Question

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

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

Answer 1

The correct option is A $27sq.cm$

Given that $AB||DC$

Given that area of triangle $AOB$ is $3$ and area of triangle $DOC$ is $12$

Let the height of triangle $AOB$ be $x$ and $COD$ be $y$

We have $\frac{1}{2}\times AB\times x=3$ and $\frac{1}{2}\times DC\times y=12$

$\Rightarrow x=\frac{6}{10},y=\frac{24}{20}$

The area of trapezium $ABCD$ is $\frac{1}{2}\times (x+y)\times (AB+CD)=\frac{1}{2}\times \frac{36}{20}\times 30=27$ sq.cm

Therefore the correct option is $A$