array}{rcl}\mathrm{Area}\mathrm{of}\u25b3\mathrm{MOP}& =& \frac{1}{2}\times \mathrm{PO}\times \mathrm{MO}\\ & =& \frac{1}{2}\times 2\times 5{\mathrm{cm}}^{2}\\ & =& 5{\mathrm{cm}}^{2}\end{array}$

$\begin{array}{rcl}\mathrm{Area}\mathrm{of}\u25b3\mathrm{MPQ}& =& \frac{1}{2}\times \mathrm{QP}\times \mathrm{MO}\\ & =& \frac{1}{2}\times 6\times 5{\mathrm{cm}}^{2}\\ & =& 15{\mathrm{cm}}^{2}\end{array}$

∴ Ratio of areas of âˆ†MNO, âˆ†MOP and âˆ†MPQ = Area of âˆ†MNO : Area of âˆ†MOP : Area of âˆ†MPQ

= 10 : 5 : 15

= 2 : 1 : 3

Hence, the correct answer is option (a).

#### Page No 270:

#### Question 12:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

In the given figure, EFGH is a parallelogram, altitudes FK and FI are 8 cm and 4cm respectively. If EF = 10 cm, then area of EFGH is

(a) 20 cm^{2}

(b) 32 cm^{2}

(c) 40 cm^{2}

(d) 80 cm^{2}

#### Answer:

Area of parallelogram EFGH = Base × Height

= HG × FI

= EF × FI

= 10 × 4 cm^{2}

= 40 cm^{2}

Hence, the correct answer is option (c).

#### Page No 271:

#### Question 13:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

In reference to a circle the value of ðœ‹ is equal to

(a) $\frac{\mathrm{area}}{\mathrm{circumference}}$

(b) $\frac{\mathrm{area}}{\mathrm{diameter}}$

(c) $\frac{\mathrm{circumference}}{\mathrm{diameter}}$

(d) $\frac{\mathrm{circumference}}{\mathrm{radius}}$

#### Answer:

Circumference of a circle is

Circumference $\times $ Diameter

Thus, the value of ðœ‹ = $\frac{\mathrm{circumference}}{\mathrm{diameter}}$

Hence, the correct answer is option (c).

#### Page No 271:

#### Question 14:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

Circumference of a circle is always

(a) more than three times of its diameter

(b) three times of its diameter

(c) less than three times of its diameter

(d) three times of its radius

#### Answer:

Circumference of a circle

Circumference of a circle

Circumference of a circle

#### Page No 271:

#### Question 15:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

Area of triangle PQR is 100 cm^{2}. If altitude QT is 10 cm, then its base PR is

(a) 20 cm

(b) 15 cm

(c) 10 cm

(d) 5 cm

#### Answer:

Given: Area of triangle

Area of triangle

$\Rightarrow 100=\frac{1}{2}\times \mathrm{PR}\times 10\phantom{\rule{0ex}{0ex}}\Rightarrow 100\times 2\times \frac{1}{10}=\mathrm{PR}\phantom{\rule{0ex}{0ex}}\Rightarrow 20=\mathrm{PR}$

Thus, its base PR is 20 cm.

Hence, the correct answer is option (a).

#### Page No 271:

#### Question 16:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

In the given figure, if PR = 12 cm, QR = 6 cm and PL = 8 cm, then QM is

(a) 6 cm

(b) 9 cm

(c) 4 cm

(d) 2 cm

#### Answer:

In right-angled triangle PLR, by using Pythagoras's theorem

$\Rightarrow {\mathrm{PR}}^{2}-{\mathrm{PL}}^{2}={\mathrm{LR}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {12}^{2}-{8}^{2}={\mathrm{LR}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 144-64={\mathrm{LR}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 80={\mathrm{LR}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{LR}=4\sqrt{5}\mathrm{cm}$

#### Page No 272:

#### Question 17:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

In the given figure âˆ† MNO is a right-angled triangle. Its legs are 6 cm and 8 cm long. Length of perpendicular NP on the side MO is

(a) 4.8 cm

(b) 3.6 cm

(c) 2.4 cm

(d) 1.2 cm

#### Answer:

Given: MNO is a right- angled triangle.

$\Rightarrow {\mathrm{MO}}^{2}={\mathrm{NO}}^{2}+{\mathrm{NM}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{MO}}^{2}={8}^{2}+{6}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{MO}}^{2}=100\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{MO}=10\mathrm{cm}$

Area of MNO = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$

$=\frac{1}{2}\times \mathrm{NO}\times \mathrm{NM}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 8\times 6\phantom{\rule{0ex}{0ex}}=24{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}$

Also, Area of MNO = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$

$\Rightarrow 24=\frac{1}{2}\times \mathrm{MO}\times \mathrm{NP}\phantom{\rule{0ex}{0ex}}\Rightarrow 24=\frac{1}{2}\times 10\times \mathrm{NP}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{NP}=\frac{24\times 2}{10}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{NP}=4.8\mathrm{cm}$

Hence, the correct answer is option (a).

#### Page No 272:

#### Question 18:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

Area of a right-angled triangle is 30 cm^{2}. If its smallest side is 5 cm, then its hypotenuse is

(a) 14 cm

(b) 13 cm

(c) 12 cm

(d) 11cm

#### Answer:

Given: Area of a right-angled triangle = 30 cm^{2}

Smallest side of the right-angled triangle = 5 cm

Consider the smallest side of the right-angled triangle be the base of the triangle such that* b* = 5 cm

Area of triangle = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$

$\Rightarrow 30=\frac{1}{2}\times 5\times \mathrm{Height}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Height}=12\mathrm{cm}$

By using Pythagoras Theorem in the given right-angled triangle,

${\mathrm{Hypotenuse}}^{2}={\mathrm{Base}}^{2}+{\mathrm{Height}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{Hypotenuse}}^{2}={5}^{2}+{12}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{Hypotenuse}}^{2}=25+144\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{Hypotenuse}}^{2}=169\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{Hypotenuse}}^{2}={13}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Hypotenuse}=13\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

Thus, the hypotenuse is 13 cm.

Hence, the correct answer is option (b).

