from a handpicked tutor in LIVE 1-to-1 classes

# In Fig. 9.25, 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 ar (ABCDE) = ar (APQ).

**Solution:**

Given, ABCDE is a pentagon

BP drawn parallel to AC meets DC produced at P

EQ drawn parallel to AD meets CD produced at Q

We have to prove that ar(ABCDE) = ar(APQ)

Given, BP || AC

AD || EQ

We know that the area of triangles on the same base and between the same parallel lines are equal.

Triangles ABC and APC are on the same base AC and between the same parallel lines AC and AB.

So, ar(ABC) = ar(APC) ----------------------- (1)

Similarly, ar(ADE) = ar(ADQ) --------------- (2)

Adding (1) and (2),

ar(ABC) + ar(ADE) = ar(APC) + ar(ADQ)

Adding ar(ACD) on both sides,

ar(ACD) + ar(ABC) + ar(ADE) = ar(APC) + ar(ADQ) + ar(ACD)

From the figure,

ar(ACD) + ar(ABC) + ar(ADE) = ar(ABCDE)

ar(APC) + ar(ADQ) + ar(ACD) = ar(APQ)

Therefore, ar(ABCDE) = ar(APQ)

**✦ Try This: **In Fig, AP ∥ BQ ∥ CR. Prove that ar(AQC) = ar(PBR)

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

**NCERT Exemplar Class 9 Maths Exercise 9.4 Problem 7**

## In Fig. 9.25, 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 ar (ABCDE) = ar (APQ).

**Summary:**

In Fig. 9.25, 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. It is proven that ar (ABCDE) = ar (APQ)

**☛ Related Questions:**

- If the medians of a ∆ ABC intersect at G, show that ar(AGB) = ar(AGC) = ar(BGC) = 1/3 ar(ABC)
- In Fig. 9.26, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are str . . . .
- In Fig. 9.27, ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD) [Hint: Join PD]

visual curriculum