The given figure shows parallelogram ABCD. Points P and Q trisects the side AB.
To prove: Area of $△$ DPQ = $\frac{1}{6}$ × Area of (parallelogram ABCD)

Open in App

Solution

We know that triangles between same parallels and on the same base have half the area of the parallelogram.
So, $A r e a o f △ D Q C = \frac{1}{2}$ (area of parallelogram ABCD )
Let 'h' be the height between the parallel lines.
$P Q = \frac{1}{3} A B$ (P and Q trisects AB)
Hence, $A r e a o f △ D P Q = \frac{1}{2} \times \frac{1}{3} \times A B \times h$
$A r e a o f △ D P Q = \frac{1}{6} (area of parallelogram ABCD)$

Suggest Corrections

Similar questions

If the given figure shows parallelogram ABCD. Points P and Q trisect the side AB, then ____.

The given figure shows parallelogram ABCD. Points P and Q trisect the side AB, then which of the following is correct?

Δ D P Q

\frac{1}{6} \times a r e a (/ /_{g m} A B C D)

The given figure shows parallelogram ABCD. Points P and Q trisect the side AB, then:

A B C D

P

Q

B C

B C

a r (△ A P Q) = a r (△ D P Q) = \frac{1}{6} a r (p a r a l l e l o g r a m A B C D)

Join BYJU'S Learning Program