#### Page No 272:

#### Question 19:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

Circumference of a circle of diameter 5 cm is

(a) 3.14 cm

(b) 31.4 cm

(c) 15.7 cm

(d) 1.57 cm

#### Answer:

Given: Diameter of the circle (*d*) = 5 cm

$\begin{array}{rcl}\mathrm{Radius}\mathrm{of}\mathrm{circle}\left(r\right)& =& \frac{d}{2}\\ & =& \frac{5}{2}\mathrm{cm}\\ & =& 2.5\mathrm{cm}\end{array}$

Circumference of the circle = 2π*r
$=2\times \frac{22}{7}\times 2.5\phantom{\rule{0ex}{0ex}}=15.71\mathrm{cm}$*

Hence, the correct answer is option (c).

#### Page No 272:

#### Question 20:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

Circumference of a circle disc is 88 cm. Its radius is

(a) 8 cm

(b) 11 cm

(c) 14 cm

(d) 44 cm

#### Answer:

Given: Circumference of the circle disc = 88 cm

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

$\mathrm{Circumference}\mathrm{of}\mathrm{the}\mathrm{circle}=2\mathrm{\pi}r\phantom{\rule{0ex}{0ex}}\Rightarrow 88=2\times \frac{22}{7}\times r\phantom{\rule{0ex}{0ex}}\Rightarrow 14=r$

Hence, the correct answer is option (c).

#### Page No 273:

#### Question 21:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

Length of tape required to cover the edges of a semicircular disc of radius 10 cm is

(a) 62.8 cm

(b) 51.4 cm

(c) 31.4 cm

(d) 15.7 cm

#### Answer:

Given: Radius of the semicircular disc (*r*) = 10 cm

Perimeter of the semi-circle ircumference of semicircle + Diameter

Circumference of semi-circle$=\frac{2\mathrm{\pi r}}{2}$

$=\frac{2\times {\displaystyle \frac{22}{7}}\times 10}{2}\phantom{\rule{0ex}{0ex}}=\frac{220}{7}$

= 31.43 cm

Tape required to cover the edges of a semicircular disc of radius 10 cm = Perimeter of the semi-circle

= Circumference of semicircle + Diameter

= 31.43 cm + (2 × 10) cm

= 31.43 cm + 20 cm

= 51.43 cm

Hence, the correct answer is option (b).

#### Page No 273:

#### Question 22:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

Area of circular garden with diameter 8 m is

(a) 12.56 m^{2}

(b) 25.12 m^{2}

(c) 50.24 m^{2}

(d) 200.96 m^{2}

#### Answer:

Given: Diameter of circular garden (*d*) = 8 m

Radius of circular garden (*r*) = $\frac{8}{2}$ = 4 m

Area of circular garden = $\mathrm{\pi}{r}^{2}$

$=\frac{22}{7}\times 4\times 4\phantom{\rule{0ex}{0ex}}=50.285{\mathrm{cm}}^{2}$

Hence, the correct answer is option (c).

#### Page No 273:

#### Question 23:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

Area of a circle with diameter ‘*m*’ radius ‘*n*’ and circumference ‘*p*’ is

(a) 2ðœ‹*n*

(b) ðœ‹*m*^{2}

(c) ðœ‹*p*^{2}

(d) ðœ‹*n*^{2}

#### Answer:

Given: Diameter = *m, * radius = *n *and circumference = *p$\mathrm{Circumference}=2\mathrm{\pi}n\phantom{\rule{0ex}{0ex}}$$\begin{array}{rcl}\mathrm{Area}\mathrm{of}\mathrm{circle}& =& \mathrm{\pi}\times {\left(\mathrm{Radius}\right)}^{2}\\ & =& \mathrm{\pi}{n}^{\mathit{2}}\\ & & \end{array}$*

Hence, the correct answer is option (d).

#### Page No 273:

#### Question 24:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

A table top is semicircular in shape with diameter 2.8 m. Area of this table top is

(a) 3.08 m^{2}

(b) 6.16 m^{2}

(c) 12.32 m^{2}

(d) 24.64 m^{2}

#### Answer:

Given: Diameter of the semicircular table top (*d*) = 2.8 m

Radius of the semicircular table top (*r*) $=\frac{2.8}{2}\mathrm{m}$ = 1.4 m

Area of the semicircular table top = $\mathrm{\pi}{r}^{2}$

$=\frac{22}{7}\times 1.4\times 1.4\phantom{\rule{0ex}{0ex}}=6.16{\mathrm{m}}^{2}$

Hence, the correct answer is option (b).

#### Page No 273:

#### Question 25:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

If 1m^{2} =* x* mm^{2} , then the value of *x* is

(a) 1000

(b) 10000

(c) 100000

(d) 1000000

#### Answer:

1 m =* *1000 mm

$1\mathrm{m}=1000\mathrm{mm}\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{squaring}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}{\left(1\mathrm{m}\right)}^{2}={\left(1000\mathrm{mm}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 1{\mathrm{m}}^{2}={\left(1000\right)}^{2}{\mathrm{mm}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 1{\mathrm{m}}^{2}=1000000{\mathrm{mm}}^{2}$

Hence, the correct answer is option (d).

#### Page No 273:

#### Question 26:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

If* p *squares of each side 1mm makes a square of side 1cm, then* p* is equal to

(a) 10

(b) 100

(c) 1000

(d) 10000

#### Answer:

Area of one square having side of

Area of the square having side of

#### Page No 273:

#### Question 27:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

12 m^{2} is the area of

(a) a square with side 12 m

(b) 12 squares with side 1m each

(c) 3 squares with side 4 m each

(d) 4 squares with side 3 m each

#### Answer:

12 squares with side

each,Area of square

#### Page No 273:

#### Question 28:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

If each side of a rhombus is doubled, how much will its area increase?

(a) 1.5 times

(b) 2 times

(c) 3 times

(d) 4 times

#### Answer:

Let the side of rhombus be

New side of rhombus

Area of rhombus

New area of rhombus

∴ Increase in the area of rhombus

Thus, the increase in the area is 3 times the original area.

Hence, the correct answer is option (c).

#### Page No 273:

#### Question 29:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

If the sides of a parallelogram are increased to twice its original lengths, how much will the perimeter of the new parallelogram?

(a) 1.5 times

(b) 2 times

(c) 3 times

(d) 4 times

#### Answer:

Let *1* be the length and be the breadth of a parallelogram.

Then, perimeter

If both sides of the parallelogram are doubled, then the breadth and length become 2*l* and , respectively.

Now, perimeter

Therefore, the perimeter of the parallelogram will be increased by 2 times.

Hence, the correct answer is option (b).

#### Page No 273:

#### Question 30:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

If radius of a circle is increased to twice its original length, how much will the area of the circle increase?

(a) 1.4 times

(b) 2 times

(c) 3 times

(d) 4 times

#### Answer:

Let $r$ be the radii of a circle.

