Unit 11: Perimeter and Area

## Exercise C: (Multiple Choice Questions and Answers 1-37)

In the Questions $1$ to $37,$ there are four options, out of which one is correct. Choose the correct one.

# Question: 1

Observe the shapes $1,\text{\hspace{0.17em}}2,\text{\hspace{0.17em}}3$ and $4$ in the figures. Which of the following statements is not correct?

a.    Shapes $1,\text{\hspace{0.17em}}3$ 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 $1,\text{\hspace{0.17em}}2$ and $4$ have the same area.

d.   Shapes $1,\text{\hspace{0.17em}}3$ and $4$ have the same perimeter.

## Solution

(a)

Shape $1$:

Area $=18×1=18$ sq. units

Shape $2$:

Area $=18×1=18$ sq. units

Shape $3$:

Area $=16×1=16$ sq. units

Shape $4$:

Area $=18×1=18$ sq. units

So option (a) is incorrect.

# Question: 2

A rectangular piece of dimensions  was cut from a rectangular sheet of paper of dimensions  (Fig. 9.14).

Area of remaining sheet of paper is

Fig. 9.14

a.

b.

c.

d.

## Solution

(c)

Given, length and breadth of the bigger rectangle as  and $5$ cm.

Area of bigger rectangle $=6\text{\hspace{0.17em}}\text{cm}×5\text{\hspace{0.17em}}\text{cm}=30\text{\hspace{0.17em}}{\text{cm}}^{2}$

Also given, length and breadth of the smaller rectangle as  and .

Area of smaller rectangle

Area of remaining sheet $=$ Area of bigger rectangle $–$ Area of smaller rectangle

# Question: 3

$36$ 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

## Solution

(b)

Area of rectangle formed $=36$ square units

We can write,             $36=6×6$

$=2×3×2×3$

$={2}^{2}×{3}^{2}$

$=4×9$

The possible pairs of sides can be:

1)              Length = 1, Breadth = 36

2)              Length = 2, Breadth = 18

3)              Length = 4, Breadth = 9

4)              Length = 12, Breadth = 3

5)              Length = 6, Breadth = 6

Now, for each pair of sides we will find the perimeter:

1)           Perimeter $=2\left(l+b\right)$

$\begin{array}{l}=2\left(1+36\right)\\ =2×37\\ =74\end{array}$

2)           Perimeter $=2\left(l+b\right)$

$\begin{array}{l}=2\left(2+18\right)\\ =2×20\\ =40\end{array}$

3)           Perimeter   $=2\left(l+b\right)$

$\begin{array}{l}=2\left(4+9\right)\\ =2×13\\ =26\end{array}$

4)           Perimeter $=2\left(l+b\right)$

$\begin{array}{l}=2\left(12+3\right)\\ =2×15\\ =30\end{array}$

5)           Perimeter $=2\left(l+b\right)$

$\begin{array}{l}=2\left(6+6\right)\\ =2×12\\ =24\end{array}$

So, the rectangle with sides 6 units and 6 units will have the least perimeter which is 24 units.

# Question: 4

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

## Solution

(b)

Given, side of square

$\because$ The wire has same length.

$\therefore$ Perimeter of square $=$ circumference of circle

$4×\text{side}=2×\pi ×r$

$4×22=2×\frac{22}{7}×r$

$r=\frac{4×22×7}{2×22}=14\text{\hspace{0.17em}}\text{cm}$

# Exercise: 5

Area of the circle obtained in Question 4 is

a.

b.

c.    $616\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{cm}}^{2}$

d.

## Solution

(c)

Area of the circle $=\pi {r}^{2}$

So, the correct option is (c)

# Question: 6

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

a.    $21$ cm

b.   $10.5$ cm

c.    $14$ cm

d.   $7$ cm.

## Solution

(d)

Given, dimensions of rectangle

According to the question,

Area of rectangle $=$ Area of circle

$l×b=\pi {r}^{2}$

$14×11=\frac{22}{7}×{r}^{2}$

${r}^{2}=\frac{14×11×7}{22}$

${r}^{2}=49$

$r=\sqrt{49}=7\text{\hspace{0.17em}}\text{cm}$

Hence, the radius of circle is .

# Question: 7

Area of shaded portion in Fig. 9.15 is

a.

b.

c.

d.

Fig. 9.15

## Solution

(d)

From the given figure,

Length of rectangle $\left(l\right)=5\text{\hspace{0.17em}}\text{cm}$

Breadth of rectangle $\left(b\right)=3+1=4\text{\hspace{0.17em}}\text{cm}$

$\begin{array}{l}=\frac{1}{2}×5×4\\ =10\text{\hspace{0.17em}}{\text{cm}}^{2}\end{array}$

# Question: 8

Area of parallelogram ABCD (Fig. 9.16) is not equal to

a.    DE $×$ DC

b.   BE $×$ AD

c.    BF $×$ DC

d.   BE $×$ BC

Fig. 9.16

## Solution

(a)

We know that, Area of parallelogram
$=$ Base $×$ Corresponding Height

So, Area of parallelogram
$\text{ABCD}=\text{AD}×\text{BE}=\text{BC}×\text{BE}$            $\left[\because \text{AD}=\text{BC}\right]$

Or, Area of parallelogram $\text{ABCD}=\text{DC}×\text{BF}$

# Question: 9

Area of triangle MNO of Fig. 9.17 is

Fig. 9.17

a.    $\frac{1}{2}\text{MN}×\text{NO}$

b.   $\frac{1}{2}\text{NO}×\text{MO}$

c.    $\frac{1}{2}\text{MN}×\text{OQ}$

d.   $\frac{1}{2}\text{NO}×\text{OQ}$

## Solution

(d)

We know that,

Area of triangle
$=\frac{1}{2}×\text{Base}×\text{Height}$

$=\frac{1}{2}\text{NO}×\text{OQ}$

# Question:10

Ratio of area of $\Delta \text{MNO}$ to the area of parallelogram MNOP in the same figure 9.17 is

a.    $2:3$

b.   $1:1$

c.    $1:2$

d.   $2:1$

## Solution

(c)

Area of $\Delta \text{MNO}:$ Area of parallelogram MNOP

$\frac{1}{2}×\text{NO}:\text{MP}$

Hence, the required ratio is $1:2$

# Question: 11

Ratio of areas of $\Delta \text{MNO}$, $\Delta \text{MOP}$ and $\Delta \text{MPQ}$ in Fig. 9.18 is

a.    $2:1:3$

b.   $1:3:2$

c.    $2:3:1$

d.   $1:2:3$

Fig. 9.18

## Solution

(a)

From the given figure,

Area of $\Delta \text{MNO}=\frac{1}{2}×\text{NO}×\text{MO}$

$=\frac{1}{2}×4×5=10\text{\hspace{0.17em}}{\text{cm}}^{2}$

Hence, required ratio is $\begin{array}{l}=10:5:15\\ =2:1:3\end{array}$

# Question: 12

In Fig. 9.19, EFGH is a parallelogram, altitudes FK and FI are 8 cm and 4cm respectively. If $\text{EF}=10$ cm, then area of EFGH is

a.

b.

c.

d.

Fig. 9.19

## Solution

(c)

In parallelogram EFGH, EF $=$ HG $=\text{1}0\text{\hspace{0.17em}}\text{cm}$  [Given]

Area of parallelogram EFGH
$=$ Base $×$ Corresponding height

= HG $×$ IF

$=10×4=40\text{\hspace{0.17em}}{\text{cm}}^{2}$

# Question: 13

In reference to a circle the value of $\pi$ is equal to

a.    $\frac{\text{Area}}{\text{Circumference}}$

b.   $\frac{\text{Area}}{\text{Diameter}}$

c.    $\frac{\text{Circumference}}{\text{Diameter}}$

d.   $\frac{\text{Circumference}}{\text{Radius}}$

## Solution

(c)

We Know that,

Circumference of a circle $=2\pi r$ (r, the radius of circle)

$⇒\text{Circumference}=\pi ×\text{Diameter}$

$⇒\pi =\frac{\text{Circumference}}{\text{Diameter}}$

# Question: 14

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

## Solution

(a)

We know that,

Circumference of a circle $=2\pi r$

Circumference $=2×3.14×r$         $\left[\because \pi =3.14\right]$

Circumference $=3.14×d$             $\left[\because d=2r\right]$

So, circumference of circle is always more than three times of its diameter.

