1979 AHSME Problems/Problem 8

Problem 8

Find the area of the smallest region bounded by the graphs of $y=|x|$ and $x^2+y^2=4$ .

$\textbf{(A) }\frac{\pi}{4}\qquad \textbf{(B) }\frac{3\pi}{4}\qquad \textbf{(C) }\pi\qquad \textbf{(D) }\frac{3\pi}{2}\qquad \textbf{(E) }2\pi$

Solution

Solution by e_power_pi_times_i

The graph of $x^2+y^2 = 4$ is a circle with radius $2$ centered at the origin. The graph of $y=|x|$ is the combined graphs of $y=x$ and $y=-x$ with a nonnegative y. Because the arguments of $y=x$ and $y=-x$ are $135^\circ$ and $45^\circ$ respectively, the angle between the graphs of $y=x$ and $y=-x$ is $90^\circ$ . Thus, the smallest region bounded by the graphs is $\frac{1}{4}\cdot2^2\cdot\pi = \boxed{\textbf{(C) } \pi}$ .

1979 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1979 AHSME Problems/Problem 8

Problem 8

Solution

See also