Then, area of circle

If radii are doubled, then radii become

.Now, area of circle

The area of the circle has increased by 4 times.

Hence, the correct answer is option (d).

#### Page No 273:

#### Question 31:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

What will be the area of the largest square that can be cut out of a circle of radius 10 cm?

(a) 100 cm^{2}

(b) 200 cm^{2}

(c) 300 cm^{2}

(d) 400 cm^{2}

#### Answer:

â€‹Given: Radius of circle = 10 cm

The diagonal of the square will be equal to the diameter of the circle.

Let the side of square be *x* units.

Now, in right-angled triangle

#### Page No 274:

#### Question 32:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

What is the radius of the largest circle that can be cut out of the rectangle measuring 10 cm in length and 8 cm in breadth?

(a) 4 cm

(b) 5 cm

(c) 8 cm

(d) 10 cm

#### Answer:

Given: Length of rectangle (*l*) = 10 cm

Breadth of rectangle (*b*) = 8 cm

It is clear that the largest circle will have diameter equals the smaller side that is $8\phantom{\rule{0ex}{0ex}}\text{cm}$.

Thus, $\mathrm{Radius}=\frac{\mathrm{Diameter}}{2}=\frac{8}{2}=4\mathrm{cm}$

Hence, the correct answer is option (a).

#### Page No 274:

#### Question 33:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

The perimeter of the figure ABCDEFGHIJ is

(a) 60 cm

(b) 30 cm

(c) 40 cm

(d) 50 cm

#### Answer:

$$

$=DE+40$

#### Page No 274:

#### Question 34:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

The circumference of a circle whose area is 81ðœ‹*r*^{2}, is

(a) 9ðœ‹*r*

(b) 18ðœ‹*r*

(c) 3ðœ‹*r*

(d) 81ðœ‹*r*

#### Answer:

Given: Area of the circle *r*^{2}

Let the area of the circle = *R*

The area of the circle = ðœ‹*R*^{2}

$\Rightarrow 81\mathrm{\pi}{r}^{2}=\mathrm{\pi}{R}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 81{r}^{\mathit{2}}\mathit{=}{R}^{\mathit{2}}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}9r\mathit{=}R$

The circumference of the circle is

#### Page No 274:

#### Question 35:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

The area of a square is 100 cm^{2}. The circumference (in cm) of the largest circle cut of it is

(a) 5 ðœ‹

(b) 10 ðœ‹

(c) 15 ðœ‹

(d) 20 ðœ‹

#### Answer:

Given: The area of a square = 100 cm^{2}

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

∴ Area of square = *x*^{2}

⇒ 100 = *x*^{2}

⇒ 10 cm = *x*

Let* r* be the radius of the circle.

Diameter of the circle = Side of the square

⇒ 2*r* = 10

⇒ *r* = 5 cm

Circumference of the circle = 2π*r
= *2π(5)

= 10π

Hence, the correct answer is option (b).

#### Page No 275:

#### Question 36:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

If the radius of a circle is tripled, the area becomes

(a) 9 times

(b) 3 times

(c) 6 times

(d) 30 times

#### Answer:

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

The area of a circle =

So, if the radius is tripled, the new radius = 3*r *units

The new area = π(3*r*)^{2} sq. units

= π (9*r*^{2}) sq. units

= 9(π*r*^{2}) sq. units

Thus, if the radius of a circle is tripled, the area becomes nine times.

Hence, the correct answer is option (a).

#### Page No 275:

#### Question 37:

**In the given Question, there are four options, out of which one is correct. Choose the correct one.**

The area of a semicircle of radius 4*r* is

(a) 8ðœ‹*r*^{2}

(b) 4ðœ‹*r*^{2}

(c) 12ðœ‹*r*^{2}

(d) 2ðœ‹*r*^{2}

#### Answer:

Given: Radius (R) = 4*r*

Area of circle = π*R*^{2}

∴ Area of semicircle = $\frac{\pi {R}^{2}}{2}$

$=\frac{\pi {\left(4r\right)}^{2}}{2}\phantom{\rule{0ex}{0ex}}=\frac{16\pi {r}^{2}}{2}\phantom{\rule{0ex}{0ex}}=8\pi {r}^{2}$

Hence, the correct answer is option (a).

#### Page No 275:

#### Question 38:

**Fill in the blank to make the statement true.**

Perimeter of a regular polygon = length of one side × ___________.

#### Answer:

Perimeter of a regular polygon $=$ length of one side ** Number of sides**.

#### Page No 275:

#### Question 39:

**Fill in the blank to make the statement true.**

If a wire in the shape of a square is rebent into a rectangle, then the __________ of both shapes remain same, but____________ may varry.

#### Answer:

If a wire in the shape of a square is rebent into a rectangle, then the ** Perimeter **of both shapes remain same, but

**may vary.**

__area__#### Page No 275:

#### Question 40:

**Fill in the blank to make the statement true.**

Area of the square MNOP of in the given figure is 144 cm^{2}. Area of each triangle is_______________.

.

#### Answer:

Given: Area of the square MNOP = 144 cm^{2}

There are 8 triangles in the given squares MNOP

∴ Area of each triangle $=\frac{1}{8}\times \mathrm{Area}\mathrm{of}\mathrm{square}$

$=\frac{1}{8}\times 144{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}=18{\mathrm{cm}}^{2}$

Thus, area of each triangle is 18 cm^{2}.

#### Page No 275:

#### Question 41:

**Fill in the blank to make the statement true.**

In the given figure, area of parallelogram BCEF is___________ cm^{2} where ACDF is a rectangle.

#### Answer:

Given = AB = 3 cm, Cd = 5 cm and FD = 10 cm

Opposite sides of a rectangle are equal.

∴ AC = FD = 10 cm

⇒ AB = ED = 3 cm

Area of parallelogram BCEF = Area of rectangle ACDF − Area of triangle ABF − Area of triangle CDE

$=\left(10\times 5\right)-\left(\frac{1}{2}\times 5\times 3\right)-\left(\frac{1}{2}\times 5\times 3\right)\phantom{\rule{0ex}{0ex}}=\left(50-15\right){\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}=35{\mathrm{cm}}^{2}$

#### Page No 275:

#### Question 42:

**Fill in the blank to make the statement true.**

To find area, any side of a parallelogram can be chosen as _________________of the parallelogram.

#### Answer:

To find an area, any side of a parallelogram can be chosen as __ Base__ of the parallelogram.