# Question: 15

Area of triangle PQR is  (Fig. 9.20). If altitude QN is , then its base PR is

a.

b.

c.

d.

Fig. 9.20

## Solution

(a)

Given, area of

We know that,

Area of triangle $=\frac{1}{2}×\text{Base}×\text{Height}$

$\therefore$ Area of $\Delta \text{QPR}=\frac{1}{2}×\text{PR}×\text{QN}$

$100=\frac{1}{2}×\text{PR}×10$

$\text{PR}=\frac{100×2}{10}=20\text{\hspace{0.17em}}\text{cm}$

# Question: 16

In Fig. 9.21, if   and  then QM is -

Fig. 9.21

a.

b.

c.

d.

## Solution

(c)

Given that,  and

Now, using Pythagoras theorem in right angled $\Delta$ PLR,

${\left(\text{Hypotenuse}\right)}^{2}={\left(\text{Perpendicular}\right)}^{2}+{\left(\text{Base}\right)}^{2}$

${\text{PR}}^{2}={\text{PL}}^{2}+{\text{LR}}^{2}$

${\text{LR}}^{2}={\text{PR}}^{2}-{\text{PL}}^{2}$

${\text{LR}}^{2}={\left(12\right)}^{2}-{\left(8\right)}^{2}$

${\text{LR}}^{2}=144-64=80$

$\text{LR}=\sqrt{80}=4\sqrt{5}\text{\hspace{0.17em}}\text{cm}$

$\text{LR}=\text{LQ}+\text{QR}$

$\text{LQ}=\text{LR}-\text{QR}=\left(4\sqrt{5}-6\right)\text{\hspace{0.17em}}\text{cm}$

Area of $\Delta \text{PLR,}$ ${\text{A}}_{1}=\frac{1}{2}×\text{LR}×\text{PL}$

$=\frac{1}{2}×\left(4\sqrt{5}\right)×8$

$=16\sqrt{5}\text{\hspace{0.17em}}{\text{cm}}^{2}$

${\text{A}}_{2}=\frac{1}{2}×\text{LQ}×\text{PL}$

$=\frac{1}{2}×\left(4\sqrt{5}-6\right)×8$

Area of $\Delta \text{PLR}=\text{Area}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\Delta \text{PLQ}+\text{Area}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\Delta \text{PQR}$

$\frac{1}{2}×\text{PR}×\text{QM}=24$

$\frac{1}{2}×12×\text{QM}=24\text{\hspace{0.17em}}\text{cm}$

$\text{QM}=4\text{\hspace{0.17em}}\text{cm}$

# Question: 17

In Fig. 9.22 $\Delta \text{MNO}$ is a right-angled triangle. Its legs are  and  long. Length of perpendicular NP on the side MO is

Fig. 9.22

a.

b.

c.

d.

## Solution

(a)

Given, $\Delta \text{MNO}$ is a right-angled triangle.

According to Pythagoras theorem,

${\text{MO}}^{2}={\text{MN}}^{2}+{\text{NO}}^{2}$

$={6}^{2}+{8}^{2}=36+64$

$\begin{array}{l}{\text{MO}}^{2}=100\\ \text{MO}=\sqrt{100}\end{array}$

$\text{MO}=10\text{\hspace{0.17em}}\text{cm}$

$\frac{1}{2}×\text{MN}×\text{NO}=\frac{1}{2}×\text{MO}×\text{NP}$

$\frac{1}{2}×6×8=\frac{1}{2}×10×\text{NP}$

# Question: 18

Area of a right-angled triangle is  If its smallest side is $5\text{\hspace{0.17em}}\text{cm},$ then its hypotenuse is

a.

b.

c.

d.   $11\text{\hspace{0.17em}}\text{cm}$

## Solution

(b)

Given, area of right-angled triangle $=30\text{\hspace{0.17em}}{\text{cm}}^{2}$

and smallest side i.e., base

Area of right-angled triangle $=\frac{1}{2}×\text{Base}×\text{Height}$

$30=\frac{1}{2}×5×\text{Height}$

Now, according to Pythagoras theorem,

${\left(\text{Hypotenuse}\right)}^{2}={\left(\text{Perpendicular}\right)}^{2}+{\left(\text{Base}\right)}^{2}$

${\left(\text{Hypotenuse}\right)}^{2}={\left(12\right)}^{2}+{\left(5\right)}^{2}$

${\left(\text{Hypotenuse}\right)}^{2}=144+25$

${\left(\text{Hypotenuse}\right)}^{2}=169$

# Question: 19

Circumference of a circle of diameter  is

a.

b.

c.

d.

## Solution

(c)

Diameter of the given circle $=5$ cm

$\therefore$ Radius $=\frac{5}{2}\text{cm}$

Now, circumference of the circle

$\begin{array}{l}=2\pi r\\ =2×\frac{22}{7}×\frac{5}{2}\end{array}$

$=\frac{110}{7}=15.7\text{\hspace{0.17em}}\text{cm}$

# Question: 20

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

a.

b.

c.

d.

## Solution

(c)

Given, circumference of the circular disc $=2\pi r$

$88=2×\frac{22}{7}×r$

$r=\frac{88×7}{2×22}$

$r=14\text{\hspace{0.17em}}\text{cm}$

Hence, the radius of the disc is .

# Question: 21

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

a.

b.

c.

d.

## Solution

(b)

In order to find the length of tape required to cover the edges of a semi-circular disc, we have to find the perimeter of semi-circle.

Perimeter of semi-circle $=$ circumference of semi-circle $+$ Diameter

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

$=\frac{22}{7}×10=\frac{220}{7}=31.4\text{\hspace{0.17em}}\text{cm}$

Length of tape required $=31.4+\left(2×10\right)=51.4\text{\hspace{0.17em}}\text{cm}$

# Question: 22

Area of circular garden with diameter  is

a.

b.

c.

d.

## Solution

(c)

Given, diameter $=8\text{\hspace{0.17em}}\text{cm}$

So, radius $=\frac{8}{2}=4\text{\hspace{0.17em}}\text{cm}$

Area of circular garden $=\pi {r}^{2}=\frac{22}{7}×4×4$

$=50.24\text{\hspace{0.17em}}{\text{m}}^{2}$

# Question: 23

Area of a circle with diameter $‘m’$,  radius $‘n’$ and circumference $‘p’$ is -

a.    $2\text{\hspace{0.17em}}\pi n$

b.   $\pi {m}^{2}$

c.    $\pi {p}^{2}$

d.   $\pi {n}^{2}$

## Solution

(d)

Given, diameter $=m,$ radius $=n$ and circumference $=p$

We know that, area of circle

$\begin{array}{l}=\pi {\left(}^{r}\\ =\pi {n}^{2}\end{array}$

# Question: 24

A table top is semicircular in shape with diameter  Area of this table top is -

a.

b.

c.

d.

## Solution

(a)

Given, diameter

Now, radius $=\frac{2.8}{2}\text{m}=1.4\text{\hspace{0.17em}}\text{m}$

Area of table top $=$ Area of semi-circle $=\frac{\pi {r}^{2}}{2}$

# Question: 25

If  then the value of $x$ is

a.    $1000$

b.   $10000$

c.    $100000$

d.   $1000000$

## Solution

(d)

Given,

[  ]

$x=1000000$

# Question: 26

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

a.    $10$

b.   $100$

c.    $1000$

d.   $10000$

## Solution

(b)

Area of $1$ square of side $1$ mm

Area of square of side $1$ cm  $=$

Number of squares with side

According to question,

$p$ $×$ Area of square of side 1 mm

$p\text{\hspace{0.17em}}{\text{mm}}^{2}=1\text{\hspace{0.17em}}{\text{cm}}^{2}$

$p\text{\hspace{0.17em}}{\text{mm}}^{2}={\left(10\text{\hspace{0.17em}}\text{mm}\right)}^{2}$

$p=100$

# Question: 27

is the area of

a.    a square with side

b.   $12$ squares with side  each

c.    $3$ squares with side  each

d.   $4$ squares with side  each

## Solution

(b)

For option (a),

Area of a square with side $12\text{\hspace{0.17em}}\text{m}=12×12=144\text{\hspace{0.17em}}{\text{m}}^{2}$

From option (b),

Area of $12$ squares with side  each $=12×1×1=12\text{\hspace{0.17em}}{\text{m}}^{2}$

For option (c),

Area of $3$ squares with side $4\text{\hspace{0.17em}}\text{m}$ each

$=3×4×4=48\text{\hspace{0.17em}}{\text{m}}^{2}$

For option (d),

Area of $4$ squares with side  Area of square side

$=4×3×3=36\text{\hspace{0.17em}}{\text{m}}^{2}$

Hence, option (b) is correct.

