Circles and Geometric Shapes | EOT

Question 21

Let $ABCD$ be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., $\angle \text{CDE} = \angle \text{ABC}$ , where $E$ is a point on the extension of side $CD$ ).

Solution

Understand the Question

A square of side length $s$ $s$ contains:
- A quarter circle of radius $s$ centered at one vertex.
- Two semicircles of diameter $s$ (radius $r = \dfrac{s}{2}$ ) constructed on two adjacent sides.
Region A is the region where the two semicircles overlap.
Region B is the region inside the quarter circle that lies outside both semicircles.
We calculate the area of Region $\text{A}$ and Region $\text{B}$ separately and verify that both are equal to $\dfrac{\pi s^2}{8} - \dfrac{s^2}{4}$ .

Step 1 · Define Dimensions and Radii

Let the side length of the square be $s$ .

Radius of the quarter circle $= s$
Diameter of each semicircle $= s \implies \text{Radius } r = \dfrac{s}{2}$

Step 2 · Calculate Area of Region A

Region $\text{A}$ is the overlap of the two semicircles. Diagram 1

Place the bottom-left vertex at $(0,0)$ . The centers of the two semicircles are at $\left(\dfrac{s}{2}, 0\right)$ and $\left(0, \dfrac{s}{2}\right)$ .

Distance $d$ between the two centers:

\begin{aligned} d &= \sqrt{(\dfrac{s}{2} - 0)^2 + \left(0 - \dfrac{s}{2}\right)^2} \\[0.6em] &= \sqrt{(\dfrac{s}{2})^2 + \left(-\dfrac{s}{2}\right)^2} \\[0.6em] &= \sqrt{\dfrac{s^2}{4} + \dfrac{s^2}{4}} \\[0.6em] &= \sqrt{\dfrac{2s^2}{4}} \\[0.6em] &= \sqrt{\dfrac{s^2}{2}} \\[0.6em] &= \dfrac{s}{\sqrt{2}} \end{aligned}

Using the formula for the area of intersection of two circles with radius $r = \dfrac{s}{2}$ :

$\text{Area} = 2r^2 \cos^{-1}(\dfrac{d}{2r}) - \left(\dfrac{d}{2}\right)\sqrt{4r^2-d^2}$

Evaluate $\dfrac{d}{2r}$ :

\begin{aligned} \dfrac{d}{2r} &= \dfrac{s/\sqrt{2}}{2(s/2)} \\[0.6em] &= \dfrac{s/\sqrt{2}}{s} \\[0.6em] &= \dfrac{1}{\sqrt{2}} \\[0.6em] &= \dfrac{\sqrt{2}}{2} \end{aligned}

Since $\cos(\dfrac{\pi}{4}) = \dfrac{\sqrt{2}}{2}$ , $\cos^{-1}\left(\dfrac{d}{2r}\right) = \dfrac{\pi}{4}\text{ radians}$ .

Evaluate $\sqrt{4r^2-d^2}$ :

\begin{aligned} \sqrt{4r^2-d^2} &= \sqrt{4(\dfrac{s}{2})^2 - \left(\dfrac{s}{\sqrt{2}} ight)^2} \\[0.6em] &= \sqrt{4\cdot\dfrac{s^2}{4} - \dfrac{s^2}{2}} \\[0.6em] &= \sqrt{s^2 - \dfrac{s^2}{2}} \\[0.6em] &= \sqrt{\dfrac{s^2}{2}} \\[0.6em] &= \dfrac{s}{\sqrt{2}} \end{aligned}

Now compute $\text{Area}(\text{A})$ :

\begin{aligned} \text{Area}(\text{A}) &= 2(\dfrac{s}{2})^2 \left(\dfrac{\pi}{4}\right) - \left(\dfrac{s/\sqrt{2}}{2}\right) (\dfrac{s}{\sqrt{2}} ight) \\[0.6em] &= 2\left(\dfrac{s^2}{4}) \left(\dfrac{\pi}{4}\right) - \left(\dfrac{s}{2\sqrt{2}}\right) \left(\dfrac{s}{\sqrt{2}} ight) \\[0.6em] &= \dfrac{s^2}{2} \cdot \dfrac{\pi}{4} - \dfrac{s^2}{4} \\[0.6em] &= \dfrac{\pi s^2}{8} - \dfrac{s^2}{4} \end{aligned}

Step 3 · Calculate Area of Region B

Area of the quarter circle of radius $s$ :

$\text{Area}(\text{Quarter Circle}) = \dfrac{1}{4} \pi s^2$

Area of one semicircle of radius $\dfrac{s}{2}$ :

\begin{aligned} \text{Area}(\text{Semicircle}) &= \dfrac{1}{2} \pi \left(\dfrac{s}{2}\right)^2 \\[0.6em] &= \dfrac{1}{2} \pi \cdot \dfrac{s^2}{4} \\[0.6em] &= \dfrac{\pi s^2}{8} \end{aligned} $$![Diagram 2](https://project-x-homi-questions-book.s3.us-east-1.amazonaws.com/public/solutions/class9/EOT_21/diagram_2.png) By the principle of inclusion-exclusion:

\begin{aligned} \text{Area}(\text{B}) &= \text{Area}(\text{Quarter Circle}) - \text{Area}(\text{Semicircle}_1) - \text{Area}(\text{Semicircle}_2) + \text{Area}(\text{A}) \[0.6em] &= \dfrac{1}{4} \pi s^2 - \dfrac{\pi s^2}{8} - \dfrac{\pi s^2}{8} + \left(\dfrac{\pi s^2}{8} - \dfrac{s^2}{4}\right) \[0.6em] &= \dfrac{\pi s^2}{4} - \dfrac{2\pi s^2}{8} + \dfrac{\pi s^2}{8} - \dfrac{s^2}{4} \[0.6em] &= \dfrac{\pi s^2}{4} - \dfrac{\pi s^2}{4} + \dfrac{\pi s^2}{8} - \dfrac{s^2}{4} \[0.6em] &= \dfrac{\pi s^2}{8} - \dfrac{s^2}{4} \end{aligned}

Step 4 · Compare the Areas

Comparing the two expressions: $\text{Area}(\text{A}) = \dfrac{\pi s^2}{8} - \dfrac{s^2}{4}$

$\text{Area}(\text{B}) = \dfrac{\pi s^2}{8} - \dfrac{s^2}{4}$

$\implies \text{Area}(\text{A}) = \text{Area}(\text{B})$

Answer

$\text{Area}(\text{A}) = \text{Area}(\text{B}) = \dfrac{\pi s^2}{8} - \dfrac{s^2}{4}$

Common Mistakes

Double-Counting Overlap: Forgetting that the two semicircles overlap in Region $\text{A}$ , so their union is $\text{Area}(\text{Semicircle}_1) + \text{Area}(\text{Semicircle}_2) - \text{Area}(\text{A})$ .
Radius vs. Diameter: Using $s$ as the radius of the semicircles instead of $\dfrac{s}{2}$ (since the side of the square is the diameter).

Question 21

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