NCERT · Class 7 · Maths

Chapter 9: Perimeter and Area

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Exercise 11.2

Textbook page 216 · 8 questions

Q1Written answer

Find the area of each of the following parallelograms.

(a)

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Worked solution

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)

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Image 3

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)

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Image 4

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)

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Image 5

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)

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Image 6

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

Q2Written answer

Find the area of each of the following triangles.

(a)

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Image 7

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Worked solution

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)

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Image 8

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)

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Image 9

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)

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Image 10

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

Q3Written answer

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

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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

Q4Written answer

Find the missing values.

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

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Worked solution

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.

Q5Written answer

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

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Worked solution

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

Q6Written answer

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

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Worked solution

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

Q7Written answer

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

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Worked solution

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

Q8Written answer

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?

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Worked solution

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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Chapter 9: Perimeter and Area

Exercise 11.2