# Question: 28

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

## Solution

(b)

Let b be the side and h be the height of a rhombus.

$\therefore$ Area of rhombus $=b×h$

If side is doubled, then side of new rhombus $=2b$

Now, area of new rhombus

Hence, the area of new rhombus will be increased by $2$ times.

# Question: 29

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

## Solution

(b)

Let the length and breadth of the parallelogram be $l$ and $b,$ respectively.

Then, perimeter

$=2\left(l+b\right)$ [ $\because$ perimeter of parallelogram

$=2×$ (length $+$ breadth)]

If both sides are increased twice, then new length and breadth will be $2l\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}2b$ respectively.

Now, new perimeter

times of original perimeter.

Hence, the perimeter of original parallelogram will increase by $2$ times.

# Question: 30

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

## Solution

(d)

Let $r$ be the radius of the circle.

$\therefore$ Area of circle $=\pi {r}^{2}$

If radius is increased to twice its original length, then new radius will be $2r.$ So, area of new circle $=\pi {\left(2r\right)}^{2}=4\pi {r}^{2}=4×\pi {r}^{2}=4$ times of original area

Hence, the area of new circle will be increased by $4$ times.

# Question: 31

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

a.

b.

c.

d.

## Solution

(b)

Given, radius of circle $=10\text{\hspace{0.17em}}\text{cm}$

The largest square that can be cut-out of a circle of radius $10\text{\hspace{0.17em}}\text{cm}$ will have its diagonal equals to the diameter of the circle.

Let the side of square be $x$ cm.

Area of the square is

Now, in right angled $\Delta \text{DAB},$

${\left(\text{BD}\right)}^{2}={\left(\text{AD}\right)}^{2}+{\left(\text{AB}\right)}^{2}$

${\left(20\right)}^{2}={x}^{2}+{x}^{2}$

$2{x}^{2}=400$

${x}^{2}=200$

${\left(\text{side}\right)}^{2}=200$

Hence, the area of the largest square is $200\text{\hspace{0.17em}}{\text{cm}}^{2}$.

# Question: 32

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

a.

b.

c.

d.

## Solution

(a)

The largest circle will have its diameter equal to the smaller side of rectangle i.e., $8\text{\hspace{0.17em}}\text{cm}$.

So, diameter $=8\text{\hspace{0.17em}}\text{cm}$

# Question: 33

The perimeter of the figure ABCDEFGHIJ is

a.

b.

c.

d.

Fig. 9.23

## Solution

(a)

We know that,

Perimeter of a closed shape $=$ sum of all its sides

So, $\text{AJ}+\text{JI}+\text{IH}+\text{HG}+\text{GF}+\text{FE}+\text{ED}+\text{CD}+\text{BC}+\text{AB}$

$=\left(\text{AJ}+\text{IH}+\text{GF}+\text{BC}\right)+3+5+2+20+4+6$

$=\text{DE}+40$

$=20+40=60\text{\hspace{0.17em}}\text{cm}$

# Question: 34

The circumference of a circle whose area is $81\pi {r}^{2},$ is

a.    $9\pi r$

b.   $18\pi r$

c.    $3\pi r$

d.   $81\pi r$

## Solution

(b)

Let the radius of circle be R

Area of circle $=\pi {\text{R}}^{2}$

$81\pi {r}^{2}=\pi {\text{R}}^{2}$

$\text{R}=\sqrt{81{r}^{2}}$

$\text{R}=9r$

Now, circumference of circle $=2\pi R=2\pi \left(9r\right)=18\pi r$

# Question: 35

The area of a square is  The circumference (in cm) of the largest circle cut of it is

a.    $5\pi$

b.   $10\pi$

c.    $15\pi$

d.   $20\pi$

## Solution

(b)

Let the side of square be a cm.

Given, area of square

Area of square $={a}^{2}$

${a}^{2}=100\text{\hspace{0.17em}}{\text{cm}}^{2}$

$a=\sqrt{100}$

$a=10\text{\hspace{0.17em}}\text{cm}$

Now, for the largest circle in the square, diameter of the circle must be equal to the side of square.

Diameter $=$ side of square

$2r=10\text{\hspace{0.17em}}\text{cm}$

$r=5\text{\hspace{0.17em}}\text{cm}$

Circumference of circle

# Question: 36

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

a.    $9$ times

b.   $3$ times

c.    $6$ times

d.   $30$ times

## Solution

(a)

Let $r$ be the radius of a circle.

$\therefore$ Area of circle $=\pi {r}^{2}$

If radius is tripled, then new radius will be $3r.$

$\therefore$ Area of new circle $=\pi {\left(3r\right)}^{2}=9{\pi }^{2}=9$ times area of original circle

Hence, the area of new circle increases by $9$ times.

# Question: 37

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

a.    $8\pi {\text{r}}^{2}$

b.   $4\pi {r}^{2}$

c.    $12\pi {r}^{2}$

d.   $2\pi {r}^{2}$

## Solution

(a)

Given, radius of semi-circle $=4r$

Area of semi-circle $=\frac{1}{2}×\pi {r}^{2}$

$=\frac{1}{2}×\pi ×{\left(4r\right)}^{2}=\frac{16}{2}\pi {r}^{2}=8\pi {r}^{2}$

In Questions 38 to 56, fill in the blanks to make the statements true.

# Question: 38

Perimeter of a regular polygon $=$ Length of one side $×$ ___________.

## Solution

Perimeter of regular polygon $=$ Length of one side $×$ Number of sides.

# Question: 39

If a wire in the shape of a square is rebent into a rectangle, then the ___________ of both shapes remain same, but ___________ may vary.

## Solution

If a wire in the shape of a square is rebent into a rectangle, then the perimeter of both shapes remain same, but area may vary.

# Question: 40

Area of the square MNOP of Fig. 9.24 is  Area of each triangle is ___________.

Fig. 9.24

## Solution

Given, area of square MNOP

Since, there are $8$ identical triangles in the given square MNOP.

Hence, area of each triangle $=\frac{1}{8}×$ Area of square MNOP

$=\frac{1}{8}×144=18\text{\hspace{0.17em}}\text{cm}$

# Question: 41

In Fig. 9.25, area of parallelogram BCEF is ___________ ${\text{cm}}^{2}$ , where ACDF is a rectangle.

Fig. 9.25

## Solution

Area of a parallelogram
BCEF $=$ Area of ACDF $-\text{\hspace{0.17em}}2×$ Area of $\Delta \text{CDE}$

$=10×5-2×\left(\frac{1}{2}×3×5\right)$

$=50-15=35\text{\hspace{0.17em}}{\text{cm}}^{2}$

# Question: 42

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

## Solution

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

# Question: 43

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

## Solution

Perpendicular dropped on the base of a parallelogram from the opposite vertex is known as the corresponding height/altitude of the base.

# Question: 44

The distance around a circle is its ___________.

## Solution

The distance around a circle is its circumference.

# Question: 45

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

## Solution

Circumference of a circle $=2\pi r$   r, the radius of circle

$\text{C}=\pi d$            since, $d=2r$

$\frac{C}{d}=\pi$

$\pi =\text{C}:d$

Here, $\pi$ is the answer.

# Question: 46

If area of a triangular piece of cardboard is  then the length of altitude corresponding to  long base is ___________ cm.

## Solution

We know that,

Area of triangle $=\frac{1}{2}×\text{Base}×\text{Height}$

$\therefore 90=\frac{1}{2}×20×h$

Length of altitude

# Question: 47

Value of $\pi$ is ___________ approximately.

## Solution

We know that, $\pi =\frac{22}{7}=3.14$

Thus, the approximate value of $\pi =3.14$

# Question: 48

Circumference $‘\text{C}’$ of a circle can be found by multiplying diameter $‘\text{d}’$ with ___________.

## Solution

$\because$ Circumference of a circle $=2\pi r$   $r$, the radius of circle

Since, diameter $\left(\text{d}\right)=2r$

So, $\text{C}=\pi ×d$

Hence, $\pi$ is the answer.

# Question: 49

Circumference $‘\text{C}’$ of a circle is equal to $2\pi ×$ ___________.

## Solution

Circumference of a circle $=2\pi ×r$   $r$, radius of circle

Hence, $r$ is the answer.

