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

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

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

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 are the part of NCERT Solutions for Class 7 Maths. Here you can find the NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2.



Ex 11.2 Class 7 Maths Question 1.

Find the area of each of the following parallelograms:

Solution:
(a) Area of a parallelogram = base × altitude = 7 cm × 4 cm
                                                                               = 28 cm2
(b) Area of a parallelogram = base × altitude = 5 cm × 3 cm
                                                                               = 15 cm2
(c) Area of a parallelogram = base × altitude = 2.5 cm × 3.5 cm
                                                                               = 8.75 cm2
(d) Area of a parallelogram = base × altitude = 5 cm × 4.8 cm
                                                                               = 24.0 cm2
(e) Area of a parallelogram = base × altitude = 2 cm × 4.4 cm
                                                                               = 8.8 cm2


Ex 11.2 Class 7 Maths Question 2.

Find the area of each of the following triangles:

Solution:
(a) Area of a triangle = ½ × b × h
½ × 4 cm × 3 cm
= 6 m2

(b) Area of a triangle = ½ × b × h
½ × 5 cm × 3.2 cm
= 8.0 cm2

(c) Area of a triangle = ½ × b × h
½ × 3 cm × 4 cm
= 6 cm2

(d) Area of a triangle = ½ × b × h
½ × 3 cm × 2 cm
= 3 cm2

 

Ex 11.2 Class 7 Maths Question 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

 

16.38 cm2

 

Solution:
(a) Area of a parallelogram = b × h
246 = 20 × h

(b) Area of a parallelogram = b × h
154.5 = b × 15


(c) Area of a parallelogram = b × h
48.72 = b × 8.4

(d) Area of a parallelogram = b × h
16.38 = 15.6 × h

Ex 11.2 Class 7 Maths Question 4.

Find the missing values:

Base

Height

Area of the triangle

15 cm

87 cm2

31.4 mm

1256 mm2

22 cm

170.5 cm2

 

Solution:
(i) Area of a triangle = ½ × b × h

So, the height =11.6 cm

(ii) Area of a triangle = ½ × b × h

So, the required base = 80 mm.

(iii) Area of a triangle = ½ × b × h

So, the required height = 15.5 cm

 

Ex 11.2 Class 7 Maths Question 5.

PQRS is a parallelogram. 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

Solution:
(a) Area of the parallelogram PQRS
= SR × QM           (
Area of a parallelogram = Base × Height)
= 12 cm × 7.6 cm
= 91.2 cm2

(b) Area of the parallelogram PQRS = PS × QN

91.2 = 8 × QN

QN = 91.2/8 = 11.4 cm

Ex 11.2 Class 7 Maths Question 6.

DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL.

Solution:
Area of the parallelogram ABCD = AB × DL
1470 cm2 = 35 cm × DL
 1470/35 = DL
DL = 42 cm
Area of the parallelogram ABCD = AD × BM
1470 cm2 = 49 cm × BM
 1470/49 = BM
BM = 30 cm
Hence, BM = 30 cm and DL = 42 cm.

 

Ex 11.2 Class 7 Maths Question 7.

∆ABC is right angled at A. AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of ∆ABC. Also find the length of AD.

Solution:
Area of right-angled triangle ABC


Ex 11.2 Class 7 Maths Question 8.

∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC, is 6 cm. Find the area of ∆ABC. What will be the height from C to AB i.e., CE?

Solution:
Area of ∆ABC = ½ × base × height



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NCERT Solutions for Maths Class 11

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