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

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 9.

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 faculty formulate these NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area to support students with their exam preparation to score good marks in Maths, with the help of solutions provided here.

NCERT Solutions For Class 7 Maths Chapter 11 Perimeter and Area – Exercise 11.2

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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 the parallelogram = 4 cm

Base of the parallelogram = 7 cm

Then,

Area of the parallelogram = Base × Height

= 7 × 4

= 28 cm²

(b)

Solution:-

From the figure,

Height of the parallelogram = 3 cm

Base of the parallelogram = 5 cm

Then,

Area of the parallelogram = Base × Height

= 5 × 3

= 15 cm²

(c)

Solution:-

From the figure,

Height of the parallelogram = 3.5 cm

Base of the parallelogram = 2.5 cm

Then,

Area of the parallelogram = Base × Height

= 2.5 × 3.5

= 8.75 cm²

(d)

Solution:-

From the figure,

Height of the parallelogram = 4.8 cm

Base of the parallelogram = 5 cm

Then,

Area of the parallelogram = Base × Height

= 5 × 4.8

= 24 cm²

(e)

Solution:-

From the figure,

Height of the parallelogram = 4.4 cm

Base of the parallelogram = 2 cm

Then,

Area of the 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 the triangle = 4 cm

Height of the height = 3 cm

Then,

Area of the triangle = ½ × Base × Height

= ½ × 4 × 3

= 1 × 2 × 3

= 6 cm²

(b)

Solution:-

From the figure,

Base of the triangle = 3.2 cm

Height of the height = 5 cm

Then,

Area of the triangle = ½ × Base × Height

= ½ × 3.2 × 5

= 1 × 1.6 × 5

= 8 cm²

(c)

Solution:-

From the figure,

Base of the triangle = 3 cm

Height of the height = 4 cm

Then,

Area of the triangle = ½ × Base × Height

= ½ × 3 × 4

= 1 × 3 × 2

= 6 cm²

(d)

Solution:-

From the figure,

Base of the triangle = 3 cm

Height of the height = 2 cm

Then,

Area of the triangle = ½ × Base × Height

= ½ × 3 × 2

= 1 × 3 × 1

= 3 cm²

3. Find the missing values.

Solution:-

(a)

From the table,

Base of the parallelogram = 20 cm

Height of the parallelogram =?

Area of the parallelogram = 246 cm²

Then,

Area of the 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 the parallelogram =?

Height of the parallelogram =15 cm

Area of the parallelogram = 154.5 cm²

Then,

Area of the 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 the parallelogram =?

Height of the parallelogram =8.4 cm

Area of the parallelogram = 48.72 cm²

Then,

Area of the 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 the parallelogram = 15.6 cm

Height of the parallelogram =?

Area of the parallelogram = 16.38 cm²

Then,

Area of the 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 the triangle =?

Base of the triangle = 15 cm

Area of the triangle = 16.38 cm²

Then,

Area of the 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 the triangle =31.4 mm

Base of the triangle =?

Area of the triangle = 1256 mm²

Then,

Area of the 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 the triangle =?

Base of the triangle = 22 cm

Area of the triangle = 170.5 cm²

Then,

Area of the 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 the ΔABC = ½ × Base × Height

= ½ × BC × AD

= ½ × 9 × 6

= 1 × 9 × 3

= 27 cm²

Now,

Area of the 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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