#### Page No 275:

#### Question 43:

**Fill in the blank to make the statement true.**

Perpendicular dropped on the base of a parallelogram from the opposite vertex is known as the corresponding_______________ of the base.

#### Answer:

Perpendicular dropped on the base of a parallelogram from the opposite vertex is known as the corresponding __ Height__ of the base.

#### Page No 275:

#### Question 44:

**Fill in the blank to make the statement true.**

The distance around a circle is its_____________ .

#### Answer:

The distance around a circle is its __ Circumference__.

#### Page No 276:

#### Question 45:

**Fill in the blank to make the statement true.**

Ratio of the circumference of a circle to its diameter is denoted by symbol_______________ .

#### Answer:

Let the radius of the circle be *r* units and the diameter be *d* units.

∴ *d *= 2*r*

Circumference of the circle (*C*) = 2π*r *= π*d$\Rightarrow \mathrm{\pi}=\frac{C}{d}$*

Thus, the Ratio of the circumference of a circle to its diameter is denoted by symbol

__.__

**π**#### Page No 276:

#### Question 46:

**Fill in the blank to make the statement true.**

If area of a triangular piece of cardboard is 90 cm^{2}, then the length of altitude corresponding to 20 cm long base is _____________cm.

#### Answer:

Given: Area of a triangular piece of cardboard = 90 cm^{2}

Base = 20 cm

Now,

Area of triangle = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$

$\Rightarrow 90=\frac{1}{2}\times 20\times \mathrm{Height}\phantom{\rule{0ex}{0ex}}\Rightarrow 90=10\times \mathrm{Height}\phantom{\rule{0ex}{0ex}}\Rightarrow 9\mathrm{cm}=\mathrm{Height}\phantom{\rule{0ex}{0ex}}$

#### Page No 276:

#### Question 47:

**Fill in the blank to make the statement true.**

Value of ðœ‹ is _________ approximately.

#### Answer:

$\mathrm{\pi}=\frac{22}{7}=3.1428$

Value of ðœ‹ is ** 3.1428** approximately.

#### Page No 276:

#### Question 48:

**Fill in the blank to make the statement true.**

Circumference ‘C’ of a circle can be found by multiplying diameter ‘*d*’ with ______________.

#### Answer:

Let the radius of the circle be *r* units and the diameter be *d* units.

∴ *d *= 2*r*

Circumference of the circle (*C*) = 2π*r *= π*d*

Thus, Circumference ‘C’ of a circle can be found by multiplying diameter ‘*d*’ with **π d.**

#### Page No 276:

#### Question 49:

**Fill in the blank to make the statement true.**

Circumference ‘C’ of a circle is equal to 2 ðœ‹ × ____________.

#### Answer:

Circumference ‘C’ of a circle is equal to 2 ðœ‹ × **Radius ( r).**

#### Page No 276:

#### Question 50:

**Fill in the blank to make the statement true.**

1 m^{2} = ___________ cm^{2}.

#### Answer:

1 m = 100 cm

Squaring both sides,

(1 m)^{2} = (100 cm)^{2}

⇒1 m^{2} = cm^{2}

Thus, 1 m^{2} = __ 10000 __cm

^{2}

#### Page No 276:

#### Question 51:

**Fill in the blank to make the statement true.**

1 cm^{2} = ___________ mm^{2}.

#### Answer:

1 cm = 10 mm

Squaring both sides

(1 cm)^{2} = (10 mm)^{2}

⇒ 1 cm^{2} = 100 mm^{2}

Thus, 1 cm^{2} =** 100 **mm

^{2}.

#### Page No 276:

#### Question 52:

**Fill in the blank to make the statement true.**

1 hectare =___________ m^{2}.

#### Answer:

1 hectare = __ 10000 __m

^{2}

#### Page No 276:

#### Question 53:

**Fill in the blank to make the statement true.**

Area of a triangle = $\frac{1}{2}$base × __________.

#### Answer:

Area of a triangle = $\frac{1}{2}$base ×** height.**

#### Page No 276:

#### Question 54:

**Fill in the blank to make the statement true.**

1 km^{2} = ___________ m^{2}.

#### Answer:

1 km = 1000 m

Squaring both sides

(1 km)^{2} = (1000 m)^{2}

⇒ 1 km^{2} = 1000000 m^{2}

Thus, 1 km^{2} = ** 1000000 **m

^{2}.

#### Page No 276:

#### Question 55:

**Fill in the blank to make the statement true.**

Area of a square of side 6 m is equal to the area of squares of___________ each side 1 cm.

#### Answer:

Let the number of squares having each side of 1 cm be *x*.

Side of bihher square = 6 m = 6 × 100 cm = 600 cm [∴ 1 m = 100 cm]

According to question,

${\left(600\right)}^{2}=x\times {\left(1\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 360000=x$

Thus, the area of a square of side 6 m is equal to the area of squares of ** 360000** each side 1 cm.

#### Page No 276:

#### Question 56:

**Fill in the blank to make the statement true.**

10 cm^{2} = _________m^{2}.

#### Answer:

$1\mathrm{cm}=\frac{1}{100}\mathrm{m}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Squaring}\mathrm{both}\mathrm{sides}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(1\mathrm{cm}\right)}^{2}={\left(\frac{1}{100}\mathrm{m}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 1{\mathrm{cm}}^{2}=\frac{1}{10000}{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 10{\mathrm{cm}}^{2}=\frac{10}{10000}{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 10{\mathrm{cm}}^{2}=\frac{1}{1000}{\mathrm{m}}^{2}$

#### Page No 276:

#### Question 57:

**State whether the statement is True or False.**

In the given figure, perimeter of (ii) is greater than that of (i), but its area is smaller than that of (i).

#### Answer:

Perimeter of the figure (ii) is greater than that of figure (i) because some part is being cut and that part has created a dip in the figure. So the perimeter is increased because now we have to add the height of the dip along with the perimeter of the figure (i).

Thus, the given statement is True.

#### Page No 277:

#### Question 58:

**State whether the statement is True or False.**

In the given figure,

(a) area of (i) is the same as the area of (ii).

(b) Perimeter of (ii) is the same as (i).

(c) If (ii) is divided into squares of unit length, then its area is 13 unit squares.

(d) Perimeter of (ii) is 18 units.

#### Answer:

In the given figure, the numbers of blocks are the same.

Therefore, the area of both figures is the same.

(b) Perimeter of (ii) is the same as (i).

Two new sides are added in figure (ii). Hence, the perimeter of figure (ii) is greater than figure (i).

