1. In the figure above, AB is the diameter of the circle, and AC = BC. What is the area of the shaded region?

(A) 4π – 2
(B) 2π – 1
(C) π
(D) π – 1
(E) π – 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:

$\large&space;\inline&space;\dpi{300}&space;Area_{shaded}=Area_{whole}-Area_{unshaded}$

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.

Areacircleπr2
Areacircle = π(22)
Areacircle = 4π

So the area of the sector we care about is simply one fourth of that, or π:

Areawholeπ

Now we just need to find the area of the unshaded part (the right triangle we created, in red):

So the area of our shaded region must be…

help! says:

what if you do this

sector ACB- area of triangle ACB

45/360 (4π) – o.5*4*2

the answer should be the same but isn’t??

The problem with that is that sector ACB isn’t 45 degrees! It’s 180!

Be careful with that, though, too. If you take the whole top half of the circle (sector ACB) then you’re not going to get only the shaded region when you take away the triangle…you’re also going to have the unshaded leftover bit on the upper left.

help! says:

OMG! I can’t believe I missed that! I guess I was really sleep deprived when I did this question…! I’m usuallly awesome at maths in school but SAT math manages to screw with my head.. 🙁

Thanks a lot!! your blog is awesome..!

pwn says:

I originally thought that the answer was 4π -2 I don’t understand why it became π
if we were supposed to do Areawhole – Areaunshaded

Well, if you subtract the area of the triangle from the area of the whole circle, you’re left with much more than just the shaded region. So first we shrink the “whole” to be only 1/4 of the circle.

kjokhslk says:

Any harder questions? Im in grade 7 and i found this pretty easy

Prospectus says:

I answered it right, I’m so happy!! I just found the area of the quarter-arc and subtracted it by the area of the 2-2-sqrt(2) triangle.