Try this beautiful problem from Geometry based on the area of Region
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?
Geometry
Triangle
Circle
But try the problem first...
Answer: (\(4\sqrt3\) - \(\frac{4\pi}{3})\)
AMC-8 (2017) Problem 25
Pre College Mathematics
First hint
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 ..........
Second Hint
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)=(\(4\sqrt3\) - \(\frac{4\pi}{3})\)