Euclidean Geometry - Circles Problem Solving

This wiki is about problem solving on circles. You need to be familiar with some (if not all) theorems on circles.

Find the area of the shaded region in the above diagram.

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The area of the shaded region is equal to the area of the semicircle with diameter $3$ plus the area of the semicircle with diameter $4$ plus the area of the right triangle minus the area of the semicircle with diameter $5$. We have

$$\begin{aligned} \text{Area}&=\dfrac{\pi}{4}\big(3^2\big)+\dfrac{\pi}{4}\big(4^2\big)+\dfrac{1}{2}(3)(4)-\dfrac{\pi}{4}(5)^2\\\\ &=\dfrac{9}{4}\pi + 4\pi + 6 - \dfrac{25}{4}\pi\\\\ &=6.\ _\square \end{aligned}$$

$BC$ is the diameter of a circle with center at $O$. If $\angle ABO=30^\circ$ and $\angle OAD=40^\circ$, find the measure of $\angle COD$.

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Draw line $AC$. Then by Thales' theorem, $\triangle BAC$ is a right triangle.

Since $\triangle BOA$ is isosceles with $OB=OA=\text{radius}$, $\angle OAB=\angle ABO=30^\circ.$

So, $\angle CAD=90^\circ-30^\circ-40^\circ=20^\circ.$

By the inscribed angle theorem,

$$\angle COD=2(\angle CAD)=2(20^\circ)=40^\circ.\ _\square$$

$DC$ is the diameter of the circle with center at $A.$ $BE$ is perpendicular to $DC$. If $DE=15$ and $EC=9$, find $BE.$

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Since $\triangle DEB \sim \triangle DBC$, we have

$$\dfrac{BE}{DE}=\dfrac{BC}{BD}\implies \dfrac{BE}{15}=\dfrac{BC}{BD}\implies BE=15\dfrac{BC}{BD}\qquad (1)$$

and

$$\dfrac{BE}{BD}=\dfrac{BC}{DC}\implies \dfrac{BE}{BD}=\dfrac{BC}{24}\implies BE=\dfrac{(BD)(BC)}{24}. \qquad (2)$$

Substituting (1) into (2) gives

$$\begin{aligned} 15\dfrac{BC}{BD}&=\dfrac{(BD)(BC)}{24}\\ 15(24)&=(BD)^2\\ BD&=\sqrt{360}\\&=6\sqrt{10}. \end{aligned}$$

Using the Pythagorean theorem, we have

$$BC=\sqrt{DC^2-BD^2}=\sqrt{24^2-\big(6\sqrt{10}\big)^2}=\sqrt{216}=6\sqrt{6}.$$

Finally,

$$BE=15\dfrac{BC}{BD}=15\left(\dfrac{6\sqrt{6}}{6\sqrt{10}}\right)=\dfrac{15\sqrt{6}}{\sqrt{10}}.$$

Rationalizing the denominator gives

$$BE=\dfrac{15\sqrt{6}}{\sqrt{10}} \cdot \dfrac{\sqrt{10}}{\sqrt{10}}=\dfrac{30}{10}\sqrt{15}=3\sqrt{15}.\ _\square$$

• Power of a Point
• Pitot's Theorem
• Cyclic Quadrilaterals
• Ptolemy's Theorem
• Tangent to Circles - Problem Solving