Question

$ABCD$ is a parallelogram. If area of $△ABL$ is $15c{m}^2$ and area of $△DCL$ is $32c{m}^2$, find the area of the parallelogram $ABCD$

• A. $94c{m}^2$
• B. $86c{m}^2$
• C. $72c{m}^2$
• D. $65c{m}^2$

Answer 1

The correct option is A $94c{m}^2$

$ae\left(\Delta ABL\right)=\frac{1}{2}\times \left(l_1\sin \theta \right)x$

$ar\left(\Delta DLC\right)=\frac{1}{2}\left(l_1\sin \theta \right)(l_2-x)$

$ar\left(ABCD\right)=l_1{l}_2\sin \theta$

As $ar\left(ABC\right)=15=\frac{1}{2}{l}_1\sin \theta \times x$

$\Rightarrow x=\frac{30}{l_1\sin \theta }$ ...(1)

$ar\left(DLC\right)=32=\frac{1}{2}\left(l_1\sin \theta \right)(l_2-x)$

From (1)

$64=l_1\sin \theta \left(\frac{l_1{l}_2\sin \theta -30}{l_1\sin \theta }\right)$

$\Rightarrow l_1{l}_2\sin \theta =94c{m}^2$

$\Rightarrow ar\left(ABCD\right)=94c{m}^2$