# Class 9 SELINA Solutions Maths Chapter 20 - Area and Perimeter of Plane Figures

## Area and Perimeter of Plane Figures Exercise Ex. 20(A)

### Solution 1

Since the sides of the triangle are 18cm,24cm and 30cm respectively.

Hence area of the triangle is

Again

Hence

### Solution 2

Let the sides of the triangle are

a=3x

b=4x

c=5x

Given that the perimeter is 144 cm.

hence

The sides are a=36 cm, b=48 cm and c=60 cm

Area of the triangle is

### Solution 3

(i)

Area of the triangle is given by

(ii)

Again area of the triangle

### Solution 4

### Solution 5

Since the perimeter of the isosceles triangle is 36cm and base is 16cm. hence the length of each of equal sides are

Here

It is given that

Let 'h' be the altitude of the isosceles triangle.

Since the altitude from the vertex bisects the base perpendicularly, we can apply Pythagoras Theorem.

Thus we have,

### Solution 6

It is given that

Let 'a' be the length of an equal side.

Hence perimeter=

### Solution 7

From ,

Area of

Area of

Now

### Solution 8

Given , AB = 8 cm, AD = 10 cm, BD = 12 cm, DC = 13 cm and DBC = 90^{o}

Hence perimeter=8+10+13+5=36cm

Area of

Area of

Now

### Solution 9

### Solution 10

(i)

(ii)

Consider the following figure.

(iii)

### Solution 11

Let the height of the triangle be x cm.

Equal sides are (x+4) cm.

According to Pythagoras theorem,

Hence perimeter=

Area of the isosceles triangle is given by

Here a=20cm

b=24cm

hence

### Solution 12

Each side of the triangle is

Hence the area of the equilateral triangle is given by

The height h of the triangle is given by

### Solution 13

The area of the triangle is given as 150sq.cm

Hence AB=15cm,AC=20cm and

### Solution 14

Let the two sides be x cm and (x-3) cm.

Now

Hence the sides are 12cm, 9cm and

The required perimeter is 12+9+15=36cm.

### Solution 15

Since AB=AC and

Now

## Area and Perimeter of Plane Figures Exercise Ex. 20(B)

### Solution 1

_{}

### Solution 2

_{}

### Solution 3

Consider the figure:

From the right triangle *ABD* we have

_{}

The area of right triangle *ABD* will be:

_{}

Again from the equilateral triangle *BCD* we have _{}

_{}

Therefore the area of the triangle *BCD* will be:

_{}

Hence the area of the quadrilateral will be:

_{}

### Solution 4

The figure can be drawn as follows:

Here *ABD* is a right triangle. So the area
will be:

_{}

Again

_{}

Now BCD is an isosceles triangle and BP is perpendicular to BD, therefore

_{}

From the right triangle DPC we have

_{}

So

_{}

Hence the area of the quadrilateral will be:

_{}

### Solution 5

Let the width be x and length 2x km.

Hence

_{}

Hence the width is 100m and length is 200m

The required area is given by

_{}

### Solution 6

Length of the laid with grass=85-5-5=75m

Width of the laid with grass=60-5-5=50m

Hence area of the laid with grass is given by

_{}

### Solution 7

Area of the rectangle is given by

_{}

Let h be the height of the triangle ,then

_{}

### Solution 8

Consider the following figure.

Thus the required area = area shaded in blue + area shaded in red

= Area ABPQ + Area TUDC + Area A'PUD' + Area QB'C'T

= 2Area ABPQ + 2Area QB'C'T

=2(Area ABPQ +Area QB'C'T)

### Solution 9

Perimeter of the garden

_{}

Again, length of the garden is given to be 120 m. hence breadth of the garden b is given by

_{}

Hence area of the field

_{}

### Solution 10

Length of the rectangle=x

Width of the rectangle=_{}

_{Hence its perimeter is given by}

_{}

_{Again it is given that the perimeter is 4400cm.}

_{Hence }

_{}

_{Length of the rectangle=1400 cm = 14 m}

### Solution 11

(i)

Breadth of the verandah=x

Length of the verandah=x+3

According to the question

_{}

(ii)

From the above equation

_{}

Hence breadth=3m

Length =3+3=6m

### Solution 12

Consider the following figure.

