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 region shaded in blue?

• 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. 14 cm. So the quadrilateral is a rhombus.
Given $\angle AOP={45}^{\circ }$.
Since the opposite sides are parallel in a rhombus, $\angle OPB=\angle OPB={45}^{\circ }$

Consider $△$OAP, OA = AP. It is an isosceles triangle and hence $\angle APO={45}^{\circ }$.
$\angle APO=\angle POB={45}^{\circ }$ 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, 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
=$\frac{{90}^{\circ }}{{360}^{\circ }}\times \pi \times {14}^2{cm}^2-\frac{1}{2}\times 14\times 14{cm}^2$
=$154-98=56$ sq cm

Since both segments are similar, the areas of both the segments are equal.
Area of the shaded region = Area of segment ADB + Area of segment AEB
= $56$ sq cm + $56$ sq cm
= $112$ sq cm