Thus, the statement is False.

(c) If (ii) is divided into squares of unit length, then its area is

Area of 1 squares $=1\times 1=1$ unit squares

Number of squares = 12

Area of 12 squares = 12 × 1 = 12 unit squares

Thus, the statement is False.

(d) Perimeter of (ii) is 18 units.

Perimeter of the figure = 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm

= 18 cm

Thus, the statement is True.

#### Page No 277:

#### Question 59:

**State whether the statement is True or False.**

If perimeter of two parallelograms are equal, then their areas are also equal.

#### Answer:

Their corresponding height and sides may be different.

So, the area cannot be equal.

Thus, the statement is False.

#### Page No 277:

#### Question 60:

**State whether the statement is True or False.**

All congruent triangles are equal in area.

#### Answer:

Congruent triangles have equal size and shape.

Therefore, their areas are also equal.

Thus, the statement is True.

#### Page No 277:

#### Question 61:

**State whether the statement is True or False.**

All parallelograms having equal areas have same perimeters. Observe all the four triangles FAB, EAB, DAB and CAB as shown in the given figure:

#### Answer:

Observing all four triangles FAB, EAB, DAB, and CAB it can be seen that it may be possible their height and base can be different, in this case not all parallelograms have equal areas.

The statement is False.

#### Page No 278:

#### Question 62:

**State whether the statement is True or False.**

All triangles have the same base and the same altitude.

#### Answer:

It is clear from the figure that all triangles have the same base AB and all the vertices lay on the same line, so the distance between vertex and base of the triangle are equal.

Thus, the statement is True.

#### Page No 278:

#### Question 63:

**State whether the statement is True or False.**

All triangles are congruent.

#### Answer:

In the given figure, all triangles have only the baseline is equal and no such other lines are equal to each other.

Thus, the statement is False

#### Page No 278:

#### Question 64:

**State whether the statement is True or False.**

All triangles are equal in area.

#### Answer:

Because the triangles between, the same parallel lines and the same base are equal in the area.

Thus, the statement is True.

#### Page No 278:

#### Question 65:

**State whether the statement is True or False.**

All triangles may not have the same perimeter.

#### Answer:

In the given figure, it is clear that not all triangles may have the same perimeter.

Thus, the statement is True.

#### Page No 278:

#### Question 66:

**State whether the statement is True or False.**

In the given figure ratio of the area of triangle ABC to the area of triangle ACD is the same as the ratio of base BC of triangle ABC to the base CD of triangle ACD.

#### Answer:

Aâ€‹rea of a triangle$=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$

Area of triangle ABC$=\frac{1}{2}\times \mathrm{BC}\times \mathrm{AC}$

Area of triangle ACD$=\frac{1}{2}\times \mathrm{CD}\times \mathrm{AC}$

The ratio of the area of triangle ABC to the area of triangle ACD$=\frac{\frac{1}{2}\times \mathrm{BC}\times \mathrm{AC}}{\frac{1}{2}\times \mathrm{CD}\times \mathrm{AC}}$

$=\frac{\mathrm{BC}}{\mathrm{CD}}$

Thus, the statement is True.

#### Page No 278:

#### Question 67:

**State whether the statement is True or False.**

Triangles having the same base have equal area.

#### Answer:

Aâ€‹rea of a triangle$=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$

From the formula, it is clear that the area of triangles depends on base and height not only on the base.

Thus, the statement is False.

#### Page No 278:

#### Question 68:

**State whether the statement is True or False.**

Ratio of circumference of a circle to its radius is always 2ðœ‹ : I.

#### Answer:

Let the radius of the circle be* r*.

∴ Circumference of the circle (*C*) = 2π*r*

Ratio of circumference of a circle to its radius = 2π*r *: *r = *2π : 1

Thus, the statement is True.

#### Page No 278:

#### Question 69:

**State whether the statement is True or False.**

5 hectare = 500 m^{2}

#### Answer:

1 hectare = 10000 m^{2}

⇒ 5 hectare = 5 × 10000 m^{2}

Thus, the statement is False.

#### Page No 278:

#### Question 70:

**State whether the statement is True or False.**

An increase in perimeter of a figure always increases the area of the figure.

#### Answer:

The perimeter is the sum of all sides of closed shapes or polygons while the area is just bounded space insides. So it is not necessary that an increase in the perimeter of a figure always increases the area of the figure.

Thus, the statement is False.

#### Page No 278:

#### Question 71:

**State whether the statement is True or False.**

Two figures can have the same area but different perimeters.

#### Answer:

The perimeter is the sum of all sides of closed shapes or polygons while the area is just bounded space inside. Therefore, it can be different.

Thus, the statement is True.

#### Page No 278:

#### Question 72:

**State whether the statement is True or False.**

Out of two figures if one has larger area, then its perimeter need not to be larger than the other figure.

#### Answer:

The perimeter is the sum of all sides of closed shapes or polygons while the area is just bounded space insides. Therefore, Out of two figures if one has a larger area, then its perimeter need not be larger than the other figure.

Thus, the statement is True.

#### Page No 278:

#### Question 73:

A hedge boundary needs to be planted around a rectangular lawn of size 72 m × 18 m. If 3 shrubs can be planted in a metre of hedge, how many shrubs will be planted in all?

#### Answer:

The length of the rectangular lawn $=72$ metres

The breadth of the rectangular lawn $=18$ metres

Perimeter of rectangle $=2$ (Length $+$ Breadth )

Perimeter of he rectangular lawn $=2(72+18)=2(90)=180$ metres

If three shrubs can be planted in a meter of the hedge.

Then,

Number of shrubs = 3 × 180 = 540

Thus, in all 540 shrubs can be planted.

#### Page No 279:

#### Question 74:

People of Khejadli village take good care of plants, trees and animals. They say that plants and animals can survive without us, but we can not survive without them. Inspired by her elders Amrita marked some land for her pets (camel and ox ) and plants. Find the ratio of the areas kept for animals and plants to the living area.

#### Answer:

Area of covered land by plants

Area of rectangular land

Area of covered land by camel

Hence, the region of land is covered by an ox in a circular area.

Hence, diameter,

Radius, *r*

Region of land covered by ox = πr^{2}

$=\frac{22}{7}\times 1.4\times 1.4{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}=6.16{\mathrm{m}}^{2}$

Total area covered by Camels, plants and ox

So, remaining land

Ratio of areas kept for animals and plants to the living areas =

We should save our environment and balance the environment.

