NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2
In NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1
Exercise 11.2 we provide NCERT Solutions for class 7 Maths in pdf from for free of cost.
Perimeter and Area Class 7 Ex 11.1
Perimeter and Area Class 7 Ex 11.3
Perimeter and Area Class 7 Ex 11.4
Perimeter and Area Class 7 MCQ
Question 1:
Find the area of each of the following parallelograms:
Answer 1:
The area of the parallelogram = base x height
base = 7 cm and height = 4 cm
Area of the parallelogram = 7 x 4 = 28 cm²
(b)
base = 5 cm and height = 3 cm
Area of the parallelogram = 5 x 3 = 15 cm²
(c)
base = 2.5 cm and height = 3.5 cm
Area of the parallelogram = 2.5 x 3.5 = 8.75 cm²
(d)
base = 5 cm and height = 4.8 cm
Area of the parallelogram = 5 x 4.8 = 24 cm²
(e)
base = 2 cm and height = 4.4 cm
Area of the parallelogram = 2 x 4.4 = 8.8 cm2
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2Question 2:
Find the area of each of the following triangles:
Answer 2:
We know that the area of triangle = (½ x base x height)
base = 4 cm and height = 3 cm
Area of triangle = (½x 4 x 3 )= 6 cm²
We know that the area of triangle = (½ x base x height)
base = 5 cm and height = 3.2 cm
Area of triangle = (½x 5 x 3.2) = 8 cm²
(c)
We know that the area of triangle = (½ x base x height)
base = 3 cm and height = 4 cm
Area of triangle = (½x 3 x 4 )= 6 cm²
(d)
We know that the area of triangle = (½ x base x height)
base = 3 cm and height = 2 cm
Area of triangle = (½x 3 x 2) = 3 cm²
Question 3:
Find the missing values:
Answer 3:
We know that the area of parallelogram = base x height
(a) when base = 20 cm and area = 246 cm²
height = (246 /20)
= 12.3 cm
(b) when, height = 15 cm and area = 154.5 cm²
base of the parallelogram = (154.5 / 15)
= 10.3 cm
© when, height = 8.4 cm and area = 48.72 cm²
base of the parallelogram = (48.72/8.4)
= 5.8 cm
(d) when, base = 15.6 cm and area = 16.38 cm² given
height of the parallelogram = (16.38/15.6)
= 1.05 cm
Question 4:
Find the missing values:
Answer 4:
We know that the area of triangle = (½x base x height)
In first horizontal row, when base = 15 cm and area = 87 cm² given
height = (87 ×2) /15
=11.6 cm
In second horizontal row, when height = 31.4 mm and area = 1256 mm² given, then
base = (1256 × 2) /31.4
=80 mm
In third horizontal row, when base = 22 cm and area = 170.5 cm² given
height = (170.5 × 2) / 22
=15.5 mm
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2
Question 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
Answer 5:
We know that area of parallelogram = base x height
Now in parallelogram PQRS,
SR = 12 cm, QM= 7.6 cm given then
Area of parallelogram PQRS = SR × QM
= (12 × 7.6)
= 91.2 cm²
when area = 91.2 cm² and PS =8cm
Area of parallelogram = base x height
→ 91.2 = 8 × QN
→ QN = (91.2 / 8)
→ QN = 11.4 cm
Question 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.
Answer 6:
We know that area of parallelogram = base x height
Now area of parallelogram = 1470 cm²
Base (AB) = 35 cm given then
Area of parallelogram ABCD = AB × DL
→1470 = 35 x DL
→DL = (1470/35 )
→DL = 42 cm
Again, when area of parallelogram = 1470 cm²
Base (AB) = 35 cm is given then
Area of parallelogram ABCD = AD×BM
1470 = 49 x BM
BM = (1470/49 )
BM = 30 cm
Thus, the lengths of DL and BM are 42 cm and 30 cm respectively.
Question 7:
ABC 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
ABC. Also, find the length of AD.
Answer 7:
In right angles triangle BAC, AB = 5 cm and AC = 12 cm
We know area of right angled triangle = (½x base x height )
= (½x AB x AC)
= (½x 5 x 12)
= 30 cm²
Now, in right angled triangle ABC,
Area of triangle ABC = (½x BC x AD)
→30 = (½x 13 x AD)
→AD = (30 × 2) /13
=( 60/13) cm
Question 8:
ABC 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
ABC. What will be the height from C to AB i.e., CE?
Answer 8:
We know that area of triangle = (½× base × height)
Now in ABC, AD = 6 cm and BC = 9 cm
Area of triangle = (½x base x height)
= (½x BC x AD)
= (½x 9 x 6 )
=27 cm²
Again, Area of triangle = (½x base x height)
→27= (½x AB x CE)
→27 = (½x 7.5 x CE)
→CE = (27 / 2)
→ CE= 7.5 cm
Thus, height from C to AB i.e., CE is 7.2 cm.
NCERT Class 7 Maths Solution Area and perimeter.
