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# Circle and Equilateral Triangle | AMC 10A, 2017| Problem No 22

Try this beautiful Problem on Triangle and Circle from AMC 10A, 2017. Problem-22. You may use sequential hints to solve the problem.

Try this beautiful Problem on Geometry based on Circle and Equilateral Triangle from AMC 10 A, 2017. You may use sequential hints to solve the problem.

## Circle and Equilateral Triangle  – AMC-10A, 2017- Problem 22

Sides $\overline{A B}$ and $\overline{A C}$ of equilateral triangle $A B C$ are tangent to a circle at points $B$ and $C$ respectively. What fraction of the area of $\triangle A B C$ lies outside the circle? $?$

,

• $\frac{4 \sqrt{3} \pi}{27}-\frac{1}{3}$
• $\frac{\sqrt{3}}{2}-\frac{\pi}{8}$
• $12$
• $\sqrt{3}-\frac{2 \sqrt{3} \pi}{9}$
• $\frac{4}{3}-\frac{4 \sqrt{3} \pi}{27}$

Geometry

Triangle

Circle

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2017 Problem-22

#### Check the answer here, but try the problem first

$\frac{4}{3}-\frac{4 \sqrt{3} \pi}{27}$

## Try with Hints

#### First Hint

Given that ABC is a equilateral triangle whose $AB$ & $AC$ are the tangents of the circle whose centre is $O$. We have to find out the fraction of the area of $\triangle A B C$ lies outside the circle

we have to find out thr ratio of the areas of Blue colour : Red colour area. Therefore we have to findout the area of the circle and Triangle ABC.

Later we have to find out red area and subtract from the Triangle ABC.

Now can you finish the problem?

#### Second Hint

Let the radius of the circle be $r$, and let its center be $O$. since $\overline{A B}$ and $\overline{A C}$ are tangent to circle $O$, then $\angle O B A=\angle O C A=90^{\circ}$, so $\angle B O C=120^{\circ} .$ Therefore, since $\overline{O B}$ and $\overline{O C}$ are equal to $r$, then $\overline{B C}=r \sqrt{3}$. The area of the equilateral triangle is $\frac{(r \sqrt{3})^{2} \sqrt{3}}{4}=\frac{3 r^{2} \sqrt{3}}{4},$ and the area of the sector we are subtracting from it is $\frac{1}{3} \pi r^{2}-\frac{1}{2} r \cdot r \cdot \frac{\sqrt{3}}{2}=\frac{\pi r^{2}}{3}-\frac{r^{2} \sqrt{3}}{4} .$

Now Can you finish the Problem?

#### Third Hint

Therefore the area outside the circle is $\frac{3 r^{2} \sqrt{3}}{4}-\left(\frac{\pi r^{2}}{3}-\frac{r^{2} \sqrt{3}}{4}\right)=r^{2} \sqrt{3}-\frac{\pi r^{2}}{3}$

Therefore the Required fraction is $\frac{r^{2} \sqrt{3}-\frac{\pi r^{2}}{3}}{\frac{3 r^{2} \sqrt{3}}{4}}$