Question

If $ar\left(ABCL\right)$ is $72c{m}^2$, find $ar\left(ADC\right)$

• A. $24c{m}^2$
• B. $48c{m}^2$
• C. $36c{m}^2$
• D. $54c{m}^2$

Answer 1

The correct option is B $48c{m}^2$

R.E.F image

Let $ABCD$ is a Parallelogram & 'L' is mid pt of DC

$ar\left(ABCD\right)=ar\left(ABCL\right)+ar\left(ADL\right)$

$l_1{l}_{2}sin\theta =72+\frac{1}{2}{l}_1\left(\frac{l_2}{2}\right)sin\theta$

$\frac{3}{4}{l}_1{l}_{2}sin\theta =72\Rightarrow l_1{l}_{2}sin\theta =96cm$

$ar\left(ADC\right)=\frac{1}{2}{l}_1{l}_{2}sin\theta =48c{m}^2$