Question

If area of parallelogram ABCD is A $c{m}^2$, then area of $△$ADC is

• A. $\frac{A}{2}$ cm${}^2$
• B. 2$A$ cm${}^2$
• C. $\frac{A}{4}$ cm${}^2$
• D. 4$A$ cm${}^2$

Answer 1

The correct option is A $\frac{A}{2}$ cm${}^2$


Let AH be the perpendicular drawn from A to CD, then AH is the height corresponding to the base CD in both, parallelogram ABCD and $△$ADC .

Area of parallelogram, A = base$\times$ height
$\Rightarrow$ A = AH$\times$CD

Area of $△$ADC = $\frac{1}{2}$ $\times$ base$\times$ height

= $\frac{1}{2}$ $\times$ AH $\times$ CD

= $\frac{1}{2}$ $\times$ (AH $\times$ CD) = $\frac{1}{2}$ $\times$ A = $\frac{A}{2}$

Therefore, area of $△$ADC is $\frac{A}{2}$ cm${}^2$.