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.

Download the PDF of NCERT Solutions For Class 7 Maths Chapter 11 Perimeter and Area – Exercise 11.2

Access other exercises of NCERT Solutions For Class 7 Chapter 11 – Perimeter and Area

Exercise 11.1 Solutions

Exercise 11.3 Solutions

Exercise 11.4 Solutions

Access answers to Maths NCERT Solutions for Class 7 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:

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.

4. Find the missing values:

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

AD = 49 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 = AD × BM

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

(30 × 2)/13 = AD

AD = 60/13

AD = 4.6 cm

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