Should you let yourself get intimidated by a shaded region questions? ABSO-EFFING-LUTELY NOT.
If I know you like I think I do, you'd probably say something like: "Thank you for insulting my intelligence with this asinine question; it's 5."
All you did was recognize that since areas just add up, and you know that the unshaded areas add up to 10, the shaded region has to make up the rest of the total area. If the total area is 15, and the unshaded part is 10, then the shaded one has to be 5. Easy, yes?
So it is with all shaded region questions:
Awhole - Aunshaded = Ashaded
Let's try an example, shall we?
- In the figure above, P is the center of the circle and also the intersection of the two right triangles. If the radius of the circle is 3, what is the area of the shaded region?
(C) 9π - 9
(D) 9π - 6
(E) 9π - 3
In order to solve for the shaded region, we need to find Awhole and Aunshaded.
The area of a circle is πr2, so Awhole = π32 = 9π.
What's the area of the unshaded bits? Note that they're both right triangles, and that each leg is a radius. In other words, we know the base and height of both triangles are 3. The area of one of the triangles is ½bh = ½(3)(3) = 4.5 Since there are two of the triangles, Aunshaded = 9.
So far so good? Now we can solve:
Awhole = 9π
- Aunshaded = 9
Ashaded = 9π - 9
That's choice (C). Easy, right? Just remember Awhole - Aunshaded = Ashaded and you'll be fine.
I know this seems painfully obvious, but it's important to remember that the SAT specializes in making it tricky to deal with concepts that are, on the surface, obvious.
Think about Rubin's Vase (pictured at the top of the post) for a minute. Are you familiar with this image (or this kind of image)? If I told you I was about to show you a picture of a vase and showed you the black and white version, you'd see a vase. But if I told you I was about to show you a picture of two faces and showed you the same image, you'd see the faces.
The writers of the SAT, of course, know this. So when they ask you to find the area of a shaded region, they're usually trying to make you focus on the most difficult thing to find directly (and sometimes, an impossible thing to find directly). That's why it's important to always keep the formula above at the front of your mind. If it's not, you'll likely find yourself choking down a dish of geometrical futility, drizzled with a balsamic disgruntlement reduction and served with a side of anguish fries. It can happen.
Just remember, the SAT would LOVE to misdirect you -- to make you focus on the part that's difficult to solve for, instead of the part that isn't. But since you're an expert pattern recognizer, you're not going to fall for it. You know that the SAT likes to make it difficult to solve directly for the shaded regions, so you're going to solve for everything else instead. Now go forth, intrepid one. Slay shaded regions, one and all.
Practice makes perfect, you know.
Note: Figure not drawn to scale.
- In the figure above, the smaller circles both have a radius of 3 and are tangent to the larger circle. If the larger circle has a radius of 7, what is the area of the shaded region?
- In the figure above, PQRS is a rectangle, M is the midpoint of SR, and N is the midpoint of PS. If NS + SM = 12 and QR = 14, what is the area of the shaded region?
Note: Figure not drawn to scale.
- In the figure above, PQRS is a square, M, N, and D are the midpoints of PS, QR, and AC, respectively, and PM = PB and BQ = QN. If
AC = 10 and CN = 3√2, what is the area of the shaded region?
(E) It cannot be determined from the information given.
Also try this previously posted question (post contains an explanation).