NCERT Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.2

Chapter 11 Perimeter and Area Exercise 11.2

Question 1: Find the area of each of the following parallelograms:
(a)


Answer:
Area of Parallelogram = height x base
             = 4 x 7
             = 28 cm²
(b)

Answer:
Area of Parallelogram = height x base
             = 3 x 5
             = 15 cm²
(c)

Answer:
Area of Parallelogram = height x base
             = 3.5 x 2.5
             = 8.75 cm²
(d)

Answer:
Area of Parallelogram = height x base
             = 4.8 x 5
             = 24 cm²
(e)

Answer:
Area of Parallelogram = height x base
             = 4.4 x 2
             = 8.8 cm²
Question 2: Find the area of each of the following triangles:
(a)

Answer:
Area of triangle = 1/2 x base x height
          = 1/2 x 4 x 3
          = 12/2
          = 6 cm²
(b)
 
Answer:
Area of triangle = 1/2 x base x height
          = 1/2 x 5 x 3.2
          = 16/2
          = 8 cm²
(c)

Answer:
Area of triangle = 1/2 x base x height
          = 1/2 x 3 x 4
          = 12/2
          = 6 cm²
(d)
 
Answer:
Area of triangle = 1/2 x base x height
          = 1/2 x 3 x 2
          = 6/2
          = 3 cm²

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 cm


16.38 cm2

Answer:
(a) Given
Area of Parallelogram = 246 cm²
Base = 20 cm
Height = ?
= Height = Area/Base
               = 246/20
               = 12.3 cm
Therefore, height of the parallelogram is 12.3 cm.

(b)
Given
Area of Parallelogram = 154.5 cm²
Base = ?
Height = 15 cm
= Base = Area/Height
               = 154.5/15
               = 10.3 cm
Therefore, base of the parallelogram is 10.3 cm.

(c) Given
Area of Parallelogram = 48.72 cm²
Base = ?
Height = 8.4 cm
= Base = Area/Height
               = 48.72/8.4
               = 5.8 cm
Therefore, base of the parallelogram is 5.8 cm.

(d) Given
Area of Parallelogram = 16.38 cm²
Base = 15.6 cm
Height = ?
= Height = Area/Base
               = 16.38/15.6
               = 1.05 cm
Therefore, height of the parallelogram is 1.05 cm.

Question 4: Find the missing values:

Base

Height

Area of Triangle

15 cm


87 cm2


31.4 mm

1256 mm2

22 cm


170.5 cm2

Answer:
(a) Given
Area of triangle = 87 cm²
Base = 15 cm
Height = ?
= Height = 2 x Area/Base
       = 2 x 87/15
       = 174/15
       = 11.6 cm
Therefore, the height of triangle is 11.6 cm.

(b) Given
Area of triangle = 1256 mm²
Base = ?
Height = 31.4 mm
= Base = 2 x Area/Height
       = 2 x 1256/31.4
       = 2512/31.4
       = 80 mm
Therefore, the base of triangle is 80 mm.

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:
(a)
Area of Parallelogram = height x base
                  = SR x QM
                  = 12 x 7.6
                  = 91.2 cm²
Therefore, the area of parallelogram PQRS is 91.2 cm².

(b)

QN = ? If PS = 8 cm
Area = 91.2 cm²
Base = 8 cm
Height = ?
= Height = Area/Base
       = 91.2/8
               = Height = 11.4 cm
Therefore, QN is 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:
Area = 1470 cm²
AB = 35 cm (base)
AD = 49 cm (base)
BM = ? (height)
DL = ? (height)
(i)
Area = 1470 cm²
Base = AB = 35 cm
Height = DL = ?
= Height = Area/Base
       = 1470/35
       = 42 cm
Therefore, DL is 42 cm.
(ii)
Area = 1470 cm²
Base = AD = 49 cm
Height = BM = ?
= Height = Area/Base
       = 1470/49
       = 30 cm
Therefore, BM is 30 cm.

Question 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.

Answer:
AB = 5 cm
BC = 13 cm
AC = 12 cm
Area of   ⃤   ABC = 1/2 x base x height
         = 1/2 x AB x AC
         = 1/2 x 5 x 12
         = 60/2
         = 30 cm²
Therefore, the area of    ⃤   ABC is 30 cm².
Area = 30 cm²
Base = BC = 13 cm
Height = AD = ?
Height = 2 x Area/Base
    = 2 x 30/13
    = 60/13
    = 4 8/13
Therefore, AD is 4 8/13 cm.

Question 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?

Answer:
AB = AC = 7.5 cm
BC = 9 cm
AD = 6 cm
Area of   ⃤   ABC= 1/2 x base x height
        = 1/2 x BC x AD
        = 1/2 x 9 x 6
        = 54/2
        = 27 cm²
Therefore, the area of   ⃤   ABC is 27 cm².
Area = 27 cm²
Base = 7.5 cm = AB
Height = ? = CE
Height = 2 x Area/Base
    = 2 x 27/7.5
    = 54/7.5
    = 54/7.5 x 10/10
    = 540/75
    = 7.2 cm
Therefore, CE is 7.2 cm.

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