Name: Figures on the Same Base
Uploaded: 2019-07-19
Duration: 6 min 34 s
Description: Two figures are said to be on one base and between the same parallels if they have a common side as the base and certain vertices lie on the line parallel to...

Question

In the parallelogram

A B C D, P

is a point on the side

A B

and

Q

is a point on the side

B C

. Prove that the area of

∆ ADQ =

the area of

∆ APD +

area of

∆ CPB

.

Accepted Answer

Given that in the parallelogram $A B C D, P$ is a point on the side $A B$ and $Q$ is a point on the side $B C$ . We have to prove that the area of $∆ ADQ =$ the area of $∆ APD +$ area of $∆ CPB$ .

Join $A Q, D Q, P D$ and $P C$ .

Since $A B C D$ is a parallelogram, we have, $A D ∥ B C$ and $A B ∥ D C$ .

We know that area of a triangle is half that of a parallelogram on the same base and between the same parallel lines.

So, we have, $ar (△ C P D) = \frac{1}{2} ar (∥^{gm} A B C D)$ -----(i) [because they lie on the same base $D C$ and are between the same parallel lines $D C$ and $A B$ ].

Again, we have, $ar (∆ A Q D) = \frac{1}{2} ar (‖^{gm} A B C D)$ -----(ii) [Because they lie on the same base $A D$ and are between the same parallel lines $∆ CPB$ and $B C$ ].

From (i) and (ii), we obtain, $ar (∆ C P D) = ar (∆ A Q D)$ -----(iii)

Again, from (ii), we obtain, $ar (∆ A Q D) = \frac{1}{2} ar (‖^{g m} A B C D)$

$\Rightarrow 2 ar (∆ A Q D) = ar (‖^{g m} A B C D)$

$\Rightarrow 2 ar (∆ A D Q) = ar (△ C P D) + ar (∆ A P D) + ar (Δ C P B)$

$\Rightarrow 2 ar (∆ A D Q) = ar (∆ A D Q) + ar (∆ A P D) + ar (Δ C P B)$ [From (iii)]

$\Rightarrow ar (Δ A D Q) = ar (Δ A P D) + ar (Δ C P B)$

Hence, proved.

In the parallelogram $A B C D, P$ is a point on the side $A B$ and $Q$ is a point on the side $B C$ . Prove that the area of $∆ ADQ =$ the area of $∆ APD +$ area of $∆ CPB$ .

Important Questions on Theorems on Area

Learn and Prepare for any exam you want !

In the parallelogram ABCD, P is a point on the side AB and Q is a point on the side BC. Prove that the area of ∆ADQ= the area of ∆APD+ area of ∆CPB.

Important Questions on Theorems on Area

Learn and Prepare for any exam you want !

In the parallelogram $A B C D, P$ is a point on the side $A B$ and $Q$ is a point on the side $B C$ . Prove that the area of $∆ ADQ =$ the area of $∆ APD +$ area of $∆ CPB$ .