Name: Interior and Exterior of a Triangle
Uploaded: 2019-07-19
Duration: 2 min 42 s
Description: When you hold a triangular slice of pizza and try to put some ketchup on it, you aim for the interior region of the triangle, or the ketchup might fall elsew...

Question

If

O

is an interior point of a parallelogram

A B C D

, prove that area of

△ A O B +

area of

△ O C D = \frac{1}{2}

area of parallelogram

A B C D

.

Accepted Answer

$∆ A O B$ and parallelogram $A B Q P$ have same base $A B$ and lie between parallel lines $A B$ and $P Q$ .

According to theorem: If a triangle and parallelogram are on the same base and between the same parallel lines, then the area of the triangle is equal to half of the area of the parallelogram.

$\Rightarrow a r e a (∆ A O B) = \frac{1}{2} a r e a (∥ g m A B Q P) - - - 1$

Similarly, we can prove that $a r e a (∆ C O D) = \frac{1}{2} a r e a (∥ g m P Q C D) - - - 2$

Adding equation $(1)$ and $(2)$ we get,

$\Rightarrow a r e a (∆ A O B) + a r e a (∆ C O D) = \frac{1}{2} a r e a (∥ g m A B Q P) + \frac{1}{2} a r e a (∥ g m P Q C D)$

$\Rightarrow a r e a (∆ A O B) + a r e a (∆ C O D) = \frac{1}{2} a r e a (∥ g m A B Q P) + \frac{1}{2} a r e a (∥ g m P Q C D) = \frac{1}{2} a r e a (∥ g m A B C D)$

$\Rightarrow a r e a (∆ A O B) + a r e a (∆ C O D) = \frac{1}{2} a r e a (∥ g m A B C D)$

Hence, proved.

If $O$ is an interior point of a parallelogram $A B C D$ , prove that area of $△ A O B +$ area of $△ O C D = \frac{1}{2}$ area of parallelogram $A B C D$ .

Important Questions on Area

Learn and Prepare for any exam you want !

If O is an interior point of a parallelogram ABCD, prove that area of △AOB+ area of △OCD=12area of parallelogram ABCD.

Important Questions on Area

Learn and Prepare for any exam you want !

If $O$ is an interior point of a parallelogram $A B C D$ , prove that area of $△ A O B +$ area of $△ O C D = \frac{1}{2}$ area of parallelogram $A B C D$ .