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# Area of The Region | AMC-8, 2017 | Problem 25

Try this beautiful problem from Geometry: The area of the region, AMC-8, 2017. You may use sequential hints to solve the problem.

Try this beautiful problem from Geometry based on the area of Region

## Find the area – AMC-8, 2017- Problem 25

In the figure shown, US and UT and are line segments each of length 2, and$\angle TUS=60^{\circ}$. Arcs TR and SR  and are each one-sixth of a circle with radius 2. What is the area of the figure shown?

• (4 – $\frac{4\pi}{3})$
• ($4\sqrt3$ – $\frac{4\pi}{3})$
• ($4\sqrt3$ – $4\pi$

### Key Concepts

Geometry

Triangle

Circle

Answer: ($4\sqrt3$ – $\frac{4\pi}{3})$

AMC-8 (2017) Problem 25

Pre College Mathematics

## Try with Hints

We have to find out the shaded region. The above diagram is not a standard geometrical figure (such as triangle, square or circle, etc.). So we can not find out the area of the shaded region by any standard formula.

Now if we extend US and UT and joined XY( shown in the above figure ) then it becomes a Triangle shape i.e $\triangle UXY$ and region XSR and region TRY are circular shapes. Now you have to find out the area of these geometrical figures…

Can you now finish the problem ……….

Given that US and UT and are line segments each of length 2 and Arcs TR and SR  and are each one-sixth of a circle with radius 2. Therefore UX=2+2=4,UY=2+2=4 and SX=2+2=4.Therefore $\triangle UXY$ is an equilateral triangle with side length 4 and area of equilateral triangle =$\frac{\sqrt 3}{4} (side)^2$ and the region XSR and region TRY are each one-sixth of a circle with radius 2.nor area of the circle =$\pi (radius)^2$

can you finish the problem……..

The area of the $\triangle UXY=\frac{\sqrt 3}{4} (4)^2=4\sqrt3$ sq.unit

Now area of (region SXR + region TRY)=$2 \times \frac{\pi (2)^2}{6}=\frac{4\pi}{3}$

Therefore the area of the region USRT=Area of $\triangle UXY$- (region SXR + region TRY)=($4\sqrt3$ – $\frac{4\pi}{3})$