Question

From the given figure, find the area of trapezium $ABCD$.

• A. $14{cm}^2$
• B. $24{cm}^2$
• C. $17{cm}^2$
• D. $18{cm}^2$

Answer 1

The correct option is A $14{cm}^2$

Given, trapezium ABCD such that $AB\parallel CD$. Draw a perpendicular from A on CD produced to meet at E succh that ABCE is a rectangle.
In $△AED$
$AE=4cm$ and $AD=5cm$
$A{D}^2=A{E}^2+E{D}^2$ (Pythagoras Theorem)
$5^2=4^2+E{D}^2$
$25-16=E{D}^2$
$ED=3cm$
Now,
$A\left(\square ABCD\right)=A\left(\square ABCD\right)-A\left(△ADE\right)$
$A\left(\square ABCD\right)=(AB\times BC)-\frac{1}{2}AE\times ED$
$A\left(\square ABCD\right)=5\times 4=\frac{1}{2}\left(3\right)\left(4\right)$
$A\left(\square ABCD\right)=20-6$
$A\left(\square ABCD\right)=14c{m}^2$