Question

$\square ABCD$ is a trapezium, $AB\parallel DC$. Diagonals of trapezium intersect to each other at point $O$:
Area $\left(\Delta AOB\right)=3$ sq. cm
Area $\left(\Delta COD\right)=12$ sq. cm
Area $\left(\square ABCD\right)=$ _____

• A. $27$ sq. cm
• B. $45$ sq. cm
• C. $36$ sq. cm
• D. $18$ sq. cm

Answer 1

The correct option is A $27$ sq. cm

Area of $\Delta AOB=\frac{1}{2}\times 10\times h_1=3$
$\Rightarrow \frac{1}{2}\times 10\times h_1=3\Rightarrow h_1=\frac{3}{5}$
Area of $\Delta COD$ = $\frac{1}{2}\times 20\times h_2=12$
$\frac{1}{2}\times 20\times h_2=12\Rightarrow h_2=\frac{12}{10}=\frac{6}{5}$
So, height of trapezium $ABCD=h_1+h_2=\frac{3}{5}+\frac{6}{5}\Rightarrow \frac{9}{5}$
$\therefore$ Area of trapezium $=\frac{1}{2}\times \left[\right(h_1+h_2)\times (10+20\left)\right]$
$=\frac{1}{2}\times \frac{9}{5}\times 30=\frac{54}{2}=27$ sq. cm.
Hence, option $A$ is correct.