# NCERT solution class 7 chapter 11 Perimeter and Area exercise 11.2 mathematics

## EXERCISE 11.2

#### Question 1:

Find the area of each of the following parallelograms: Area of parallelogram = Base × Height

(a) Height= 4 cm

Base = 7 cm

Area of parallelogram = 7 × 4 = 28 cm2

(b) Height= 3 cm

Base = 5 cm

Area of parallelogram = 5 × 3 = 15 cm2

(c) Height= 3.5 cm

Base = 2.5 cm

Area of parallelogram = 2.5 × 3.5 = 8.75 cm2

(d) Height= 4.8 cm

Base = 5 cm

Area of parallelogram = 5 × 4.8 = 24 cm2

(e) Height= 4.4 cm

Base = 2 cm

Area of parallelogram = 2 × 4.4 = 8.8 cm2

#### Question 2:

Find the area of each of the following triangles:  (a) Base = 4 cm, height= 3 cm

Area = 6 cm2

(b) Base = 5 cm, height= 3.2 cm

Area = 8 cm2

(c) Base = 4 cm, height= 3 cm

Area = 6 cm2

(d)Base = 3 cm, height= 2 cm

Area = = 3 cm2

#### Question 3:

Find the missing values:

 So No Base Height Area of parallelogram a. 20 cm – 246 cm2 b. – 15 cm 154.5 cm2 c. – 8.4 cm 48.72 cm2 d. 15.6 cm – 16.38 cm2

Area of parallelogram = Base × Height

(a) b = 20 cm

h = ?

Area = 246 cm2

20 × h = 246 Therefore, the height of such parallelogram is 12.3 cm.

(b) b = ?

h = 15 cm

Area = 154.5 cm2

b × 15 = 154.5

b = 10.3 cm

Therefore, the base of such parallelogram is 10.3 cm.

(c) b = ?

h = 8.4 cm

Area = 48.72 cm2

b × 8.4 = 48.72 Therefore, the base of such parallelogram is 5.8 cm.

(d) b = 15.6 cm

h = ?

Area = 16.38 cm2

15.6 × h = 16.38 Therefore, the height of such parallelogram is 1.05 cm.

#### Question 4:

Find the missing values:

 Base Height Area of triangle 15 cm _______ 87 cm2 _______ 31.4 mm 1256 mm2 22 cm _______ 170.5 cm2 (a) b = 15 cm

h = ?

Area =  Therefore, the height of such triangle is 11.6 cm.

(b) b = ?

h = 31.4 mm

Area =  Therefore, the base of such triangle is 80 mm.

(c) b = 22 cm

h = ?

Area =  Therefore, the height of such triangle is 15.5 cm.

#### Question 5:

PQRS is a parallelogram (see the given figure). 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 (a) Area of parallelogram = Base × Height = SR × QM

= 7.6 × 12 = 91.2 cm2

(b) Area of parallelogram = Base × Height = PS × QN = 91.2 cm2

QN × 8 = 91.2 #### Question 6:

DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (see the given figure). If the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL. Area of parallelogram = Base × Height = AB × DL

1470 = 35 × DL Also, 1470 = AD × BM

1470 = 49 × BM #### Question 7:

ΔABC is right angled at A (see the given figure). 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. Area =  = 30 cm2 #### Question 8:

ΔABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (see the given figure). 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?    