Problem of the Day


In the figure above, point B has an x-coordinate of 0.5 and a y-coordinate that is higher than any other point on the semi-circle. If \overline{AB} lies on the line y=2x+1, what is the area of the shaded region, in square units?

A. \displaystyle \frac{2\pi}{3}

B. \displaystyle \pi-\frac{\sqrt{3}}{8}

C. \displaystyle \pi-1

D. \displaystyle 2\pi-\frac{3}{2}

E. \displaystyle \frac{3\pi}{8}

Reveal Answer

Answer

[latexpage]

C. $\displaystyle \pi-1$

See the Solution

Solution

[latexpage]

If $B$ is the highest point on the semi-circle, it must be directly above the center. Because the center is on the $x$-axis, the $y$-coordinate of $B$ is equal to the length of the radius. $B$ lies on the line $y=2x+1$, so we can plug in $0.5$ for $x$ to get $y=2$. The area of the semi-circle would then be $\displaystyle \frac{\pi2^2}{2}=2\pi$. The area of the shaded region is equal to half of the area of the semi-circle, $\pi$, minus the area of a right triangle. The right triangle can be drawn in by dropping a perpendicular line from $B$ to the $x$-axis. The height of this triangle is the same as the length of the radius, so we just need to find the length of the base to calculate the area. Note that the third point of the triangle is the same as the $x$-intercept of $y=2x+1$. Find this by plugging zero in for $y$ and solving the equation. This gives us the $x$-intercept: $x=-0.5$. The base of the triangle goes from $x=-0.5$ to $x=0.5$, so its length is 1. The area of the triangle is therefore $\frac{1}{2}\cdot2\cdot1=1$, so the area of the shaded region is $\pi-1$.


One Response

  1. Pete says:

    What a great problem!