These NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.2
Question 1.
Find the area of each of the following parallelograms:
Answer:
Here, Base(b) = 7cm
Height (h) = 4 cm
∴ Area of the parallelogram
= b × h sq. units
= 7 × 4 cm2
= 28 cm2
(b) Here Base (b) = 5 cm
Height (h) = 3 cm
∴ Area of the parallelogram
= b x h sq. units
= 5 x 3 cm2 =15 cm2
(c) Here Base (b) = 2.5 cm
Height (h) = 3.5 cm
∴ Area of the parallelogram
= b x h sq. units
= 2.5 x 3.5 cm2
= 8.75 cm2
(d) Here Base (b) = 5 cm
Height (h) = 4.8 cm
∴ Area of the parallelogram
= b x h sq. units
= 5 x 4.8 cm2
= 24 cm2
(e) Here Base (b) = 2 cm
Height (h) =4.4 cm
∴ Area of the parallelogram
= b x h sq. units
= 2 x 4.4 cm2
= 8.8 cm2
Question 2.
Find the area of each of the following triangles:
Answer:
(a) Here, Base (b) = 4cm
Height (h) = 3 cm
Area of the triangle
= \(\frac { 1 }{ 2 }\) x b x h sq.m
= \(\frac { 1 }{ 2 }\) x 4 x 3 cm2
= 6 cm2
(b) Here base (b) = 5 cm
Height (h) = 3.2 cm
Area of the triangle
= \(\frac { 1 }{ 2 }\) x b x h sq. units
= \(\frac { 1 }{ 2 }\) x 5 x 3.2 cm2
= 8 cm2
(c) Here base (b) = 3 cm
Height = 4 cm
Area of the triangle = \(\frac { 1 }{ 2 }\) x b x h sq. units
= \(\frac { 1 }{ 2 }\) x 3 x 4 cm2
= 6 cm2
(d) Here base (b) = 3 cm
Height (h) = 2m
Area of the triangle
= \(\frac { 1 }{ 2 }\) x b x h sq units
= \(\frac { 1 }{ 2 }\) x 3 x 2 cm2 = 3 cm2
Question 3.
Find the missing values:
| Base | Height | Area of the Parallelogram |
| (a) 20 cm | 246 cm; | |
| (b) | 15 cm | 154.5 cm2 |
| (c) | 8.4 cm | 48.72 cm2 |
| (d) 15.6 cm | 16.38 cm2 |
Answer:
(a) Here, base of the parallelogram
(b) = 20 cm
Let the height ‘h’
Area of the Parallelogram = 246 cm2
b x h = 246
20 x h = 246
h = \(\frac{246}{20}\)
\(\frac{123}{10}\) = 12.3 cm
∴ The missing value height = 12.3 cm
(b) Here, height (h) = 15 cm
Let the base of the parallelogram be ‘b’
Area of a parallelogram = 154.5 cm2
b x h = 154.5 cm2
b = \(\frac{154.5}{15}\) = 12.3
= \(\frac{1545}{15 \times 10}\) = 12.3cm
= \(\frac{103}{10}\)10.3
∴ The missing value base = 10.3 cm.
(c) Here, height (h) = 8.4 cm
Let the base of the parallelogram be ‘b’
Area of the parallelogram = 48.72 cm2
b × h = 48.72
b × 8.4 =48.72
b = \(\frac{48.72}{8.4}=\frac{48.72}{8.4}\)
= 5.8 cm
Thus, the missing value base = 5.8 cm.
(d) Here, base (b) = 15.6 cm
Let the height of the parallelogram
be ‘h’
Area of the parallelogram = 16.38 cm2
b × h = 16.38
15.6 × h = 16.38
h = \(\frac{16.38}{15.6}\)
= \(\frac{163.8}{156}\)
= 1.05 cm
Thus, the missing value (height)
= 1.05 cm.
Question 4.
