Question

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

Answer 1

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, $\angle OPB=\angle AOP={45}^{\circ }$
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, $\angle AOB=\angle POB+\angle AOP={45}^{\circ }+{45}^{\circ }={90}^{\circ }$. (1 mark)

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 (1 mark)
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 (1 mark)