# Question: 50

___________ ${\text{cm}}^{2}.$

## Solution

We know that, $1\text{\hspace{0.17em}}\text{m}=100\text{\hspace{0.17em}}\text{cm}$

On squaring both sides, we get

# Question: 51

___________ ${\text{mm}}^{2}.$

## Solution

We know that,

On squaring both sides, we get

# Question: 52

$1$ hectare $=$ ___________ ${\text{m}}^{2}$.

# Question: 53

Area of a triangle $=\frac{1}{2}×\text{\hspace{0.17em}}\text{base}×$ ___________.

## Solution

Area of triangle $=\frac{1}{2}×\text{Base}×\text{Height}\text{.}$

# Question: 54

___________ ${\text{m}}^{2}.$

## Solution

We know that,

On squaring both sides, we get

# Question: 55

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

## Solution

Let number of squares having side

According to the question,

Area of square with side  $×$ Area of square with side

$\because$ area of square $=$ (side)2

$\therefore {\left(6\text{\hspace{0.17em}}\text{m}\right)}^{2}=x×{\left(1\text{\hspace{0.17em}}\text{cm}\right)}^{2}$           [ $\because 1\text{\hspace{0.17em}}\text{m}=100\text{\hspace{0.17em}}\text{cm}$ ]

$⇒360000\text{\hspace{0.17em}}{\text{cm}}^{2}=x\text{\hspace{0.17em}}{\text{cm}}^{2}$

$⇒x=360000$

# Question: 56

___________ ${\text{m}}^{2}.$

## Solution

We know that  m

$\therefore$

$10\text{\hspace{0.17em}}{\text{cm}}^{2}=\frac{1}{1000}{\text{m}}^{2}$

In Questions 57 to 72, state whether the statements are True or False.

# Question: 57

In Fig. 9.26, perimeter of (ii) is greater than that of (i), but its area is smaller than that of (i).

(i)                                    (ii)

Fig. 9.26

## Solution

True

We know that, perimeter is the sum of all sides of any polygon.

We also know that, area is the space covered in the polygon.

So, by observing the figures we can say that, perimeter of (ii) is greater than (i) and area is less than that of (i).

# Question: 58

In Fig. 9.27,

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

Fig. 9.27

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.

## Solution

1. True

Since the number of blocks in fig (i) and fig (ii) are same, area of both figures is same.

1. False

Because $2$ new sides are added in (ii).

So, the perimeter of (ii) is greater than (i).

1. False

$\because$ Area of $1$ square $=1×1=1$ unit squares

and number of squares $=12$

So, total area $=\text{12}×1=12$ unit squares

1. True

Perimeter is the sum of all sides.

When we add the sides of the figure, we get 18 units.

So, the perimeter is $18$ units.

# Question: 59

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

## Solution

False

Even if the perimeter of two parallelograms are equal, their corresponding sides and height may be different. So, their area cannot be equal.

# Question: 60

All congruent triangles are equal in area.

## Solution

True

Congruent triangles have equal shape and size. Hence, their areas are also equal.

# Question: 61

All parallelograms having equal areas have same perimeters.

## Solution

False

It is not necessary that all parallelograms having equal areas have same perimeters as their base and height may be different.

Observe the figure and answer the statements 62 to 65 as true or false:

# Question: 62

Observe all the four triangles FAB, EAB, DAB and CAB as shown in Fig. 9.28:

Fig. 9.28

All triangles have the same base and the same altitude.

## Solution

True

It is clear from the figure that all triangles have same base AB and all the vertices lie on the same line, so the distance between vertex and base of triangle (i.e. length of altitude) are equal.

# Question: 63

All triangles are congruent.

## Solution

False

It is clear from the Fig. 9.28,that the lengths of sides other than the base in all the triangles are not equal. Hence, not all triangles are congruent.

# Question: 64

All triangles are equal in area.

## Solution

True

This is true because the triangles on same base and between same parallel lines have equal in area.

# Question: 65

All triangles may not have the same perimeter.

## Solution

True

It is clear from the Fig. 9.28,that the lengths of sides other than the base in all the triangles are not equal. Hence, all triangles may not have the same perimeter.

# Question: 66

In Fig. 9.29 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.

Fig. 9.29

## Solution

True

Area of $\Delta \text{ABC}$ : Area of $\Delta \text{ACD}$ $=\frac{1}{2}×\text{BC}×\text{AC}:\frac{1}{2}×\text{CD}×\text{AC}$

$=\text{BC}:\text{CD}$

# Question: 67

Triangles having the same base have equal area.

## Solution

False

$\because$ Area of triangle $=\frac{1}{2}×\text{Base}×\text{Height}$

So, area of triangle does not only depend on base, it also depends on height.

Hence, if triangles have equal base and equal height, then only their areas are equal.

# Question: 68

Ratio of circumference of a circle to its radius is always $2\pi :1.$

## Solution

True

We know that, the circumference C of a circle of radius r, is $\text{C}=2\pi r$

$\because$ Circumference : Radius $=2\pi r:r=2\pi :1$

# Question: 69

$5$ hectare

## Solution

False

As we know that, $1$ hectare $=10000\text{\hspace{0.17em}}{\text{m}}^{2}$

So, $5$ hectare $=5×10000\text{\hspace{0.17em}}{\text{m}}^{2}=50000\text{\hspace{0.17em}}{\text{m}}^{2}$

# Question: 70

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

## Solution

False

Perimeter is the sum of sides of any polygon and area is space covered by a polygon.

# Question: 71

Two figures can have the same area but different perimeters.

## Solution

True

Perimeter of (i)

Perimeter of (ii)

Area of  figure (i)

Area of  figure (ii)

So, two figures can have the same area but different perimeters.

See the Q.58

# Question: 72

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

## Solution

True

Perimeter of figure (i) $=4×4=16$ cm

Perimeter of figure (ii) $=18$ cm

Though the first figure has the larger area, its perimeter is less than the second one.

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

# Question: 73

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

## Solution

Given, length of rectangular lawn  and breadth of rectangular lawn

$\because$ Perimeter of rectangle $=2×\left(l\text{+b}\right)$

$\therefore$ Perimeter of rectangular lawn $=2\left(72+18\right)=2\left(90\right)=180\text{\hspace{0.17em}}\text{m}$

No. of shrubs to be planted in 1 m of hedge $=3$

No. of shrubs to be planted in $180$ m of hedge

$=3×180=540$

# 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 cannot 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.

Fig. 9.30

## Solution

Area of total rectangular land $=15\text{\hspace{0.17em}}\text{m}×10\text{\hspace{0.17em}}\text{m}=150\text{\hspace{0.17em}}{\text{m}}^{\text{2}}$

Area of land covered by plants $=9\text{\hspace{0.17em}}\text{m}×1\text{\hspace{0.17em}}\text{m}=9\text{\hspace{0.17em}}{\text{m}}^{\text{2}}$

Area of land covered by camel $=5\text{\hspace{0.17em}}\text{m}×3\text{\hspace{0.17em}}\text{m}=15\text{\hspace{0.17em}}{\text{m}}^{\text{2}}$

$\therefore$ Region of land covered by ox is circular area.

So, diameter, $d=2.8\text{\hspace{0.17em}}\text{m}$

Radius $=\frac{d}{2}=\frac{2.8}{2}=1.4\text{\hspace{0.17em}}\text{cm}$

Area of land covered by ox

$=\pi {r}^{2}=\frac{22}{7}×1.4×1.4=6.16\text{\hspace{0.17em}}{\text{m}}^{2}$

Total area covered by plants, camel, and ox $=9+15+6.16\text{\hspace{0.17em}}{\text{m}}^{2}$

Living area $=\left(150-30.16\right)\text{\hspace{0.17em}}{\text{m}}^{2}=119.84\text{\hspace{0.17em}}{\text{m}}^{2}$

Ratio of areas kept for animals & plants to the living area

$=30.16:119.84=3016:11984=377:1498$

# Question: 75

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

## Solution

Let breadth of the rectangle $=x$

Then, length of the rectangle $=5x-4$

Perimeter of the rectangle $=2\left(l+b\right)$

$40=2\left(l+b\right)$

$40=2\left(5x-4+x\right)$

$40=2\left(6x-4\right)$

$12x-8=40$

$12x=40+8$

$12x=48$

$x=\frac{48}{12}=4$

So, breadth $=x=4\text{\hspace{0.17em}}\text{m}$ and length $=5x-4=5×4-4=16\text{\hspace{0.17em}}\text{m}$

Area of rectangle $=l×b=4×16=64\text{\hspace{0.17em}}{\text{m}}^{2}$

Hence, the area of rectangle is $64\text{\hspace{0.17em}}{\text{m}}^{2}$

# Question: 76

A wall of a room is of dimensions  It has a window of dimensions  and a door of dimensions  Find the area of the wall which is to be painted.

