Ncert Solutions For Class 7 Maths Ex 11.2

NCERT Solutions For Class 7 Maths Ex 11.2 PDF Free Download

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 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

 

ncert sol class 7 math ch 11 ex 2
ncert sol class 7 math ch 11 ex 2
ncert sol class 7 math ch 11 ex 2
ncert sol class 7 math ch 11 ex 2
ncert sol class 7 math ch 11 ex 2
ncert sol class 7 math ch 11 ex 2
ncert sol class 7 math ch 11 ex 2
ncert sol class 7 math ch 11 ex 2
ncert sol class 7 math ch 11 ex 2
ncert sol class 7 math ch 11 ex 2
ncert sol class 7 math ch 11 ex 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)

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Image 2

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

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

(a)

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

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

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

Height = 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

Height = 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

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

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.

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Image 12

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.

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Image 13

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?

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Image 14

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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