Question

Two circles intersect as shown in the diagram below. The radii of the circles are $14cm$ each and $\angle AOP={45}^{\circ }$​ What is the area of the shaded region?

• A. $132.5\pi sq.cm$
• B. $156sq.cm$
• C. $112sq.cm$
• D. $56sq.cm$

Answer 1

The correct option is C

$112sq.cm$

Observe the quadrilateral OAPB, all the sides are of equal length i.e. $14cm.$ So, the quadrilateral is a rhombus.
Given $\angle AOP={45}^{\circ }$,
Since the opposite sides are parallel in a rhombus, $\angle AOP=\angle OPB={45}^{\circ }$
In $△OAP,OA=AP.$ It is an isosceles triangle and hence, $\angle APO={45}^{\circ }$.
Also, $\angle APO=\angle POB={45}^{\circ }$ [Internal opposite angles of the parallel sides AP and OB]
Therefore, $\angle AOB=\angle POB+\angle AOP={45}^{\circ }+{45}^{\circ }={90}^{\circ }$.

Now, join the points A and B and let this line segment intersect OP at C.
Area of the shaded region = Area of segment ADB + Area of segment AEB.
Area of segment ADB = Area of the sector OADB - Area of triangle OAB
$=\frac{{90}^{\circ }}{{360}^{\circ }}\times \pi \times {14}^2{cm}^2-\frac{1}{2}\times 14\times 14{cm}^2$
$=154-98=56sq.cm$
Since, both the segments are similar, the areas of both the segments are equal.
$\Rightarrow$Area of the shaded region = Area of segment ADB + Area of segment AEB
$=56+56sq.cm$
$=112sq.cm$