#### Page No 279:

#### Question 75:

The perimeter of a rectangle is 40 m. Its length is four metres less than five times its breadth. Find the area of the rectangle.

#### Answer:

Let the breadth of the rectangle be $x$, then the length of the rectangle = 5*x*−4.

Perimeter of rectangle

(*l *+ *b*) $$

Breadth $4$ metres

Hence, length

So, the area of rectangle

#### Page No 279:

#### Question 76:

A wall of a room is of dimensions 5 m × 4 m. It has a window of dimensions 1.5 m × 1m and a door of dimensions 2.25 m × 1m. Find the area of the wall which is to be painted.

#### Answer:

Length of the room

Breadth of the room

Area of the room

Also,Length of the window

Breadth of the window

Area of the window

Now,

Length of the door

Breadth of the door

Area of the door

Now, area of the wall to be painted $=$

= 20 − (1.5+2.25)

#### Page No 279:

#### Question 77:

Rectangle MNOP is made up of four congruent rectangles . If the area of one of the rectangles is 8 m^{2} and breadth is 2 m, then find the perimeter of MNOP.

#### Answer:

Area = 8 m^{2} and Breadth

Now, perimeter of rectangle MNOP = MN + NC + CD + DO + PO + PF + FA + MA

= 4 + 2 + 4 + 2 + 4 + 2 + 4 + 2 m

= 24 m

Thus, the perimeter of MNOP is 24 metres.

#### Page No 280:

#### Question 78:

In the given figure, area of âˆ† AFB is equal to the area of parallelogram ABCD. If altitude EF is 16 cm long, find the altitude of the parallelogram to the base AB of length 10 cm. What is the area of âˆ†DAO, where O is the mid point of DC?

#### Answer:

Given, Area of triangle $=$ Area of parallelogram ABCD

$\frac{1}{2}\times \mathrm{AB}\times \mathrm{EF}=\mathrm{CD}\times \mathrm{EG}$

Let

$\frac{1}{2}\times 10\times 16=10\times h\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\times 16=h\phantom{\rule{0ex}{0ex}}\Rightarrow 8\mathrm{cm}=h$

In triangle DAO,

DO

Area of triangle

#### Page No 281:

#### Question 79:

Ratio of the area of âˆ† WXY to the area of âˆ† WZY is 3 : 4. If the area of âˆ† WXZ is 56 cm^{2} and WY = 8 cm, find the lengths of XY and YZ.

#### Answer:

Area of triangle

$\Rightarrow \frac{1}{2}\times \mathrm{WY}\times \mathrm{XZ}=56\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\times 8\times \mathrm{XZ}=56\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{XZ}=14\mathrm{cm}$

Area of triangle WXY: Area of triangle WZY = 3 : 4

$\Rightarrow \frac{{\displaystyle \frac{1}{2}}\times \mathrm{WY}\times \mathrm{XY}}{\frac{1}{2}\times \mathrm{YZ}\times \mathrm{WY}}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{XY}}{\mathrm{YZ}}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{XY}}{14-\mathrm{XY}}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow 4\mathrm{XY}=42-3\mathrm{XY}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{XY}=6\mathrm{cm}$

$\text{m}$

Thus,

and#### Page No 281:

#### Question 80:

Rani bought a new field that is next to one she already owns. This field is in the shape of a square of side 70 m. She makes a semi circular lawn of maximum area in this field.

(i) Find the perimeter of the lawn.

(ii) Find the area of the square field excluding the lawn.

#### Answer:

Given: Side of a square field $70\phantom{\rule{0ex}{0ex}}\text{m}$

∴ The diameter of semi-circular lawn = Side of the square field

Radius of semi circular lawn

Perimeter of semi circular lawn

Thus, the perimeter of the lawn

(ii) Area of square

∴ Required area

#### Page No 281:

#### Question 81:

In the given figure, find the area of parallelogram ABCD if the area of shaded triangle is 9 cm^{2}.

#### Answer:

Given: Area of the shaded portion

Base of triangle

Area of triangle =

Area of parallelogram

#### Page No 282:

#### Question 82:

Pizza factory has come out with two kinds of pizzas. A square pizza of side 45 cm costs â‚¹ 150 and a circular pizza of diameter 50 cm costs â‚¹160 . Which pizza is a better deal?

#### Answer:

Given: Side of square pizza

Diameter of circular pizza

Radius

Area of square pizza

Area of circular pizza

=1964.28 cm^{2}

∴ Price of $=\frac{2053}{150}=\u20b913.5$

square pizzaPrice of $1$ circular pizza

Thus, the best deal is circular pizza.

#### Page No 282:

#### Question 83:

Three squares are attached to each other as shown in the given figure. Each square is attached at the mid point of the side of the square to its right. Find the perimeter of the complete figure.

#### Answer:

The perimeter of complete figure

#### Page No 283:

#### Question 84:

In the given figure, ABCD is a square with AB = 15 cm. Find the area of the square BDFE.

#### Answer:

Given:

Diagonal of square

BDEF∴ Area of the square BDEF$=15{\left(\sqrt{2}\right)}^{2}$${\mathrm{}}^{}$

$=225\times 2{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}=450{\mathrm{cm}}^{2}$

Thus, the area of the square BDFE is 450 cm^{2}.

#### Page No 283:

#### Question 85:

In the given triangles of in the given figure, perimeter of âˆ†ABC = perimeter of âˆ†PQR. Find the area of âˆ†ABC.

#### Answer:

Given: Perimeter of triangle

Perimeter of triangleThe perimeter of the triangle

The perimeter of the triangle

#### Page No 283:

#### Question 86:

Altitudes MN and MO of parallelogram MGHK are 8 cm and 4 cm long respectively. One side GH is 6 cm long. Find the perimeter of MGHK.

#### Answer:

Given: MN = 8 cm, MO = 4 cm and GH = 6 cm

Area of parallelogram MGHK $=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$

$=\frac{1}{2}\times \mathrm{GH}\times \mathrm{MN}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times 6\times 8\phantom{\rule{0ex}{0ex}}=24{\mathrm{cm}}^{2}$

Also, area of parallelogram MGHK $=\frac{1}{2}\times \mathrm{HK}\times \mathrm{MO}$

$\Rightarrow 24=\frac{1}{2}\times \mathrm{HK}\times 4\phantom{\rule{0ex}{0ex}}\Rightarrow 24=2\mathrm{HK}\phantom{\rule{0ex}{0ex}}\Rightarrow 12=\mathrm{HK}$

Since opposite sides of a parallelogram are equal.