Find the missing values:
| Base | Height | Area of Triangle |
| 15 cm | 87cm2 | |
| 3.14 mm | 1256mm2 | |
| 22 cm | 170.5cm2 |
Answer:
(i) Let the height of the triangle be ‘h’
Here base (b) = 15 cm
Area of the triangle = 87 cm2
\(\frac { 1 }{ 2 }\) × b × h =87 2
\(\frac { 1 }{ 2 }\) × 15 × h =87
h = \(\frac{87 \times 2}{15}\)
= \(\frac{29 \times 2}{5}\)
= \(\frac{58}{5}\)
= 11.6 cm
∴ The missing value height = 11.6 cm
(ii) Here Height = 31.4 mm
Let the base be “b”
Area of a triangle = 1256 mm2
\(\frac { 1 }{ 2 }\) × b × h = 1256 2
\(\frac { 1 }{ 2 }\) × b × 31.4 = 1256
b = \(\frac{1256 \times 2}{31.4}\)
b = \(\frac{1256 \times 2 \times 10}{314}\)
= 4 × 2 × 10 mm
= 80 mm.
∴ The missing value base = 80 mm.
(iii) Let the height of the triangle be ‘h’
base (b) = 22 cm.
Area of the triangle = 170.5 cm2
\(\frac { 1 }{ 2 }\) × b × h = 170.5
\(\frac { 1 }{ 2 }\) × 22 × h = 170.5
11 × h = 170.5
h = \(\frac{170.5}{11}\)
= 15.5 cm
∴ The missing value height = 15.5 cm.
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
Answer:
Here, Base (SR) = 12 cm
Corresponding height (QM) = 7.6 cm
(a) Area of the parallelogram
= b × h sq units
= 12 × 7.6 cm2
= 91.2 cm2
(b) Base of the parallelogram (PS)
= 8 cm
Area of the parallelogram = 91.2 cm2
b × h = 91.2
8 × h = 91.2
h = \(\frac{91.2}{8}\) = 11.4 cm
The height QN =11.4 cm.
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.
Answer:
Area of the parallelogram ABCD
= Base x× height
= AD × BM
Area of a parallelogram
= 1470cm2
∴ AD × BM = 1470cm2
49 × BM = 1470
BM = \(\frac{1470}{49}\)
BM = \(\frac{210}{7}\) = 30 cm
Area of the parallelogram ABCD = 1470cm2
AB × DL = 1470
35 × DL = 1470
DL = \(\frac{1470}{35}\)
= \(\frac{210}{7}\) = 42 cm
∴ Length of BM = 30 cm
Length of DL = 42 cm]
Question 7.
ΔABC 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 ΔABC. Also, find the length of AD.
Answer:
Area of AABC = \(\frac { 1 }{ 2 }\) × b × h sq. units 2 H
(base = 5 cm and height =12 cm)
= \(\frac { 1 }{ 2 }\) × 5 × 12 cm2 = 5 × 6 cm2 = 30 cm2
Area of the AABC = \(\frac { 1 }{ 2 }\) x base x height
30 = \(\frac { 1 }{ 2 }\) × 13 × AD
AD = \(\frac{30 \times 2}{13}\) cm = \(\frac{60}{13}\) cm
Hence, length of AD = \(\frac{60}{13}\) cm = 4.6 cm.
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?
Answer:
Here base BC = 9 cm
Corresponding height AD = 6 cm
∴ Area of ∆ABC
= \(\frac { 1 }{ 2 }\) × base × height
= \(\frac { 1 }{ 2 }\) × BC × AD
= \(\frac { 1 }{ 2 }\) × 9 × 6 cm2 = 27 cm2
Let the height from C to AB be ‘h’
= \(\frac { 1 }{ 2 }\) × AB × h = 27
= \(\frac { 1 }{ 2 }\) × 7.5 × h = 27
h = \(\frac{27 \times 3}{7.5}\)
= \(\frac{27 \times 2 \times 10}{75}\)
= 7.2 cm
∴ The height CE = 7.2 cm.