Question

Area of the trapezium is:

• A. $19.5c{m}^2$
• B. $7.8c{m}^2$
• C. $15.6c{m}^2$
• D. $9.75c{m}^2$

Answer 1

Answer: (D)

$9.75c{m}^2$

$\Rightarrow$ Draw a line segment $MN=6cm$.

$\Rightarrow$ Construct an angle of ${60}^{\circ }$ at point $N$.

$\Rightarrow$ Cut off $NO=3cm$.

$\Rightarrow$ Draw a line through $O$ which is parallel to $MN$.

$\Rightarrow$ Take $M$ as a center and radius $4cm$, draw an arc on parallel line which formed point $p$.

$\Rightarrow$ Join $MP$. Draw $OQ\perp MN$.

$\Rightarrow$ Thus, $MNOP$ is required trapezium.

$\Rightarrow$ We get, $l\left(OP\right)=b=1.5cm$, $l\left(OQ\right)=h=2.6cm$ and $l\left(MN\right)=a=6cm$ -- ( 1 )

$\Rightarrow$ Area of trapezium $MNOP=\frac{1}{2}\times h(a+b)$

$\Rightarrow$ $\frac{1}{2}\times 2.6(6+1.5)$ [From ( 1 )]

$\Rightarrow$ $\frac{1}{2}\times 19.5$

$\therefore$ Area of trapezium $MNOP=9.75c{m}^2$