∴ MK = GH = 6 cm and MG = HK = 12 cm

∴ Perimeter of MGHK = MK + HK + GH + MG

= 6 + 12 + 6 + 12

= 36 cm

Thus, the perimeter of MGHK is 36 cm.

#### Page No 284:

#### Question 87:

In the given figure, area of âˆ†PQR is 20 cm^{2} and area of âˆ†PQS is 44 cm^{2}. Find the length RS, if PQ is perpendicular to QS and QR is 5 cm.

#### Answer:

Given: Area of âˆ†PQR = 20 cm^{2}, area of âˆ†PQS = 44 cm^{2}, QR = 5 cm and PQ is perpendicular to QS and QR.

Area of âˆ†PQR = 20 cm^{2}

$\Rightarrow \frac{1}{2}\times \mathrm{QR}\times \mathrm{QP}=20\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\times 5\times \mathrm{QP}=20\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{QP}=8$

Area of âˆ†PQS = 44 cm^{2}

$\Rightarrow \frac{1}{2}\times \mathrm{QS}\times \mathrm{QP}=44\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\times \mathrm{QS}\times 8=44\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{QS}=11$

Now, RS = QR − RS

⇒ RS = (11 − 6) cm

⇒ RS = 5 cm

Thus, length of RS = 6 cm.

#### Page No 284:

#### Question 88:

Area of an isosceles triangle is 48 cm^{2}. If the altitudes corresponding to the base of the triangle is 8 cm, find the perimeter of the triangle.

#### Answer:

Given: Area of an isosceles triangle = 48 cm^{2}

Altitudes corresponding to the base of the triangle is 8 cm.

Consider $\u25b3$ABC to be the isosceles triangle with area 48 cm^{2} such that AD = 8 cm is perpendicular to the base BC.

Now, area of $\u25b3$ABC = $\frac{1}{2}\times \mathrm{BC}\times \mathrm{AD}$

$\Rightarrow 48=\frac{1}{2}\times \mathrm{BC}\times 8\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=12{\mathrm{cm}}^{2}$

$\because \u25b3\mathrm{ADC}\cong \mathrm{ADB}\left(\mathrm{By}\mathrm{RHS}\mathrm{Congruency}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{DC}=\mathrm{BD}\left(\mathrm{By}\mathrm{CPCT}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{DC}=\mathrm{BD}=\frac{\mathrm{BD}}{2}=\frac{12}{2}=6\mathrm{cm}$

In $\u25b3$ADC, by Pythagoras Theorem

${\mathrm{AC}}^{2}={\mathrm{AD}}^{2}+{\mathrm{DC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}={6}^{2}+{8}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}=36+68\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}=100\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}={10}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=10\mathrm{cm}$

⇒ AB = AC = 10 cm

∴ Perimeter of $\u25b3$ABC = AB + BC + CA

= 10 + 12 + 10

= 32 cm

Thus, the perimeter of the triangle is 32 cm.

#### Page No 284:

#### Question 89:

Perimeter of a parallelogram shaped land is 96 m and its area is 270 square metres. If one of the sides of this parallelogram is 18 m, find the length of the other side. Also, find the lengths of altitudes *l* and *m* .

#### Answer:

ans

#### Page No 285:

#### Question 90:

Area of a triangle PQR right-angled at Q is 60 cm^{2} . If the smallest side is 8cm long, find the length of the other two sides.

#### Answer:

ans

#### Page No 285:

#### Question 91:

In the given figure a rectangle with perimeter 264 cm is divided into five congruent rectangles. Find the perimeter of one of the rectangles.

#### Answer:

ans

#### Page No 285:

#### Question 92:

Find the area of a square inscribed in a circle whose radius is 7 cm .

[**Hint:** Four right-angled triangles joined at right angles to form a square]

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#### Page No 285:

#### Question 93:

Find the area of the shaded portion in question 92.

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#### Page No 286:

#### Question 94:

**Find the area enclosed by the given figure :**

#### Answer:

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#### Page No 286:

#### Question 95:

**Find the area enclosed by the given figure :**

#### Answer:

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#### Page No 286:

#### Question 96:

**Find the area enclosed by the given figure :**

#### Answer:

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#### Page No 286:

#### Question 97:

**Find the area enclosed by the given figure :**

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#### Page No 287:

#### Question 98:

**Find the area of the shaded region:**

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#### Page No 287:

#### Question 99:

**Find the area of the shaded region:**

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#### Page No 287:

#### Question 100:

A circle with radius 16 cm is cut into four equal parts and rearranged to form another shape as shown in the given figure:

Does the perimeter change? If it does change, by how much does it increase or decrease?

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#### Page No 288:

#### Question 101:

A large square is made by arranging a small square surrounded by four congruent rectangles as shown in the given figure. If the perimeter of each of the rectangle is 16 cm, find the area of the large square.

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#### Page No 288:

#### Question 102:

ABCD is a parallelogram in which AE is perpendicular to CD. Also AC = 5 cm, DE = 4 cm, and the area of âˆ† AED = 6 cm^{2}. Find the perimeter and area of ABCD.

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#### Page No 289:

#### Question 103:

Ishika has designed a small oval race track for her remote control car. Her design is shown in the given figure. What is the total distance around the track? Round your answer to the nearest whole cm.

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#### Page No 290:

#### Question 104:

A table cover of dimensions 3 m 25 cm × 2 m 30 cm is spread on a table. If 30 cm of the table cover is hanging all around the table, find the area of the table cover which is hanging outside the top of the table. Also find the cost of polishing the table top at â‚¹ 16 per square metre.

#### Answer:

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#### Page No 290:

#### Question 105:

The dimensions of a plot are 200 m × 150 m. A builder builds 3 roads which are 3 m wide along the length on either side and one in the middle. On either side of the middle road he builds houses to sell. How much area did he get for building the houses?

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#### Page No 290:

#### Question 106:

A room is 4.5 m long and 4 m wide. The floor of the room is to be covered with tiles of size 15 cm by 10 cm. Find the cost of covering the floor with tiles at the rate of â‚¹ 4.50 per tile.

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#### Page No 290:

#### Question 107:

Find the total cost of wooden fencing around a circular garden of diameter 28 m, if 1m of fencing costs â‚¹ 300.

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#### Page No 290:

#### Question 108:

Priyanka took a wire and bent it to form a circle of radius 14 cm. Then she bent it into a rectangle with one side 24 cm long. What is the length of the wire? Which figure encloses more area, the circle or the rectangle?

