 # NCERT Solutions for Class 7 Maths Exercise 11.2 Chapter 11 Perimeter and Area

NCERT Solutions for Class 7 Maths Exercise 11.2 Chapter 11 Perimeter and Area in simple PDF are available here. Triangles as parts of rectangles, generalising for other congruent parts of rectangles, area of a parallelogram and area of a triangle are the topics covered in this exercise of NCERT Solutions for Class 7 Maths Chapter 11. Our expert professors formulate these NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area to support you with your exam preparation to score good marks in Maths, with the help of solutions provided here.

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### Access Answers to NCERT Class 7 Maths Chapter 11 – Perimeter and Area Exercise 11.2

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

(a) Solution:-

From the figure,

Height of parallelogram = 4 cm

Base of parallelogram = 7 cm

Then,

Area of parallelogram = base × height

= 7 × 4

= 28 cm2

(b) Solution:-

From the figure,

Height of parallelogram = 3 cm

Base of parallelogram = 5 cm

Then,

Area of parallelogram = base × height

= 5 × 3

= 15 cm2

(c) Solution:-

From the figure,

Height of parallelogram = 3.5 cm

Base of parallelogram = 2.5 cm

Then,

Area of parallelogram = base × height

= 2.5 × 3.5

= 8.75 cm2

(d) Solution:-

From the figure,

Height of parallelogram = 4.8 cm

Base of parallelogram = 5 cm

Then,

Area of parallelogram = base × height

= 5 × 4.8

= 24 cm2

(e) Solution:-

From the figure,

Height of parallelogram = 4.4 cm

Base of parallelogram = 2 cm

Then,

Area of parallelogram = base × height

= 2 × 4.4

= 8.8 cm2

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

(a) Solution:-

From the figure,

Base of triangle = 4 cm

Height of height = 3 cm

Then,

Area of triangle = ½ × base × height

= ½ × 4 × 3

= 1 × 2 × 3

= 6 cm2

(b) Solution:-

From the figure,

Base of triangle = 3.2 cm

Height of height = 5 cm

Then,

Area of triangle = ½ × base × height

= ½ × 3.2 × 5

= 1 × 1.6 × 5

= 8 cm2

(c) Solution:-

From the figure,

Base of triangle = 3 cm

Height of height = 4 cm

Then,

Area of triangle = ½ × base × height

= ½ × 3 × 4

= 1 × 3 × 2

= 6 cm2

(d) Solution:-

From the figure,

Base of triangle = 3 cm

Height of height = 2 cm

Then,

Area of triangle = ½ × base × height

= ½ × 3 × 2

= 1 × 3 × 1

= 3 cm2

3. Find the missing values:

 S.No. Base Height Area of the 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

Solution:-

(a)

From the table,

Base of parallelogram = 20 cm

Height of parallelogram =?

Area of the parallelogram = 246 cm2

Then,

Area of parallelogram = base × height

246 = 20 × height

Height = 246/20

Height = 12.3 cm

∴Height of the parallelogram is 12.3 cm.

(b)

From the table,

Base of parallelogram =?

Height of parallelogram =15 cm

Area of the parallelogram = 154.5 cm2

Then,

Area of parallelogram = base × height

154.5 = base × 15

Base = 154.5/15

Base = 10.3 cm

∴Base of the parallelogram is 10.3 cm.

(c)

From the table,

Base of parallelogram =?

Height of parallelogram =8.4 cm

Area of the parallelogram = 48.72 cm2

Then,

Area of parallelogram = base × height

48.72 = base × 8.4

Base = 48.72/8.4

Base = 5.8 cm

∴Base of the parallelogram is 5.8 cm.

(d)

From the table,

Base of parallelogram = 15.6 cm

Height of parallelogram =?

Area of the parallelogram = 16.38 cm2

Then,

Area of parallelogram = base × height

16.38 = 15.6 × height

Height = 16.38/15.6

Height = 1.05 cm

∴Height of the parallelogram is 1.05 cm.

 S.No. Base Height Area of the Parallelogram a. 20 cm 12.3 cm 246 cm2 b. 10.3 cm 15 cm 154.5 cm2 c. 5.8 cm 8.4 cm 48.72 cm2 d. 15.6 cm 1.05 16.38 cm2

4. Find the missing values:

 Base Height Area of Triangle 15 cm 87 cm2 31.4 mm 1256 mm2 22 cm 170.5 cm2

Solution:-

(a)

From the table,

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

∴Height of the triangle is 11.6 cm.

(b)

From the table,

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

∴Base of the triangle is 80 mm or 8 cm.

(c)

From the table,

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

∴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 Fig 11.23

Solution:-

From the question it is given that,

SR = 12 cm, QM = 7.6 cm

(a) We know that,

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. Fig 11.24

Solution:-

From the question it is given that,

Area of the parallelogram = 1470 cm2

AB = 35 cm

Then,

We know that,

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. Fig 11.25

Solution:-

From the question it is given that,

AB = 5 cm, BC = 13 cm, AC = 12 cm

Then,

We know that,

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:-

From the question it is given that,

AB = AC = 7.5 cm, BC = 9 cm, AD = 6cm

Then,

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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