## Solution

Length of the room $=5\text{\hspace{0.17em}}\text{m}$

Breadth of the room $=4\text{\hspace{0.17em}}\text{m}$

$\therefore$ Area of the room $=l×b=5×4=20\text{\hspace{0.17em}}{\text{m}}^{\text{2}}$

Also, length of the window  and breadth of the window $=1\text{\hspace{0.17em}}\text{m}$         [given]

$\therefore$ Area of the window $=l×b=1.5×1=1.5\text{\hspace{0.17em}}{\text{m}}^{\text{2}}$

Now, length of the door  and

breadth of the door $=1\text{\hspace{0.17em}}\text{m}$

$\therefore$ Area of the door

Now, area of the wall to be painted $=$ Area of the room $-$ (Area of the window $+$ Area of the door) $=20-\left(1.5+2.25\right)=20-3.75=16.25\text{\hspace{0.17em}}{\text{m}}^{\text{2}}$

# Question: 77

Rectangle MNOP is made up of four congruent rectangles (Fig. 9.31).If the area of one of the rectangles is  and breadth is  then find the perimeter of MNOP.

Fig. 9.31

## Solution

Area of $1$ smaller rectangle

Breadth of smaller rectangle $=2\text{\hspace{0.17em}}\text{m}$

We know that, Area of the rectangle $=l×b$

$l×b=8$

$l×2=8$

$l=4\text{\hspace{0.17em}}\text{m}$

Perimeter of MNOP $=\text{MN}+\text{NC}+\text{CD}+\text{DO}+\text{PO}+\text{PF}+\text{FA}+\text{MA}$

$=4+2+4+2+4+2+4+2=24\text{\hspace{0.17em}}\text{m}$

So, the perimeter of MNOP is .

# Question: 78

In Fig. 9.32, area of $\Delta \text{AFB}$ is equal to the area of parallelogram ABCD. If altitude EF is  long, find the altitude of the parallelogram to the base AB of length  What is the area of $\Delta \text{DAO},$ where O is the mid point of DC?

Fig. 9.32

## Solution

Given,

Area of $\Delta \text{AFB}=$ Area of parallelogram ABCD

$\frac{1}{2}×\text{AB}×\text{EF}=\text{CD}×\left(\text{corresponding}\text{\hspace{0.17em}}\text{height}\right)$

Let corresponding height be $h$

Then,

$\frac{1}{2}×10×16=10×h$

$h=8$

In

Area of $\Delta \text{DAO}=\frac{1}{2}×\text{OD}×h$

$=\frac{1}{2}×5×8=20\text{\hspace{0.17em}}{\text{cm}}^{2}$

# Question: 79

Ratio of the area of $\Delta \text{WXY}$ to the area of $\Delta \text{WZY}$ is $3:4$ (Fig. 9.33). If the area of $\Delta \text{WXZ}$ is  and WY $=8$ cm, find the lengths of XY and YZ.

Fig. 9.33

## Solution

We know that, area of a triangle $=\frac{1}{2}×b×h$

Given, area of $\Delta \text{WXZ}=56\text{\hspace{0.17em}}{\text{cm}}^{2}$

$\frac{1}{2}×\text{WY}×\text{XZ}=56$

$\frac{1}{2}×8×\text{XZ}=56$

$\text{XZ}=14\text{\hspace{0.17em}}\text{cm}$

Area of $\Delta \text{WXY}:\text{Area}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{ΔWZY}=3:4$

$\frac{\text{Area}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{ΔWXY}}{\text{Area}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{ΔWZY}}=\frac{3}{4}$

$\frac{\frac{1}{2}×\text{WY}×\text{XY}}{\frac{1}{2}×\text{YZ}×\text{WY}}=\frac{3}{4}$

$\frac{\text{XY}}{\text{YZ}}=\frac{3}{4}$

$\frac{\text{XY}}{\text{XZ}-\text{XY}}=\frac{3}{4}$

$\frac{\text{XY}}{14-\text{XY}}=\frac{3}{4}$

$4\text{XY}=42-3\text{XY}$

$7\text{XY}=42$

$\text{XY}=6\text{\hspace{0.17em}}\text{cm}$

So, $\text{YZ}=\text{XZ}-\text{XY}=14-6$

$\text{YZ}=8\text{\hspace{0.17em}}\text{cm}$

Hence, $\text{XY}=6\text{\hspace{0.17em}}\text{cm}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{YZ}=8\text{\hspace{0.17em}}\text{cm}.$

# Question: 80

Rani bought a new field that is next to one she already owns (Fig. 9.34). This field is in the shape of a square of side  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.

Fig. 9.34

## Solution

(i)        Given, side of the square field $=70\text{\hspace{0.17em}}\text{m}$

Diameter of semi-circular lawn $=70\text{\hspace{0.17em}}\text{m}$

From the figure we understand that, Diameter of semi-circle $=$ side of square

Therefore, Radius $=\frac{70}{2}=35\text{\hspace{0.17em}}\text{cm}$

Perimeter of lawn $=\pi r+2r=\frac{22}{7}×35+2×35$

$=110+70=180\text{\hspace{0.17em}}\text{m}$

(ii)    Area of square $=$ side $×$ side

$=70×70=4900\text{\hspace{0.17em}}{\text{m}}^{2}$

Required area $=$ Area of square $-$ Area of semi-circle

$=4900-\frac{1}{2}×\pi ×{\left(35\right)}^{2}$

$=4900-\frac{1}{2}×\frac{22}{7}×35×35=4900-1925$

# Question: 81

In Fig. 9.35, find the area of parallelogram ABCD if the area of shaded triangle is

Fig. 9.35

## Solution

Given, area of shaded triangle $=9\text{\hspace{0.17em}}{\text{cm}}^{2}$

And base of the triangle $=3\text{\hspace{0.17em}}\text{cm}$

We know that, Area of triangle $=\frac{1}{2}×\text{base}×\text{height}$

Therefore, $9=\frac{1}{2}×3×h$

$\frac{18}{3}=h$

$h=6\text{\hspace{0.17em}}\text{cm}$

Area of parallelogram $=$ height $×$ base of parallelogram

$=\text{AE}×\text{BC}$

# Question: 82

Pizza factory has come out with two kinds of pizzas. A square pizza of side  costs $Rs\text{15}0$ and a circular pizza of diameter  costs $Rs\text{16}0$ (Fig. 9.36). Which pizza is a better deal?

Fig. 9.36

## Solution

Given, side of square pizza $=45\text{\hspace{0.17em}}\text{cm}$

Area of a square pizza $={\left(\text{side}\right)}^{2}={\left(45\right)}^{2}=2025\text{\hspace{0.17em}}{\text{cm}}^{2}$

Diameter of circular pizza $=50\text{\hspace{0.17em}}\text{cm}$

Radius $=\frac{50}{2}=25\text{\hspace{0.17em}}\text{cm}$

Now, area of the circular pizza $=\frac{22}{7}×25×25$

$=\frac{22}{7}×625$

$=\frac{13750}{7}=1964.28\text{\hspace{0.17em}}{\text{cm}}^{2}$

Price of $1$ sq. cm. pizza $=\frac{2025}{150}=\text{\hspace{0.17em}}Rs13.5$

Price of $1$ cm circular pizza $=\frac{1964.24}{160}=\text{\hspace{0.17em}}Rs12.27$

Since price of  of circular pizza is lesser, hence circular pizza is a better deal.

# Question: 83

Three squares are attached to each other as shown in Fig. 9.37. Each square is attached at the mid-point of the side of the square to its right. Find the perimeter of the complete figure.

Fig. 9.37

## Solution

Perimeter of the complete figure

Side of first square

Side of second square

Side of third square

$=6+6+6+3+1.5+1.5+1.5+3+3+1.5=33\text{\hspace{0.17em}}\text{m}$

# Question: 84

In Fig. 9.38, ABCD is a square with AB $=15$ cm. Find the area of the square BDFE.

