Question

A parallelogram and a Triangle have a common base. If area of the Triangle is 20cm? then find area of parallelogram?

Answer 1

We have given that the area of triangle is 20 cm² and So, a parallelogram is also there which is having same base as of triangle. We have to find out the area of the parallelgram from this given data.

First of all, Both are having same base and so that same height too.

Why same height ? Reason is simple , let we consider the parallelgram ABCD and ∆ ACD having same base. Then,

If vertex also lie on the same side then the height of the //gm will going to be equal to that of the height of the triangle. See figure in attachment.

Lets do it :

As we come to know that,

$:\implies \sf Area \: of \: Triangle = 20 \: cm^{2}$

$:\implies \sf Both \: have \: same \: Base \: i.e, = DC$

$:\implies \sf Having \: same \: height \: too = AE$

Now,

Formulae used to solve this -

${\boxed{\sf{ Area \: of \: Triangle = \dfrac{1}{2} \times Base \times Height}}}$

${\boxed{\sf{ Area \: of \: Parallelogram = Base \times Height }}}$

Then,

The area of parallegram is :

$\dashrightarrow\:\:\sf Area \: of \: Parallelogram = 2 \times Area \: of \: the \: triangle$

$\dashrightarrow\:\:\sf Area \: of \: Parallelogram = 2 \times \dfrac{1}{2} \times Base \times Height$

Triangle area, put = 20 cm²

$\dashrightarrow\:\:\sf Area \: of \: Parallelogram = 2 \times 20$

$\dashrightarrow\:\:\sf Area \: of \: Parallelogram = 40 \: cm^{2}$

$\dashrightarrow\:\: \underline{ \boxed{\sf Area \: of \: Parallelogram = 40 \: cm^{2} }}$