Thus, the area of the shaded portion

=Area(ABCD) + Area(EFGH) - Area(IJKL)…(1)

### Solution 13

First we have to calculate the area of the hall.

_{}

_{}

We need to find the cost of carpeting of 80 cm = 0.8 m wide carpet, if the rate of carpeting is Rs. 25. Per metre.

Then

### Solution 14

Let a be the length of each side of the square.

Hence

_{}

Hence

_{}

And

_{}

### Solution 15

Consider the following figure.

(i)

The length of the lawn = 30 - 2 - 2 = 26 m

The breadth of the lawn = 12 - 2 = 10 m

(ii)

The orange shaded area in the figure is the required area.

Area of the flower bed is calculated as follows:

_{}

### Solution 16

_{}

_{}

Number of tiles required

_{}

_{}

Fraction of floor uncovered=_{}

### Solution 17

Since

_{}

Hence the distance between the shorter sides is 16m.

### Solution 18

At first we have to calculate the area of the triangle having sides, 28cm, 26cm and 30cm.

Let the area be S.

By Heron's Formula,

Area of a Parallelogram = 2 × Area of a triangle

= 2 × 336

= 672 cm^{2}

We know that,

Area of a parallelogram = Height × Base

⇒ 672 = Height × 26

⇒ Height = 25.84 cm

∴ the distance between its shorter sides is 25.84 cm.

### Solution 19

(i)

We know that

_{}

Here A=216sq.cm

AC=24cm

BD=?

Now

_{}

(ii)

Let a be the length of each side of the rhombus.

_{}

(iii)

Perimeter of the rhombus=4a=60cm.

### Solution 21

Let a be the length of each side of the rhombus.

_{}

We know that,

_{}

### Solution 22

The diagram is redrawn as follows:

Here

AF=1.2m,EF=0.3m,DC=0.6m,BK=1.8-0.6-0.3=0.9m

Hence

_{}

### Solution 23

Here we found two geometrical figure, one is a triangle and other is the trapezium.

Now

_{}

_{}

hence area of the whole figure=150+240=630sq.m

### Solution 24

We can divide the field into three triangles and one trapezium.

Let A,B,C be the three triangular region and D be the trapezoidal region.

Now

_{}

_{}

_{}

_{}

Area of the figure=Area of A+ Area of B+ Area of C+ Area of D

=3900+1250+312.5+2062.5

=7525sq.m

### Solution 25

Let x be the width of the footpath.

Then

_{}

Again it is given that area of the footpath is 360sq.m.

Hence

_{}

Hence width of the footpath is 3m.

### Solution 26

Area of the square is 484.

Let a be the length of each side of the square.

Now

_{}

Hence length of the wire is=4x22=88m.

(i)

Now this 88m wire is bent in the form of an equilateral triangle.

_{}

(ii)

Let x be the breadth of the rectangle.

Now

_{}

Hence area=16x28=448m_{2}

### Solution 27

(i)

Again

_{}

_{}

(ii)

_{}

(iii)

For the triangle EBC,

S=19cm

Let h be the height.

(iv)

In the given figure, we can observe that the non-parallel sides are equal and hence it is an isosceles trapezium.

Therefore, let us draw DE and CF perpendiculars to AB.

Thus, the area of the parallelogram is given by

### Solution 28

Let b be the breadth of rectangle. then its perimeter

Again

Hence its length is 20cm and width is 15cm.

### Solution 29

Let b be the width of the rectangle.

_{}

Again perimeter of the rectangle is 104m.

