Concise Class-9 Area and Perimeter of Plane Figure ICSE Mathematics Solutions Chapter-20. We provide step by step Solutions of Exercise / lesson-20 Area and Perimeter of Plane Figure for ICSE Class-9th Concise Selina Mathematics by R K Bansal.

Our Solutions contain all type Questions with Exe-20 A, Exe-20 B , Exe-20 C and Exe-20 D to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9th Mathematics .

## Concise Class-9 Area and Perimeter of Plane Figure ICSE Mathematics Solutions Chapter-20

-: Select Topics :-

Exe-20 A,

Exe-20 B,

Exe-20 C,

Exe-20 D,

### Exe-20 A, Concise Class-9 Area and Perimeter of Plane Figure ICSE Mathematics Solutions

Question 1

Find the area of a triangle whose sides are 18 cm, 24 cm and 30 cm.

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

s = 18+24+302
= 36

Hence the area of the triangle is

= 216 sq cm

again

A =1/2 base x height

hence

216 = 1/2 base x height

216 = 1/2 x 30 x h

h = 14.4 cm

Question 2

The length of the sides of a triangle are in the ratio 3: 4: 5. Find the area of the triangle if its perimeter is 144 cm.

Let the sides of the triangle are

a=3x

b=4x

c=5x

Given that the perimeter is 144 cm.

hence

⇒ 3x + 4x + 5x =144

⇒ x = 144/ 12

s = (a+b+c )/2

⇒ 12x/2

⇒ 6x =72

⇒ x = 12

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

Area of the triangle is

= 864 sq cm

Question 3

ABC is a triangle in which AB = AC = 4 cm and A = 90o. Calculate:

(i) The area of ABC,

(ii) The length of perpendicular from A to BC.

(i)

Area of the triangle is given by

(ii)

Again area of the triangle

Question 4

The area of an equilateral triangle is    sq. cm. Find its perimeter.

Solution 4

Question 5

Find the area of an isosceles triangle with perimeter is 36 cm and base is 16 cm.

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,

Question 6

The base of an isosceles triangle is 24 cm and its area is 192 sq. cm. Find its perimeter.

Solution 6

It is given that

Let ‘a’ be the length of an equal side.

Hence perimeter=

Question 7

The given figure shows a right-angled triangle ABC and an equilateral triangle BCD. Find the area of the shaded portion.

Solution 7

From ,

Area of

Area of

Now

Question 8

Find the area and the perimeter of quadrilateral ABCD, given below; if AB = 8 cm, AD = 10 cm, BD = 12 cm, DC = 13 cm and DBC = 90o

Solution 8

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

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

Area of

Area of

Now

Question 9

The base of a triangular field is three times its height. If the cost of cultivating the field at 36.72 per 100 m2 is 49,572; find its base and height.

Solution 9

Question 10

The sides of a triangular field are in the ratio 5 : 3 : 4 and its perimeter is 180 m. Find:

(i) its area.

(ii) altitude of the triangle corresponding to its largest side.

(iii) the cost of leveling the field at the rate of Rs. 10 per square metre.

Solution 10

(ii)

Consider the following figure.

Question 11

Each of equal sides of an isosceles triangle is 4 cm greater than its height. IF the base of the triangle is 24 cm; calculate the perimeter and the area of the triangle.

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

Question 12

Calculate the area and the height of an equilateral triangle whose perimeter is 60 cm.

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

Question 13

In triangle ABC; angle A = 90o, side AB = x cm, AC = (x + 5) cm and area = 150 cm2. Find the sides of the triangle.

Solution 13

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

Hence AB=15cm, AC=20cm and

Question 14

If the difference between the sides of a right angled triangle is 3 cm and its area is 54 cm2; find its perimeter.

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.

Question 15

AD is altitude of an isosceles triangle ABC in which AB = AC = 30 cm and BC = 36 cm. A point O is marked on AD in such a way that BOC = 90o. Find the area of quadrilateral ABOC.

Solution 15
Since AB=AC and

Now

### Concise Class-9 Area and Perimeter of Plane Figure ICSE Mathematics Solutions  Ex. 20(B)

Question 1

Find the area of a quadrilateral one of whose diagonals is 30 cm long and the perpendiculars from the other two vertices are 19 cm and 11 cm respectively.

Solution 1

Question 2

The diagonals of a quadrilateral are 16 cm and 13 cm. If they intersect each other at right angles; find the area of the quadrilateral.

Solution 2

Question 3

Calculate the area of quadrilateral ABCD, in which ABD = 90o, triangle BCD is an equilateral triangle of side 24 cm and AD = 26 cm.

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:

Question 4

Calculate the area of quadrilateral ABCD in which AB = 32 cm, AD = 24 cm A = 90o and BC = CD = 52 cm.

Solution 4

The figure can be drawn as follows:

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

Again

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