- In the figure above, AB is the diameter of the circle, and AC = BC. What is the area of the shaded region?
(A) 4 – 2
(B) 2 – 1
(E) – 2
Answer and explanation after the jump…
As is usually the case with shaded region problems, the easiest way (and in this case, the only way) to solve for the shaded region is to solve for the regions around it first. The relationship we want to keep in mind is:
In this particular case, we’re going to have to do a little legwork to figure out what out “whole” is before we get down to business.
Let’s start by dropping a vertical from the top of our isosceles triangle (and noting that in doing so, we’re drawing a radius, so it’s got a length of 2):
That vertical is of course perpendicular to AB, and creates a right angle that nicely frames the area we’re looking to solve for. So the Areawhole we’re looking for here is actually only the part of the circle marked off by that right angle. Since a circle has 360 degrees of arc and we’re only dealing with 90 of them, we’re dealing with one fourth of the circle.
Areacircle = 4
Now we just need to find the area of the unshaded part (the right triangle we created, in red):
Areaunshaded = 1/2 (2)(2)
Areaunshaded = 2
So the area of our shaded region must be…