2017 AMC 8 Problems/Problem 25
Problem
In the figure shown, and are line segments each of length 2, and . Arcs and are each one-sixth of a circle with radius 2. What is the area of the region shown?
Solution 1
Let the centers of the circles containing arcs and be and , respectively. Extend and to and , and connect point with point . We can clearly see that is an equilateral triangle, because the problem states that . We can figure out that and because they are of a circle. The area of the figure is equal to minus the combined area of the sectors of the circles (in red). Using the area formula for an equilateral triangle, where is the side length of the equilateral triangle, is The combined area of the sectors is , which is Thus, our final answer is
Solution 2
In addition to the given diagram, we can draw lines and The area of rhombus is half the product of its diagonals, which is . However, we have to subtract off the circular segments. The area of those can be found by computing the area of the circle with radius 2, multiplying it by , then finally subtracting the area of an equilateral triangle with a side length 2 from the sector. The sum of the areas of the circular segments is The area of rhombus minus the circular segments is
~PEKKA
Video Solutions
- Happytwin
~savannahsolver
Solution 3 & Video
First, we must make it like an equilateral triangle. If you add two of the same arcs, you will get an equilateral triangle. The goal is to find ONE of the arcs and multiply it by two. Then subtract it from the equilateral triangle. If you look at the triangle, T and S are midpoints of the equilateral triangle. So now we can find the arc.These arcs are so now you multiply by 2 so . And subtract. Equilateral's area is . So now you do . That is the answer and thank you.
~ math.is.hard.to.understand
Here's a video to understand it (credits to ~ pi_is_3.14)
https://youtu.be/j3QSD5eDpzU?t=1350
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
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