Fig. 9.38

## Solution

Given, ABCD is a square and AB $=15\text{\hspace{0.17em}}\text{cm}$

Diagonal of square $\text{ABCD}=\sqrt{2a}=\sqrt{2}×15=15\sqrt{2}\text{\hspace{0.17em}}\text{cm}$  (a, the side of square ABCD)

Diagonal of square ABCD is the side of square BDEF

Area of the square $\text{BDFE}={\left(\text{side}\right)}^{2}$

$={\left(15\sqrt{2}\right)}^{2}=15×15×\sqrt{2}×\sqrt{2}$

$=225×2=450\text{\hspace{0.17em}}{\text{cm}}^{2}$

# Question: 85

In the given triangles of Fig. 9.39, perimeter of $\Delta \text{ABC}$$=$ perimeter of $\Delta \text{PQR}.$ Find the area of $\Delta \text{ABC}.$

Fig. 9.39

## Solution

Given, Perimeter of $\Delta \text{ABC}=\text{Perimeter}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\Delta \text{PQR}$

$\text{AB}+\text{BC}+\text{CA}=\text{PQ}+\text{QR}+\text{RP}$

$\text{AB}+5+13=6+10+14$

$\text{AB}+18=30$

Area of $\Delta \text{ABC}=\frac{1}{2}×\text{base}×\text{height}$

$=\frac{1}{2}×5×12=5×6=30\text{\hspace{0.17em}}{\text{cm}}^{2}$

# Question: 86

Altitudes MN and MO of parallelogram MGHK are $8$ cm and $4$ cm long respectively (Fig. 9.40). One side GH is $6$ cm long. Find the perimeter of MGHK.

Fig. 9.40

## Solution

Given, MGHK is a parallelogram where

We know that, area of parallelogram $=$ base $×$ height

Area of parallelogram MGHK, when base is GH

$=\text{GH}×\text{MN}$

$=6×8\text{\hspace{0.17em}}{\text{cm}}^{2}=48\text{\hspace{0.17em}}{\text{cm}}^{2}$

Area of parallelogram MGHK, when base is HK

$=\text{HK}×\text{MO}$

$48=\text{HK}×4$

$\text{HK}=12\text{\hspace{0.17em}}\text{cm}$

In parallelogram, opposite sides are equal

So,

Perimeter of parallelogram $\text{MGHK}=\left(6+6+12+12\right)\text{\hspace{0.17em}}\text{cm}=36\text{\hspace{0.17em}}\text{cm}$

# Question: 87

In Fig. 9.41, area of $\Delta \text{PQR}$ is  and area of $\Delta \text{PQS}$ is $44$ ${\text{cm}}^{2}.$ Find the length RS, if PQ is perpendicular to QS and QR is $5\text{\hspace{0.17em}}\text{cm}.$

Fig. 9.41

## Solution

Given, area of $\Delta \text{PQR}=20\text{\hspace{0.17em}}{\text{cm}}^{2}$

area of $\Delta \text{PQS}=44\text{\hspace{0.17em}}{\text{cm}}^{2}$

Area of triangle $=\frac{1}{2}×\text{base}×\text{height}$

Area of $\Delta \text{PQR}=\frac{1}{2}×\text{PQ}×\text{QR}$

$20=\frac{1}{2}×\text{PQ}×5$

$\frac{20×2}{5}=\text{PQ}$

$\text{PQ}=8\text{\hspace{0.17em}}\text{cm}$

Area of $\Delta \text{PQS}=\frac{1}{2}×\text{PQ}×\text{QS}$

$44=\frac{1}{2}×8×\text{QS}$

$\text{QS}=\frac{44×2}{8}=11\text{\hspace{0.17em}}\text{cm}$

Now, $\text{RS}=\text{QS}-\text{QR}=11-5=6\text{\hspace{0.17em}}\text{cm}$

# Question: 88

Area of an isosceles triangle is  If the altitudes corresponding to the base of the triangle is  find the perimeter of the triangle.

## Solution

Given, area of $\Delta \text{ABC}=48\text{\hspace{0.17em}}{\text{cm}}^{2}$ and altitude $=8\text{\hspace{0.17em}}\text{cm}$

$\Delta \text{ABC}$ is an isosceles triangle, where $\text{AB}=\text{AC}$

area of $\Delta \text{ABC}=\frac{1}{2}×\text{BC}×\text{AD}$

$48=\frac{1}{2}×\text{BC}×\text{AD}$

$\frac{1}{2}×\text{BC}×8=48$

$\text{BC}=\frac{48×2}{8}$

$\text{BC}=12\text{\hspace{0.17em}}\text{cm}$

In an isosceles triangle, $\text{BD}=\text{DC}=6\text{\hspace{0.17em}}\text{cm}$

Using Pythagoras theorem in right angled $\Delta \text{ADB},$

${\text{AB}}^{2}={\text{BD}}^{2}+{\text{AD}}^{2}$

${\text{AB}}^{2}={6}^{2}+{8}^{2}$

$=36+64$

${\text{AB}}^{2}=100$

$\text{AB}=10\text{\hspace{0.17em}}\text{cm}$

Perimeter of triangle $=\text{AB}+\text{AC}+\text{BC}=\text{AB}+\text{AB}+\text{BC}$

$=10+10+12=32\text{\hspace{0.17em}}\text{cm}$

# Question: 89

Perimeter of a parallelogram shaped land is  and its area is $270$ square meters. If one of the sides of this parallelogram is $18\text{m},$ find the length of the other side. Also, find the lengths of altitudes l and m (Fig. 9.42).

Fig. 9.42

## Solution

Perimeter of parallelogram ABCD $=96\text{\hspace{0.17em}}\text{m}$

Area of parallelogram ABCD $=270\text{\hspace{0.17em}}{\text{m}}^{2}$

Perimeter of parallelogram $\text{ABCD}=\text{AB}+\text{BC}+\text{CD}+\text{DA}$

$96=18+\text{AD}+18+\text{AD}$

$96=36+2\text{AD}$

$2\text{AD}=60$

$\text{AD}=30\text{\hspace{0.17em}}\text{cm}$

So, $\text{AD}=\text{BC}=30\text{\hspace{0.17em}}\text{cm}$

Now, area of parallelogram $\text{ABCD}=\text{Base}×\text{Corresponding}\text{\hspace{0.17em}}\text{Height}$

$270=\text{AB}×\text{DE}$

$270=18×\text{DE}$

$\frac{270}{18}=\text{DE}$

$\text{DE}=15\text{\hspace{0.17em}}\text{cm}$

Also, area of parallelogram $\text{ABCD}=\text{AD}×\text{BF}$

$270=30×l$

$l=\frac{270}{30}=9\text{\hspace{0.17em}}\text{m}$

Hence, altitudes of parallelogram $l=9\text{\hspace{0.17em}}\text{cm}$ and $\text{m}=15\text{\hspace{0.17em}}\text{cm}$

# Question: 90

Area of a triangle PQR right-angled at Q is  (Fig. 9.43). If the smallest side is $8\text{cm}$ long, find the length of the other two sides.

Fig. 9.43

## Solution

Area of

Area of $\Delta \text{PQR}=\frac{1}{2}×\text{PQ}×\text{QR}$

$60=\frac{1}{2}×8×\text{QR}$

$\text{QR}=\frac{60×2}{8}$

$\text{QR}=15\text{\hspace{0.17em}}\text{cm}$

In right angled  (using Pythagoras theorem)

${\text{PR}}^{2}={8}^{2}+{15}^{2}$

$=64+225$

${\text{PR}}^{2}=289=\sqrt{289}=17\text{\hspace{0.17em}}\text{cm}$

Hence, the length of two sides are  and .

# Question: 91

In Fig. 9.44 a rectangle with perimeter  is divided into five congruent rectangles. Find the perimeter of one of the rectangles.

Fig. 9.44

## Solution

Let $l\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}b$ be the length of each smaller rectangle

Given, perimeter of the bigger rectangle $=264\text{\hspace{0.17em}}\text{cm}$

According to the figure, $4l+5b=264\dots \left(\text{i}\right)$

& $2l=3b\dots \left(\text{ii}\right)$

Put the value of $3\text{b}$ from eq. (ii) in eq.(ii),

$2\left(2l\right)+5b=264$

$2×3b+5b=264$

$6b+5b=264$

$11b=264$

$b=\frac{264}{11}=24\text{\hspace{0.17em}}\text{cm}$

$l=\frac{3b}{2}=\frac{3×24}{2}=36\text{\hspace{0.17em}}\text{cm}$

Perimeter of a rectangle $=2\left(l+b\right)$

Hence, perimeter of smaller rectangle

$=2\left(36+24\right)=120\text{\hspace{0.17em}}\text{cm}$

# Question: 92

Find the area of a square inscribed in a circle whose radius is  (Fig. 9.45).

