Question

In given figure $ABCD$ is a trapezium with $AB||DC$, $AB$ $=$ $18$ cm, $DC$ $=$ $32$ cm and distance between $AB$ and $DC$ $=$ $14$ cm. If arcs of equal radii $7$ cm with centres $A,B,C$ and $D$ have been drawn, then the area of the shaded region of the figure is

• A. $154$ $c{m}^2$
• B. $196$ $c{m}^2$
• C. $350$ $c{m}^2$
• D. $392$ $c{m}^2$

Answer 1

Answer: (B)

$196$ $c{m}^2$

Given-

$ABCD$ is a trapezium. $AB\parallel DC$ and $AB=18cm$, $DC=32cm$.

There are $4$ arcs centring the vertices $A,B,C,D$ with radius $r=7cm$.

To find out - area of shaded region

Solution-

Area of trapezium $=\frac{1}{2}\times$ sum of the parallel sides $\times$ distance between the parallel sides.$=\frac{1}{2}\times (AB+CD)\times 14=\frac{1}{2}\times (18+32)\times 14{cm}^2=350{cm}^2$.

Let us take the angles of the trapezium as $a$ at $A,b$ at $B,c$ at $C$ and $d$ at $D$.

Now the given arcs form $4$ sectors.

Together they form a circle of radius $r=7cm$ as $a+b+c+d={360}^o$.

$\therefore$ The area of the sectors=area of circle with radius $(r=7cm)$ $=\pi \times r^2=\frac{22}{7}\times 7^2=154c{m}^2$.

$\therefore$ Area of shaded region $=$ area of trapezium $-$ area of circle

$=(350-154)=196c{m}^2.$