## What are we learning ?

**Competency in Focus:**Â 2D Geometry (Areas related to circle)

This problem from American Mathematics Contest 8 (AMC 8, 2017) is based on calculation of areas related to circle. It is Question no. 25 of the AMC 8 2017 Problem series.

## First look at the knowledge graph:-

## Next understand the problem

In the figure shown, $\overline{US}$ and $\overline{UT}$ are line segments each of length 2, and $m\angle TUS = 60^\circ$. Arcs $TR$ and $SR$ are each one-sixth of a circle with radius 2. What is the area of the region shown? $\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}$

##### Source of the problem

American Mathematical Contest 2017, AMC 8 Problem 25

##### Key Competency

### Finding the area of a triangle and sector of a circle. (Area related to circles)

##### Difficulty Level

5/10

##### Suggested Book

## Start with hintsÂ

Do you really need a hint? Try it first!

C0nstruction : Let $X$ and $Y$ are the centres of the scetors $ST$ and $TR$ Now Let us join $SX$ and $TY$ What do you think? Will the points $U,S,\textbf{ and}\quad X$ be in a straightline?

$U,S,\textbf{ and}\quad X$ will be in a straight line because $\angle STU =60^{\circ}$ And angle of aÂ circle is $360$Â i.e., $\angle SXR = \angle TYR = 60^{\circ}$ [Since sector($SXR$)=$\frac{1}{6}circle$] Then $UXY$ will make an equilateral triangle.

So after construction the figure will look like this : Therefore, The required area = Area of $\triangle UXY$ – $2 \times$ Area of the sector $SXR$.

Area of equilateral triangle $\triangle UXY= 4\sqrt{3}$ And the are of sector $SXR= \frac{2\pi}{3}$ ANS : $4\sqrt{3}-\frac{4\pi}{3}$

Area of an equilateral triangle =$\frac{a^2\sqrt{3}}{4}$ [where $a$ is a sied of the triangle] Area of a sector of a circle of angle $\theta$ = $\frac{\theta}{360}\pi r^2$ [where $r$ is the radius of the circle]

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