2016 AMC 10B Problems/Problem 21
Contents
Problem
What is the area of the region enclosed by the graph of the equation
Solution 1
Without loss of generality (WLOG) note that if a point in the first quadrant satisfies the equation, so do its corresponding points in the other three quadrants. Therefore, we can assume that , which implies that and , and multiply by at the end.
We can rearrange the equation to get , which describes a circle with center and radius It's clear we now want to find the union of four equal areas.
There are several ways to find the area, but note that if you connect to its other three respective points in the other three quadrants, you get a square of area , along with four half-circles of diameter , for a total area of which is .
Solution 2
Another way to solve this problem is using cases. Though this may seem tedious, we only have to do one case since the area enclosed is symmetrical. The equation for this figure is To make this as easy as possible, we can make both and positive. Simplifying the equation for and being positive, we get the equation
Using the complete the square method, we get
Therefore, the origin of this section of the shape is at
Using the equation we can also see that the radius has a length of .
With this shape we see that this shape can be cut into a right triangle and a semicircle. The length of the hypotenuse of the triangle is so using special right triangles, we see that the area of the triangle is . The semicircle has the area of .
But this is only case. There are cases in total so we have to multiply by .
After multiplying, our answer is:
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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