Circle problem

It is assumed that $AB || CD$ .

Let $A$ be the origin and $AD = a$ lie on the positive $x$ -axis. The circle on the left then has equation $x^{2} + y^{2} = 1$ and the circle on the right has equation $(x - a)^{2} + y^{2} = 1$ . The uppermost point of intersection of the two circles lies at $(\frac{a}{2}, \sqrt{1 - (\frac{a}{2})^{2}})$ .

The area of the upper shaded region will then be

$2*\displaystyle\int_{0}^{\frac{a}{2}} (1 - \sqrt{1 - x^{2}}) dx$ ,

and the area of the lower shaded region will be

$2*\displaystyle\int_{\frac{a}{2}}^{1} \sqrt{1 - x^{2}} dx$ .

For these two areas to be equal we must then have that

$\displaystyle\int_{0}^{\frac{a}{2}} (1 - \sqrt{1 - x^{2}}) dx = \int_{\frac{a}{2}}^{1} \sqrt{1 - x^{2}} dx$

$\Longrightarrow \displaystyle\int_{0}^{\frac{a}{2}} dx = \int_{0}^{1} \sqrt{1 - x^{2}} dx \Longrightarrow \dfrac{a}{2} = \dfrac{\pi}{4} \Longrightarrow a = \dfrac{\pi}{2}$ .

So the area of quadrilateral $ABCD$ is $AB * AD = 1 * a = \frac{\pi}{2} = \boxed{1.57}$

to $2$ decimal places.

the area would be sum of two quarters and the upper shaded region...but due to overlapping of the quarters, the lower shaded portion needs to be deducted...so area= sum of two quarters(( 3.14 /2 ) since radius is 1) + upper shaded region - lower shaded region...as the areas of shaded regions are equal so we are finally left with the area of the two quarters.

The area of each quarter of the circles added together is (Pi/4) * 2=Pi/2. This includes a double counting of the shaded area inside the circles, but that double counting exactly equals the shaded area outside the circles as given in the instructions. So the area of the quadrilateral is just Pi/2 = 1.57.

Let x be the area of one shaded region. By analyzing the geometry, We can say that, Area of Quadrilateral ABCD is = Pi/4 +Pi/4 (Quarter of a unit circle - whole circle area = Pi as radius is 1) -x(subtract the area of intersection)+x (add area of above intersection which is same as x)=Pi/2=1.57