# The area of the parallelogram ABCD is 90 cm² (see Fig.9.13). Find

a. ar (ABEF)

b. ar (ABD

c. ar (BEF)

**Solution:**

Given, ABCD is a parallelogram

Area of parallelogram = 90 cm².

(i) We have to find the area of ABEF.

We know that the parallelograms on the same base and between the same parallels are equal in areas.

Parallelograms ABCD and ABEF are on the same base AB and between the same parallels AB and CF.

So, ar (ΔBEF) = ar (ABCD)

Therefore, ar(ABEF) = 90 cm².

(ii) We have to find the area of (ABD)

We know that, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half of the area of the parallelogram.

The triangle ABD and parallelogram ABCD are on the same base AB and between the same parallels AB and CD.

So, ar (ΔABD) = 1/2 ar (ABCD)

= 1/2 × 90

= 45 cm²

Therefore, ar(ABD) = 45 cm²

(iii) we have to find the area of BEF

We know that, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half of the area of the parallelogram.

The triangle BEF and parallelogram ABEF are on the same base EF and between the same parallels AB and EF.

So, ar (ΔBEF) = 1/2 ar (ABEF)

= 1/2 × 90

= 45 cm²

Therefore, ar(BEF) = 45 cm²

**✦ Try This: **A point D is taken on the side BC of a △ABC such that BD = 2DC. Prove that ar(△ABD) = 2 ar(△ADC).

**☛ Also Check: **NCERT Solutions for Class 9 Maths Chapter 9

**NCERT Exemplar Class 9 Maths Exercise 9.3 Problem 3**

## The area of the parallelogram ABCD is 90 cm² (see Fig.9.13). Find (i) ar (ABEF), (ii) ar (ABD), (iii) ar (BEF)

**Summary:**

The area of the parallelogram ABCD is 90 square cm (see Fig.9.13). The (i) ar (ABEF) is cm², (ii) ar (ABD) is 45 cm², (iii) ar (BEF) is 45 cm²

**☛ Related Questions:**

- In ∆ ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Fig. 9.14), . . . .
- ABCD is a square. E and F are respectively the midpoints of BC and CD. If R is the mid-point of EF ( . . . .
- O is any point on the diagonal PR of a parallelogram PQRS (Fig. 9.16). Prove that ar (PSO) = ar (PQO . . . .

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