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 cm^{2} 

b. 
15 cm 
154.5 cm^{2} 

c. 
8.4 cm 
48.72 cm^{2} 

d. 
15.6 cm 
16.38 cm^{2} 
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 cm^{2} 

31.4 mm 
1256 mm^{2} 

22 cm 
170.5 cm^{2} 
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:
(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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