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?

Area of the region

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

Key Concepts


Geometry

Triangle

Circle

Check the Answer


But try the problem first…

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

Source
Suggested Reading

AMC-8 (2017) Problem 25

Pre College Mathematics

Try with Hints


First hint

shaded region

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.

shaded portion

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 ……….

Second Hint

shaded portion

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……..

Final Step

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)=(\(4\sqrt3\) – \(\frac{4\pi}{3})\)

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