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 figure given below, $O$ is any point inside a parallelogram $ABCD$ . Prove that area of $∆ OBC +$ area of $∆ OAD = \frac{1}{2}$ area of $‖ gmABCD$ .

Accepted Answer

Let us draw a line $P Q$ through the point $O$ such that $P Q$ is parallel to both $A D$ and $B C$ .

$A P Q D$ is a parallelogram because $A D ∥ P Q$ and $A P ∥ D Q$ .

Now, $△ O A D$ and parallelogram $A P Q D$ are on the same base $A D$ and the same parallel lines $A D$ and $P Q$ .

So, $ar △ O A D = \frac{1}{2} (ar ∥^{gm} A P Q D)$ -----(i) [If a triangle and a parallelogram lie on the same base and between the same parallel lines, then the area of the triangle is half the area of the parallelogram].

Now, $△ O B C$ and parallelogram $B C Q P$ are on the same base $B C$ and the same parallel lines $B C$ and $P Q$ .

So, $ar △ O B C = \frac{1}{2} (ar ∥^{gm} B C Q P)$ -----(ii) [If a triangle and a parallelogram lie on the same base and between the same parallel lines, then the area of the triangle is half the area of the parallelogram].

Adding (ii) and (i), we obtain, $ar ∆ O B C + ar ∆ O A D = \frac{1}{2} (ar ‖^{gm} B C Q P) + \frac{1}{2} (ar ‖ {}^{gm}A P Q D)$

$ar ∆ O B C + ar ∆ O A D = a r ‖^{g m} A B C D$ .

Hence, proved.

In the figure given below, $O$ is any point inside a parallelogram $ABCD$ . Prove that area of $∆ OBC +$ area of $∆ OAD = \frac{1}{2}$ area of $‖ gmABCD$ .

Important Questions on Theorems on Area

Learn and Prepare for any exam you want !

In the figure given below, O is any point inside a parallelogram ABCD. Prove that area of ∆OBC+ area of ∆OAD=12 area of ‖gmABCD.

Important Questions on Theorems on Area

Learn and Prepare for any exam you want !

In the figure given below, $O$ is any point inside a parallelogram $ABCD$ . Prove that area of $∆ OBC +$ area of $∆ OAD = \frac{1}{2}$ area of $‖ gmABCD$ .