# ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD

a. is a rectangle

b. is always a rhombus

c. is a parallelogram

d. need not be any of (A), (B) or (C)

**Solution:**

ABCD need not be a rectangle, parallelogram and rhombus

If ABCD is a square, the diagonal AC divides it into two parts of equal area

Therefore, ABCD need not be any of (A), (B) or (C).

**✦ Try This: **The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 7 cm and 5 cm is :

It is given that

Length of rectangle = 7 cm

Breadth of rectangle = 5 cm

Consider E, F, G and H as the mid-points of sides AB, BC, CD and AD

EFGH is a rhombus

Diagonals are EG and HF

So EF = BC = 7 cm

HF = AB = 5 cm

We know that

Area of rhombus = Product of diagonals/ 2

By further calculation

= (7 × 5)/2

= 35/2

= 17.5 cm²

Therefore, the figure obtained is a rhombus of area 17.5 cm².

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

**NCERT Exemplar Class 9 Maths Exercise 9.1 Problem 8**

## ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD a. is a rectangle, b. is always a rhombus, c. is a parallelogram, d. need not be any of (A), (B) or (C)

**Summary:**

ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD need not be any of (A), (B) or (C)

**☛ Related Questions:**

- If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of . . . .
- ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (Fig. 9.6). E and F are the mid-poin . . . .
- If P is any point on the median AD of a ∆ ABC, then ar (ABP) ≠ ar (ACP). Is the given statement true . . . .

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