# If P is any point on the median AD of a ∆ ABC, then ar (ABP) ≠ ar (ACP). Is the given statement true or false and justify your answer.

**Solution:**

We know that

ar (ABD) = ar (ACD)

ar (PBD) = ar (PCD)

So we get

ar (ABP) = ar (ACP)

Therefore, the statement is false.

**✦ Try This: **The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 4 cm and 2 cm is :

It is given that

Length of rectangle = 4 cm

Breadth of rectangle = 2 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 = 4 cm

HF = AB = 2 cm

We know that

Area of rhombus = Product of diagonals/ 2

By further calculation

= (4 × 2)/2

= 8/2

= 4 cm²

Therefore, the figure obtained is a rhombus of area 4 cm².

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

**NCERT Exemplar Class 9 Maths Exercise 9.2 Sample Problem 1**

## If P is any point on the median AD of a ∆ ABC, then ar (ABP) ≠ ar (ACP). Is the given statement true or false and justify your answer.

**Summary:**

The statement “If P is any point on the median AD of a ∆ ABC, then ar (ABP) ≠ ar (ACP)” is false

**☛ Related Questions:**

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