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#### Page No 290:

#### Question 109:

How much distance, in metres, a wheel of 25 cm radius will cover if it rotates 350 times?

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#### Page No 291:

#### Question 110:

A circular pond is surrounded by a 2 m wide circular path. If outer circumference of circular path is 44 m, find the inner circumference of the circular path. Also find area of the path.

#### Answer:

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#### Page No 291:

#### Question 111:

A carpet of size 5 m × 2 m has 25 cm wide red border. The inner part of the carpet is blue in colour. Find the area of blue portion. What is the ratio of areas of red portion to blue portion?

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#### Page No 291:

#### Question 112:

Use the in the given figure showing the layout of a farm house:

(a) What is the area of land used to grow hay?

(b) It costs â‚¹ 91 per m^{2} to fertilise the vegetable garden. What is the total cost?

(c) A fence is to be enclosed around the house. The dimensions of the house are 18.7 m ×12.6 m. At least how many metres of fencing are needed?

(d) Each banana tree required 1.25 m^{2} of ground space. How many banana trees can there be in the orchard?

#### Answer:

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#### Page No 292:

#### Question 113:

Study the layout given below in the given figure and answer the questions:

(a) Write an expression for the total area covered by both the bedrooms and the kitchen.

(b) Write an expression to calculate the perimeter of the living room.

(c) If the cost of carpeting is â‚¹ 50/m^{2}, write an expression for calculating the total cost of carpeting both the bedrooms and the living room.

(d) If the cost of tiling is â‚¹ 30/m^{2}, write an expression for calculating the total cost of floor tiles used for the bathroom and kitchen floors.

(e) If the floor area of each bedroom is 35 m^{2}, then find *x*.

#### Answer:

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#### Page No 292:

#### Question 114:

A 10 m long and 4 m wide rectangular lawn is in front of a house. Along its three sides a 50 cm wide flower bed is there as shown in the given figure. Find the area of the remaining portion.

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#### Page No 293:

#### Question 115:

A school playground is divided by a 2 m wide path which is parallel to the width of the playground, and a 3 m wide path which is parallel to the length of the ground. If the length and width of the playground are 120 m and 80 m respectively, find the area of the remaining playground.

#### Answer:

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#### Page No 293:

#### Question 116:

In a park of dimensions 20 m × 15 m, there is a L shaped 1m wide flower bed as shown in the given figure. Find the total cost of manuring for the flower bed at the rate of Rs 45 per m^{2}.

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#### Page No 293:

#### Question 117:

Dimensions of a painting are 60 cm × 38 cm. Find the area of the wooden frame of width 6 cm around the painting as shown in the given figure.

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#### Page No 294:

#### Question 118:

A design is made up of four congruent right triangles as shown in the given figure. Find the area of the shaded portion.

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#### Page No 294:

#### Question 119:

A square tile of length 20 cm has four quarter circles at each corner as shown in the given figure (i). Find the area of shaded portion. Another tile with same dimensions has a circle in the centre of the tile in the given figure (ii). If the circle touches all the four sides of the square tile, find the area of the shaded portion. In which tile, area of shaded portion will be more? (Take ðœ‹ = 3.14)

#### Answer:

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#### Page No 294:

#### Question 120:

A rectangular field is 48 m long and 12 m wide. How many right triangular flower beds can be laid in this field, if sides including the right angle measure 2 m and 4 m, respectively?

#### Answer:

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#### Page No 295:

#### Question 121:

Ramesh grew wheat in a rectangular field that measured 32 metres long and 26 metres wide. This year he increased the area for wheat by increasing the length but not the width. He increased the area of the wheat field by 650 square metres. What is the length of the expanded wheat field?

#### Answer:

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#### Page No 295:

#### Question 122:

In the given figure, triangle AEC is right-angled at E, B is a point on EC, BD is the altitude of triangle ABC, AC = 25 cm, BC = 7 cm and AE = 15 cm. Find the area of triangle ABC and the length of DB.

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#### Page No 295:

#### Question 123:

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#### Page No 296:

#### Question 124:

Calculate the area of shaded region in the given figure, where all of the short line segments are at right angles to each other and 1 cm long.

#### Answer:

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#### Page No 296:

#### Question 125:

The plan and measurement for a house are given in the given figure. The house is surrounded by a path 1m wide.

Find the following:

(i) Cost of paving the path with bricks at rate of â‚¹ 120 per m^{2}.

(ii) Cost of wooden flooring inside the house except the bathroom at the cost of â‚¹ 1200 per m^{2}.

(iii) Area of Living Room.

#### Answer:

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#### Page No 297:

#### Question 126:

Architects design many types of buildings. They draw plans for houses, such as the plan shown in the given figure:

An architect wants to install a decorative moulding around the ceilings in all the rooms. The decorative moulding costs â‚¹ 500/metre.

(a) Find how much moulding will be needed for each room.

(i) family room (ii) living room (iii) dining room (iv) bedroom 1 (v) bedroom 2

(b) The carpet costs â‚¹ 200/m^{2}. Find the cost of carpeting each room.

(c) What is the total cost of moulding for all the five rooms.

#### Answer:

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#### Page No 297:

#### Question 127:

ABCD is a given rectangle with length as 80 cm and breadth as 60 cm. P, Q, R, S are the mid points of sides AB, BC, CD, DA respectively. A circular rangoli of radius 10 cm is drawn at the centre as shown in the given figure. Find the area of shaded portion.

#### Answer:

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#### Page No 298:

#### Question 128:

4 squares each of side 10 cm have been cut from each corner of a rectangular sheet of paper of size 100 cm × 80 cm. From the remaining piece of paper, an isosceles right triangle is removed whose equal sides are each of 10 cm length. Find the area of the remaining part of the paper.

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#### Page No 298:

#### Question 129:

A dinner plate is in the form of a circle. A circular region encloses a beautiful design as shown in the given figure. The inner circumference is 352 mm and outer is 396 mm. Find the width of circular design.

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#### Page No 298:

#### Question 130:

The moon is about 384000 km from earth and its path around the earth is nearly circular. Find the length of path described by moon in one complete revolution. (Take ðœ‹ = 3.14)

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#### Page No 299:

#### Question 131:

A photograph of Billiard/Snooker table has dimensions as $\frac{1}{10}$th of its actual size as shown in the given figure:

The portion excluding six holes each of diameter 0.5 cm needs to be polished at rate of â‚¹ 200 per m^{2}. Find the cost of polishing.

#### Answer:

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