Question

In the adjoining figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that area of ABCDE = area of ∆APQ.

Answer 1

Answer:

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Answer 2

area of ΔAPQ = area of ABCDE.

Explanation:

Given:

ABCDE is a pentagon.

BP || AC and EQ || AD.

To prove:

Area of (ABCDE) = area of (APQ)

Solution:

We know that, triangles on the same base and between the same parallels are equal in area.

Therefore, ΔADQ and ΔADE lie on the same base AD and between the same parallels AD and EQ.

So, area of ΔADQ = area of ΔADE -------(1)

Similarly, ΔACP and ΔACB lie on the same base AC and between the same parallels AC and BP.

So, area of ΔACP = area of ΔACB --------(2)

Let's add equation 1 & 2. We get:

area of ΔADQ + ΔACP = area ΔADE + ΔACB

Let's add "area of ΔACD on both sides of the equation. We get:

area of ΔADQ + ΔACP + ΔACD = area of ΔADE + ΔACB +ΔACD

⇒ area of ΔAPQ = area of ABCDE.

Hence proved.