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

Fig. 9.45

## Solution

ABCD is a square                      (given)

According to the question,

Area of square $\text{ABCD}=4×\text{Area}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\Delta \text{AOB}$

$=4×\left(\frac{1}{2}×\text{AO}×\text{BO}\right)$

$=2×7×7=98\text{\hspace{0.17em}}{\text{cm}}^{2}$

Hence, area of inscribed square is $98\text{\hspace{0.17em}}{\text{cm}}^{2}$

# Question: 93

Find the area of the shaded portion in question $92.$

## Solution

Area of shaded option $=$ area of circle $-$ area of square

$=\pi {r}^{2}-98$

$=\frac{22}{7}×7×7-98=154-98=56\text{\hspace{0.17em}}{\text{cm}}^{2}$

In Questions 94 to 97 find the area enclosed by each of the following figures:

# Question: 94

Fig. 9.46

## Solution

Given shape contains a rectangle and a semi-circle.

Area of rectangle $\begin{array}{l}=l×b\\ =\left(10.2×1.5\right)\text{\hspace{0.17em}}{\text{cm}}^{2}\\ =15.3\text{\hspace{0.17em}}{\text{cm}}^{2}\end{array}$

Here, diameter of semi-circle $=\left(10.2-3.9\right)\text{\hspace{0.17em}}\text{cm}=6.3\text{\hspace{0.17em}}\text{cm}$

So, radius $=\frac{\text{diameter}}{2}=\frac{6.3}{2}=3.15\text{\hspace{0.17em}}\text{cm}$

Area of semi-circle $=\frac{1}{2}\pi {r}^{2}=\frac{22}{7}×\frac{1}{2}×3.15×3.15$

$=15.59\text{\hspace{0.17em}}{\text{cm}}^{2}$

Total area $=$ area of rectangle $+$ area of semi-circle

$=15.3+15.59=30.89\text{\hspace{0.17em}}{\text{cm}}^{2}$

# Question: 95

Fig. 9.47

## Solution

For rectangle, $l=13\text{\hspace{0.17em}}\text{cm}$ and $b=4\text{\hspace{0.17em}}\text{cm}$

Area of rectangle $=l×b=13×4=52\text{\hspace{0.17em}}{\text{cm}}^{2}$

For triangle, base $\left(b\right)=13-8=5\text{\hspace{0.17em}}\text{cm}$

And height $\left(h\right)=\left(16-4\right)\text{\hspace{0.17em}}\text{cm}=12\text{\hspace{0.17em}}\text{cm}$

Area of triangle $\begin{array}{l}=\frac{1}{2}×b×h\\ =\frac{1}{2}×5×12=30\text{\hspace{0.17em}}{\text{cm}}^{2}\end{array}$

Total area enclosed by the shape $=\left(52+30\right)\text{\hspace{0.17em}}{\text{cm}}^{2}=82\text{\hspace{0.17em}}{\text{cm}}^{2}$

# Question: 96

Fig. 9.48

## Solution

For rectangle, $l=15\text{\hspace{0.17em}}\text{cm},b=3\text{\hspace{0.17em}}\text{cm}$

Area of rectangle $=l×b=15×3=45\text{\hspace{0.17em}}{\text{cm}}^{2}$

According to questions

Now, base of triangle

For triangle, base $\left(b\right)=5\text{\hspace{0.17em}}\text{cm}$ and height $\left(h\right)=4\text{\hspace{0.17em}}\text{cm}$

Area of triangle $=\frac{1}{2}×b×h=\frac{1}{2}×5×4=10\text{\hspace{0.17em}}{\text{cm}}^{2}$

Total area enclosed by the shape $=\left(45+10\right)\text{\hspace{0.17em}}{\text{cm}}^{2}=55\text{\hspace{0.17em}}{\text{cm}}^{2}$

# Question: 97

Fig. 9.49

## Solution

Given shape contains a semi-circle and a triangle

Area of semi-circle $=\frac{1}{2}\pi {r}^{2}$

$=\frac{1}{2}×\frac{22}{7}×10×10=\frac{1100}{7}\text{\hspace{0.17em}}{\text{cm}}^{2}$

Area of triangle $=\frac{1}{2}×b×h=\frac{1}{2}×20×7=70\text{\hspace{0.17em}}{\text{cm}}^{2}$

Hence, total area enclosed by the shape = area of semicircle + area of triangle

$=\frac{1100}{7}+70=\frac{1100+490}{7}=\frac{1590}{7}=227.14\text{\hspace{0.17em}}{\text{cm}}^{2}$

# Question: 98

Find the areas of the shaded region:

Fig. 9.50

## Solution

Let the radius of smaller circle be $r$ and bigger circle be $\text{R}$

From the figure, $r=\frac{7}{2}\text{\hspace{0.17em}}\text{cm}$ and $\text{R}=\frac{7}{2}+7=\frac{21}{2}\text{\hspace{0.17em}}\text{cm}$

Area of the shaded region $=$ area of bigger circle – area of smaller circle

$=\pi {\text{R}}^{2}-\pi {r}^{2}$

$=\pi \left({\text{R}}^{2}-{r}^{2}\right)=\pi \left(\frac{21}{2}×\frac{21}{2}-\frac{7}{2}×\frac{7}{2}\right)$

$=\pi \left(\frac{441}{4}-\frac{49}{4}\right)=\frac{22}{7}×\frac{392}{4}=308\text{\hspace{0.17em}}{\text{cm}}^{2}$

Hence, the area of shaded region is $308\text{\hspace{0.17em}}{\text{cm}}^{2}$.

# Question: 99

Find the area of the shaded region:

Fig. 9.51

## Solution

Diameter of bigger circle $=14\text{\hspace{0.17em}}\text{cm}$

Radius of bigger circle $=\frac{14}{2}=7\text{\hspace{0.17em}}\text{cm}$

So, area of bigger circle $=\pi {r}^{2}=\frac{22}{7}×7×7=154\text{\hspace{0.17em}}{\text{cm}}^{2}$

Diameter of smaller circle $=\frac{7}{4}\text{\hspace{0.17em}}\text{cm}$

Radius of the smaller circle $=\frac{7}{4×2}=\frac{7}{8}\text{\hspace{0.17em}}\text{cm}$

Area of two smaller circles $=2×\pi {r}^{2}=2×\frac{22}{7}×\frac{7}{8}×\frac{7}{8}$

$=\frac{77}{16}={\text{cm}}^{2}$

Area of shaded region $=$ area of bigger circle $-$ area of two smaller circles

$=154-\frac{77}{16}=\frac{154×16-77}{16}$

$=\frac{2464-77}{16}=\frac{2387}{16}=149\frac{3}{16}\text{\hspace{0.17em}}{\text{cm}}^{2}$

Hence, area of shaded region is $149\frac{3}{16}\text{\hspace{0.17em}}{\text{cm}}^{2}$

# Question: 100

A circle with radius  is cut into four equal parts and rearranged to form another shape as shown in Fig. 9.52: Does the perimeter change? If it does change, by how much does it increase or decrease?

Fig. 9.52

## Solution

Yes, the perimeter changes.

Perimeter of the circle in first fig includes 4 arcs of equal length.

While in the second fig, perimeter includes 4 arcs as well as radius twice.

So, The perimeter is increased by $2r=2×16=32\text{\hspace{0.17em}}\text{cm}.$

# Question: 101

A large square is made by arranging a small square surrounded by four congruent rectangles as shown in Fig. 9.53. If the perimeter of each of the rectangle is  find the area of the large square.

Fig. 9.53

## Solution

Let the length & breadth of rectangle be  respectively

Given, perimeter of $1$ rectangle $=16\text{\hspace{0.17em}}\text{cm}$

$2\left(l+b\right)=16$

$l+b=8\text{\hspace{0.17em}}\text{cm}$

Since, the side of large square is $\left(l+b\right)$

Hence, area $={\left(\text{side}\right)}^{2}={\left(l+b\right)}^{2}={\left(8\right)}^{2}=64\text{\hspace{0.17em}}{\text{cm}}^{2}$

# Question: 102

ABCD is a parallelogram in which AE is perpendicular to CD (Fig. 9.54). Also   and the area of  Find the perimeter and area of ABCD.

