In Fig. 9.32, area of ∆ AFB is equal to the area of parallelogram ABCD. If altitude EF is 16 cm long, find the altitude of the parallelogram to the base AB of length 10 cm. What is the area of ∆DAO, where O is the midpoint of DC?
Solution:
Given, the area of ∆ AFB is equal to the area of parallelogram ABCD.
The length of altitude EF is 16 cm
We have to find the altitude of the parallelogram to the base AB of length 10 cm.
Area of triangle = 1/2 × base × height
Area of parallelogram = base × corresponding height
Area of ∆ AFB = area of parallelogram ABCD
1/2 × AB × EF = AB × EG
1/2 × 10 × 16 = 10 × EG
10 × 8 = 10 × EG
EG = 8 cm
Therefore, the corresponding altitude to the base AB is 8 cm.
We have to find the area of ∆DAO, where O is the midpoint of DC.
Since O is the midpoint of DC
DC = DO + CO
DO = CO = 5 cm
Area of ∆ DAO = 1/2 × DO × height
= 1/2 × 5 × 8
= 5(4)
= 20 cm²
Therefore, the area of ∆ DAO is 20 cm².
✦ Try This: The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find the area of the whole land.
☛ Also Check: NCERT Solutions for Class 7 Maths Chapter 11
NCERT Exemplar Class 7 Maths Chapter 9 Problem 78
In Fig. 9.32, area of ∆ AFB is equal to the area of parallelogram ABCD. If altitude EF is 16 cm long, find the altitude of the parallelogram to the base AB of length 10 cm. What is the area of ∆DAO, where O is the midpoint of DC?
Summary:
In Fig. 9.32, the area of ∆ AFB is equal to the area of parallelogram ABCD. If altitude EF is 16 cm long, the altitude of the parallelogram to the base AB of length 10 cm is 8 cm. The area of ∆DAO, where O is the midpoint of DC, is 20 cm²
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