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# 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

**☛ Related Questions:**

- Ratio of the area of ∆ WXY to the area of ∆ WZY is 3 : 4 (Fig. 9.33). If the area of ∆ WXZ is 56 cm² . . . .
- Find the perimeter of the lawn. Rani bought a new field that is next to one she already owns (Fig. 9 . . . .
- Find the area of the square field excluding the lawn. Rani bought a new field that is next to one sh . . . .

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