Fig. 9.54

## Solution

Area of $\Delta \text{AED}=6\text{\hspace{0.17em}}{\text{cm}}^{2}&\text{AC}=5\text{\hspace{0.17em}}\text{cm}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{DE}=4\text{\hspace{0.17em}}\text{cm}$

Area of $\Delta \text{AED}=\frac{1}{2}×\text{DE}×\text{AE}$

$=\frac{1}{2}×4×\text{AE}=6$

$\text{AE}=\frac{6×2}{4}$

$\text{AE}=3\text{\hspace{0.17em}}\text{cm}$

In right angled

So,        ${\left(\text{EC}\right)}^{2}={\left(\text{AC}\right)}^{2}-{\left(\text{AE}\right)}^{2}$      using Pythagoras theorem

${\left(\text{EC}\right)}^{2}={5}^{2}-{3}^{2}=25-9$

$\text{EC}=\sqrt{16}$

$\text{EC}=4\text{\hspace{0.17em}}\text{cm}$

$\text{DE}+\text{EC}=\text{DC}$

$\text{DC}=4+4=8\text{\hspace{0.17em}}\text{cm}$

Now, in right angled

${\text{AD}}^{2}={3}^{2}+{4}^{2}=9+16$

$\text{AD}=\sqrt{25}$

$\text{AD}=5\text{\hspace{0.17em}}\text{cm}$

Perimeter of parallelogram $\text{ABCD}=2\left(l+b\right)=2\left(\text{DC}+\text{AD}\right)$

$=2\left(8+5\right)=2×13=26\text{\hspace{0.17em}}\text{cm}$

Area of parallelogram $\text{ABCD}=b×h=\text{DC}×\text{AE}$

$=8×3=24\text{\hspace{0.17em}}{\text{cm}}^{2}$

# Question: 103

Ishika has designed a small oval race track for her remote control car. Her design is shown in the figure 9.55. What is the total distance around the track? Round your answer to the nearest whole cm.

Fig. 9.55

## Solution

Total distance around the track $=$ length of $2$ parallel strips $+$ length of two semi-circles

$=2×52+2×\pi ×16$

$=104+2×3.14×16$

$=104+100.5009$

$=205\text{\hspace{0.17em}}\text{cm}\text{\hspace{0.17em}}\text{(approx}\text{.}\right)$

# Question: 104

A table cover of dimensions  is spread on a table. If  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 $Rs\text{16}$ per square metre.

## Solution

Length of table cloth

Area of the table cloth $=3.25×2.30$

Area of the table cloth on the top of table $=\left(3.25-0.60\right)×\left(2.30-0.60\right)$

Area of hanging table cloth $=$ Area of the table cloth $-$ Area of the table cloth on the top of table $=\left\{3.25×2.30\right\}-\left\{\left(3.25-0.60\right)×\left(2.30-0.60\right)\right\}$

$=7.4750-4.5050$

Since the area of the table top = 4.5050 sq. metre

Cost of polishing table top $=4.5050×16$

$=\text{\hspace{0.17em}}Rs72.08$

# Question: 105

The dimensions of a plot are  A builder builds $3$ roads which are  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?

## Solution

Dimension of plot $=200\text{\hspace{0.17em}}\text{m}×150\text{\hspace{0.17em}}\text{m}$

& width of road $=3\text{\hspace{0.17em}}\text{m}$

Total area available for houses $=$ area of plot $-$ area of $3$ roads

$=200×150-3×\left(3×200\right)$

$=30000-1800=28200\text{\hspace{0.17em}}{\text{m}}^{2}$

# Question: 106

A room is  long and  wide. The floor of the room is to be covered with tiles of size  by  Find the cost of covering the floor with tiles at the rate of $Rs\text{4}.\text{5}0$ per tile.

## Solution

Length of room $=4.5\text{\hspace{0.17em}}\text{m},$ width of room $=4\text{\hspace{0.17em}}\text{m}$ and size of tiles $15\text{\hspace{0.17em}}\text{cm}×10\text{\hspace{0.17em}}\text{cm}$

Area of room

$=18×{\left(100\right)}^{2}\text{\hspace{0.17em}}{\text{cm}}^{2}=180000\text{\hspace{0.17em}}{\text{cm}}^{2}$

Area of $1$ tile $=15×10=150\text{\hspace{0.17em}}{\text{cm}}^{2}$

So, number of tiles $=\frac{\text{area}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{room}}{\text{area}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}1\text{\hspace{0.17em}}\text{tile}}=\frac{180000}{150}=1200$

Cost of covering the floor with tiles at the rate of $Rs4.50$ per tile $=\text{\hspace{0.17em}}Rs4.50×1200=\text{\hspace{0.17em}}Rs5400$

# Question: 107

Find the total cost of wooden fencing around a circular garden of diameter  if  of fencing costs $Rs\text{3}00.$

## Solution

Given, diameter of the circular garden $=28\text{\hspace{0.17em}}\text{m}$

Length of the fencing $=$ circumference of circle

$=\pi d=\frac{22}{7}×28=88\text{\hspace{0.17em}}\text{m}$

Total cost of fencing $=88×300=\text{\hspace{0.17em}}`26400$

# Question: 108

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

## Solution

Given, radius of circle $\left(r\right)=14\text{\hspace{0.17em}}\text{cm}$ and length of rectangle $\left(l\right)=24\text{\hspace{0.17em}}\text{cm}$

Length of the wire $=$ circumference of the circle $=2\pi r=2×\frac{22}{7}×14=88\text{\hspace{0.17em}}\text{m}$

Let $b$ be the width of rectangle

Since, the wire is re-bent in the form of rectangle,

Perimeter of rectangle $=$ circumference of the circle

$2\left(l+b\right)=2\pi r$

$2\left(24+b\right)=88$

$24+b=44$

$b=44-24=20\text{\hspace{0.17em}}\text{cm}$

Area of circle $=\pi {r}^{2}=\frac{22}{7}×{\left(14\right)}^{2}=616\text{\hspace{0.17em}}{\text{cm}}^{2}$

Area of rectangle $=l×b=24×20=480\text{\hspace{0.17em}}{\text{cm}}^{2}$

Hence, the circle enclosed more area than rectangle.

# Question: 109

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

## Solution

Radius of wheel $\left(r\right)=25\text{\hspace{0.17em}}\text{cm}=\frac{25}{100}\text{\hspace{0.17em}}\text{m}=\frac{1}{4}\text{\hspace{0.17em}}\text{m}$

Distance travelled in $1$ rotation = circumference of circle

$=2\pi r=2×\frac{22}{7}×\frac{1}{4}=\frac{1}{4}=\frac{11}{7}\text{m}$

Distance travelled in $350$ rotation $=\frac{11}{7}×\text{35}0=550\text{\hspace{0.17em}}\text{m}$

Hence, the wheel covers a distance of $550$ m.

# Question: 110

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

## Solution

Let $\text{R}$ and $r$ be the radius of outer circle and inner circle respectively.

Circumference of outer circle is .

$2×\frac{22}{7}×\text{R}=44$

Since, $\begin{array}{l}r=\left(\text{R}-2\right)\text{\hspace{0.17em}}\text{m}\\ =\left(7-2\right)\text{\hspace{0.17em}}\text{m}\\ =5\text{\hspace{0.17em}}\text{m}\end{array}$

Inner circumference of the circular path $=2\pi r=2×\frac{22}{7}×5=31.43\text{\hspace{0.17em}}\text{m}\text{\hspace{0.17em}}\left(\text{approx}\text{.}\right)$

Area of the path $=$ area of outer circle $-$ area of inner circle $=\pi \left({\text{R}}^{2}-{r}^{2}\right)$

$\begin{array}{l}=\frac{22}{7}\left({7}^{2}-{5}^{2}\right)=\frac{22}{7}×14\\ =44\text{\hspace{0.17em}}{\text{m}}^{2}\end{array}$

# Question: 111

A carpet of size  has  wide red border. The inner part of the carpet is blue in colour (Fig. 9.56). Find the area of blue portion. What is the ratio of areas of red portion to blue portion?

Fig. 9.56

## Solution

Length of the carpet $=5$ m

Breadth of the carpet $=2$ m

Given, size of carpet $=5\text{\hspace{0.17em}}\text{m}×2\text{\hspace{0.17em}}\text{m}$

& width of border $=25\text{\hspace{0.17em}}\text{cm}=\frac{25}{100}×\text{m}=0.25\text{\hspace{0.17em}}\text{m}$

Area of carpet