Hence

_{}

Hence

length=32m

width=20m.

### Solution 30

Let a be the length of the sides of the square.

According to the question

_{}

Hence sides of the square are 12cm each and

Length of the rectangle =2a=24cm

Width of the rectangle=a+6=18cm.

### Solution 31

### Solution 32

Length of the wall=45+2=47m

Breath of the wall=30+2=32m

Hence area of the inner surface of the wall is given by

### Solution 33

Let a be the length of each side.

_{}

Hence length of the wire=96cm

(i)

For the equilateral triangle,

_{}

(ii)

Let the adjacent side of the rectangle be x and y cm.

Since the perimeter is 96 cm, we have,

Hence

_{}

Hence area of the rectangle is = 26 x 22 = 572 sq.cm

### Solution 34

Let 'y' and 'h' be the area and the height of the first parallelogram respectively.

Let 'height' be the height of the second parallelogram

base of the first parallelogram=_{}cm

base of second parallelogram=_{}cm

_{}

### Solution 35

_{}

### Solution 36

### Solution 20

Let a be the length of each side of the rhombus.

_{}

(i)

It is given that,

AC=24cm

We have to find BD.

Now

_{}

Hence the other diagonal is 10cm.

(ii)

## Area and Perimeter of Plane Figures Exercise Ex. 20(C)

### Solution 1

Let r be the radius of the circle.

(i)

_{}

_{}

= 88 cm

(ii)

_{}

= 616 cm^{2}

### Solution 2

### Solution 3

Let r be the radius of the circle.

_{}

### Solution 4

Circumference of the first circle

_{}

Circumference of the second circle

_{}

Let r be the radius of the resulting circle.

_{}

### Solution 5

Circumference of the first circle

_{}

Circumference of the second circle

_{}

Let r be the radius of the resulting circle.

_{}

Hence area of the circle

_{}

### Solution 6

Let the area of the resulting circle be r.

_{}

Hence the radius of the resulting circle is 20cm.

### Solution 8

Area of the rectangle is given by

_{}

For the largest circle, the radius of the circle will be half of the sorter side of the rectangle.

Hence r=21cm.

_{}

_{}

Hence

_{}

### Solution 9

Area of the square is given by

_{}

Since there are four identical circles inside the square.

Hence radius of each circle is one fourth of the side of the square.

_{}

Area
remaining in the cardboard is _{}

### Solution 10

Let the radius of the two circles be 3r and 8r respectively.

_{}

According to the question

_{}

Hence radius of the smaller circle is _{}

Area of the smaller circle is given by

### Solution 11

Let the diameter of the three circles be 3d, 5d and 6d respectively.

Now

_{}

_{}

### Solution 12

_{}

### Solution 13

Let r be the radius of the circular park.

_{}

Radius of the outer circular region including the path is given by

Area of that circular region is

Hence area of the path is given by

### Solution 14

### Solution 15

### Solution 16

_{}

### Solution 17

_{}

_{}

_{ }

### Solution 18

Time interval is

Area covered in one 60 minutes=_{}

Hence area swept in 35 minutes is given by

### Solution 19

Let R and r be the radius of the big and small circles respectively.

### Solution 20

From the given data, we can calculate the area of the outer circle and then the area of inner circle and hence the width of the shaded portion.

### Solution 21

Let the radius of the field is *r* meter.

Therefore circumference of the field will be: _{}meter.

Now the cost of fencing the circular field is 52,800 at rate 240 per meter.

Therefore

_{}

Thus the radius of the field is 35 meter.

Now the area of the field will be:

_{}

Thus the cost of ploughing the field will be:

_{}

### Solution 22

Let r and R be the radius of the two circles.

_{}

Putting the value of r in (2)

_{}

Hence the radius of the two circles is 3cm and 7cm respectively.

### Solution 23

### Solution 24

Since the diameter of the circle is the diagonal of the square inscribed in the circle.

