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.
 NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1
 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 11.3
 NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4
Ex 11.2 Class 7 Maths Question 1.
Find the area of each of the following parallelograms:(a) Area of a parallelogram = base × altitude = 7 cm × 4 cm
= 28 cm^{2}
(b) Area of a parallelogram = base × altitude = 5 cm × 3 cm
= 15 cm^{2}
(c) Area of a parallelogram = base × altitude = 2.5 cm × 3.5 cm
= 8.75 cm^{2}
(d) Area of a parallelogram = base × altitude = 5 cm × 4.8 cm
= 24.0 cm^{2}
(e) Area of a parallelogram = base × altitude = 2 cm × 4.4 cm
= 8.8 cm^{2}
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 m^{2}
(b)
Area of a triangle = ½ × b × h
= ½ × 5 cm × 3.2 cm
= 8.0 cm^{2}
(c)
Area of a triangle = ½ × b × h
= ½ × 3 cm × 4 cm
= 6 cm^{2}
(d)
Area of a triangle = ½ × b × h
= ½ × 3 cm × 2 cm
= 3 cm^{2}
Ex 11.2 Class 7 Maths Question 3.
Find the missing values:
S. No. 
Base 
Height 
Area of the parallelogram 
(a) 
20 cm 

246 cm^{2} 
(b) 

15 cm 
154.5 cm^{2} 
(c) 

8.4 cm 
48.72 cm^{2} 
(d) 
15.6 

16.38 cm^{2} 
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 cm^{2} 
— 
31.4 mm 
1256 mm^{2} 
22 cm 
— 
170.5 cm^{2} 
Solution:
(i) Area of
a triangle = ½ × b × h
(ii) Area of a triangle
= ½ × b × h
So, the required base = 80 mm.
(iii)
Area of a triangle = ½ × b × h
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 cm^{2}
(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 cm^{2}, AB = 35 cm and AD = 49 cm, find the length of BM and DL.Solution:
Area of the parallelogram ABCD = AB × DL
⇒ 1470 cm^{2} = 35 cm × DL
⇒ 1470/35 = DL
∴ DL = 42 cm
Area of the parallelogram ABCD = AD × BM
1470 cm^{2} = 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 rightangled 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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