Question

In the Fig. $O$ is any point inside a parallelogram $ABCD$. Prove that

$ar(∆OAB)+ar(∆OCD)=\frac{1}{2}ar\left(\right||gmABCD)$

Answer 1

Step 1: Construct the required figure.

$O$ is a point inside a parallelogram $ABCD$.

Now, for calculating the areas of triangles, we draw $EOF\left|\right|AB$.

So, the figure will be,

Step 2: Proving that $ar(∆OAB)+ar(∆OCD)=\frac{1}{2}ar\left(\right||gmABCD)$.

It is known that, 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.

Since $∆AOB$and parallelogram $ABFE$ have the same base $AB$ and lie between the same parallel lines $AB$ and $EF$.

So, according to the theorem,

$ar(∆AOB)=\frac{1}{2}ar\left(\right||gmABEF)$ …(1)

Similarly,

$ar(∆COD)=\frac{1}{2}ar\left(\right||gmEFCD)$ …(2)

Now, adding both equations (1) and (2),

$ar(∆AOB)+ar(∆COD)=\frac{1}{2}ar\left(\right||gmABEF)+\frac{1}{2}ar\left(\right||gmEFCD)$

Therefore, from the figure,

$ar(∆OAB)+ar(∆OCD)=\frac{1}{2}ar\left(\right||gmABCD)$

Hence proved.

It is proved that $ar(∆OAB)+ar(∆OCD)=\frac{1}{2}ar\left(\right||gmABCD)$.