Let a be the length of the sides of the square.

Hence

_{}

Hence the area of the square is 98sq.cm._{}

### Solution 25

### Solution 7

Area of the circle having radius 85m is

_{}

Let r be the radius of the circle whose area is 49times of the given circle.

_{}

Hence circumference of the circle

_{}

## Area and Perimeter of Plane Figures Exercise Ex. 20(D)

### Solution 1

Let the sides of the triangle be

a = 12x

b = 5x

c = 13x

Given that the perimeter = 450 cm

⇒ 12x + 5x + 13x = 450

⇒ 30x = 450

⇒ x = 15

Hence, the sides of a triangle are

a = 12x = 12(15) = 180 cm

b = 5x = 5(15) = 75 cm

c = 13x = 13(15) = 195 cm

Now,

### Solution 2

Let the sides of the triangle be

a = 26 cm, b = 28 cm and c = 30 cm

Now,

### Solution 3

Construction: Draw CM ⊥ AB

In right-angled triangle CMB,

BM^{2}
= BC^{2} - CM^{2} = (15)^{2} - (9)^{2} = 225
- 81 = 144

⇒ BM = 12 m

Now, AB = AM + BM = 23 + 12 = 35 m

### Solution 4

Let the sides of two squares be a and b respectively.

Then,
area of one square, S_{1} = a^{2}

And,
area of second square, S_{2} = b^{2}

Given,
S_{1} + S_{2} = 400 cm^{2}

⇒ a^{2} + b^{2} =
400 cm^{2} …..(1)

Also, difference in perimeter = 16 cm

⇒ 4a - 4b = 16 cm

⇒ a - b = 4

⇒ a = (4 + b)

Substituting the value of 'a' in (1), we get

(4
+ b)^{2} + b^{2} = 400

⇒ 16 + 8b + b^{2} + b^{2}
= 400

⇒ 2b^{2} + 8b - 384 = 0

⇒ b^{2} + 4b - 192 = 0

⇒ b^{2} + 16b - 12b - 192 =
0

⇒ b(b + 16) - 12(b + 16) = 0

⇒ (b +16)(b - 12) = 0

⇒ b + 16 = 0 or b - 12 = 0

⇒ b = -16 or b = 12

Since, the side of a square cannot be negative, we reject -16.

Thus, b = 12

⇒ a = 4 + b = 4 + 12 = 16

Hence, the sides of a square are 16 cm and 12 cm respectively.

### Solution 5

Diagonal of a square = 24 cm

### Solution 6

### Solution 7

### Solution 8

Let the radii of two circles be r_{1}
and r_{2} respectively.

Sum of the areas of two circles = 130π sq. cm

⇒ πr_{1}^{2} + πr_{2}^{2}
= 130π

⇒ r_{1}^{2} + r_{2}^{2}
= 130 ….(i)

Also, distance between two radii = 14 cm

⇒ r_{1} + r_{2} =
14

⇒ r_{1} = (14 - r_{2})

Substituting
the value of r_{1} in (i), we get

(14
- r_{2})^{2} + r_{2}^{2} = 130

⇒ 196 - 28r_{2} + r_{2}^{2}
+ r_{2}^{2} = 130

⇒ 2r_{2}^{2} - 28r_{2}
+ 66 = 0

⇒ r_{2}^{2} - 14r_{2}
+ 33 = 0

⇒ r_{2}^{2} - 11r_{2}
- 3r_{2} + 33 = 0

⇒ r_{2 }(r_{2} -
11) - 3 (r_{2} - 11) = 0

⇒ (r_{2} - 11) (r_{2}
- 3) = 0

⇒ r_{2} = 11 or r_{2}
= 3

⇒ r_{1} = 14 - 11 = 3 or r_{1}
= 14 - 3 = 11

Thus, the radii of two circles are 11 cm and 3 cm respectively.

### Solution 9

### Solution 10

### Solution 11