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# Mathematics – Class 7 – Chapter 11 – Perimeter and Area – Exercise 11.2 – NCERT Exercise Solution

1. Find the area of each of the following parallelograms:

1. Find the area of each of the following parallelograms:

(a)

Solution:

According to the figure,

Height of parallelogram = 4 cm

Base of parallelogram = 7 cm

Now,

Area of parallelogram = base × height

= 7 × 4

= 28 cm2

(b)

Solution:

According to the figure,

Height of parallelogram = 3 cm

Base of parallelogram = 5 cm

Now,

Area of parallelogram = base × height

= 5 × 3

= 15 cm2

(c)

Solution:

According to the figure,

Height of parallelogram = 3.5 cm

Base of parallelogram = 2.5 cm

Now,

Area of parallelogram = base × height

= 2.5 × 3.5

= 8.75 cm2

(d)

Solution:

According to the figure,

Height of parallelogram = 4.8 cm

Base of parallelogram = 5 cm

Now,

Area of parallelogram = base × height

= 5 × 4.8

= 24 cm2

(e)

Solution:

According to the figure,

Height of parallelogram = 4.4 cm

Base of parallelogram = 2 cm

Now,

Area of parallelogram = base × height

= 2 × 4.4

= 8.8 cm2

2. Find the area of each of the following triangles:

(a)

Solution:

According to the figure,

Base of triangle = 4 cm

Height of height = 3 cm

Now,

Area of triangle = ½ × base × height

= ½ × 4 × 3

= 1 × 2 × 3

= 6 cm2

(b)

Solution:

According to the figure,

Base of triangle = 3.2 cm

Height of height = 5 cm

Now,

Area of triangle = ½ × base × height

= ½ × 3.2 × 5

= 1 × 1.6 × 5

= 8 cm2

(c)

Solution:

According to the figure,

Base of triangle = 3 cm

Height of height = 4 cm

Now,

Area of triangle = ½ × base × height

= ½ × 3 × 4

= 1 × 3 × 2

= 6 cm2

(d)

Solution:

According to the figure,

Base of triangle = 3 cm

Height of height = 2 cm

Now,

Area of triangle = ½ × base × height

= ½ × 3 × 2

= 1 × 3 × 1

= 3 cm2

3. Find the missing values:

Solution:

(a)

According to the table,

Given,

Base of parallelogram = 20 cm

Height of parallelogram =?

Area of the parallelogram = 246 cm2

Now,

Area of parallelogram = base × height

246 = 20 × height

Height = 246/20

Height = 12.3 cm

Hence, Height of the parallelogram is 12.3 cm.

(b)

According to the table,

Given,

Base of parallelogram =?

Height of parallelogram =15 cm

Area of the parallelogram = 154.5 cm2

Now,

Area of parallelogram = base × height

154.5 = base × 15

Base = 154.5/15

Base = 10.3 cm

Hence, Base of the parallelogram is 10.3 cm.

(c)

According to the table,

Given,

Base of parallelogram =?

Height of parallelogram =8.4 cm

Area of the parallelogram = 48.72 cm2

Now,

Area of parallelogram = base × height

48.72 = base × 8.4

Base = 48.72/8.4

Base = 5.8 cm

Hence, Base of the parallelogram is 5.8 cm.

(d)

According to the table,

Given,

Base of parallelogram = 15.6 cm

Height of parallelogram =?

Area of the parallelogram = 16.38 cm2

Now,

Area of parallelogram = base × height

16.38 = 15.6 × height

Height = 16.38/15.6

Height = 1.05 cm

Hence, Height of the parallelogram is 1.05 cm.

4. Find the missing values:

Solution:

(1)

According to the table,

Given,

Height of triangle =?

Base of triangle = 15 cm

Area of the triangle = 16.38 cm2

Then,

Area of triangle = ½ × base × height

87 = ½ × 15 × height

Height = (87 × 2)/15

Height = 174/15

Height = 11.6 cm

Hence, Height of the triangle is 11.6 cm.

(2)

According to the table,

Given,

Height of triangle =31.4 mm

Base of triangle =?

Area of the triangle = 1256 mm2

Then,

Area of triangle = ½ × base × height

1256 = ½ × base × 31.4

Base = (1256 × 2)/31.4

Base = 2512/31.4

Base = 80 mm = 8 cm

Hence, Base of the triangle is 80 mm or 8 cm.

(3)

According to the table,

Given,

Height of triangle =?

Base of triangle = 22 cm

Area of the triangle = 170.5 cm2

Then,

Area of triangle = ½ × base × height

170.5 = ½ × 22 × height

170.5 = 1 × 11 × height

Height = 170.5/11

Height = 15.5 cm

Hence, Height of the triangle is 15.5 cm.

5. PQRS is a parallelogram (Fig 11.23). QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find:

(a) The area of the parallelogram PQRS (b) QN, if PS = 8 cm

Solution:

(a) The area of the parallelogram PQRS

According to the question,

Given,

SR = 12 cm

QM = 7.6 cm

Now,

Area of the parallelogram = base × height

= SR × QM

= 12 × 7.6

= 91.2 cm2

(b)

Area of the parallelogram = base × height

91.2 = PS × QN

91.2 = 8 × QN

QN = 91.2/8

QN = 11.4 cm

6. DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (Fig 11.24). If the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL.

Solution:

According to the question,

Given,

Area of the parallelogram = 1470 cm2

AB = 35 cm

Now,

Area of the parallelogram = base × height

1470 = AB × BM

1470 = 35 × DL

DL = 1470/35

DL = 42 cm

And,

Area of the parallelogram = base × height

1470 = 49 × BM

BM = 1470/49

BM = 30 cm

7. ΔABC is right angled at A (Fig 11.25). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, Find the area of ΔABC. Also find the length of AD.

Solution:

According to the question,

Given,

AB = 5 cm

BC = 13 cm

AC = 12 cm

Now,

Area of the ΔABC = ½ × base × height

= ½ × AB × AC

= ½ × 5 × 12

= 1 × 5 × 6

= 30 cm2

Now,

Area of ΔABC = ½ × base × height

30 = ½ × AD × BC

30 = ½ × AD × 13

8. ΔABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26). The height AD from A to BC, is 6 cm. Find the area of ΔABC. What will be the height from C to AB i.e., CE?

Solution:

According to the question it is given that,

AB = AC = 7.5 cm

BC = 9 cm

Now,

Area of ΔABC = ½ × base × height

= ½ × BC × AD

= ½ × 9 × 6

= 1 × 9 × 3

= 27 cm2

Now,

Area of ΔABC = ½ × base × height

27 = ½ × AB × CE

27 = ½ × 7.5 × CE

(27 × 2)/7.5 = CE

CE = 54/7.5

CE = 7.2 cm

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