I was just noodling around with Geometer’s Sketchpad today, you know—like a totally normal person, and I made this figure, which I thought would make an awesome challenge question. Wasn’t even planning to post a challenge question, and then BAM! Inspiration.

In the figure above, equilateral triangle DEF is partially obscured by a semicircle with center B and a radius of 5. What is the area of the shaded region?

First correct answer in the comments gets a Math Guide! (Usual contest rules apply.) I’m trying something new this time, though. If I did the settings right, comments will not appear until the contest is over. That way, you can’t see everyone else’s comments before posting yours.

Also, don’t freak out if you’re stumped. My challenge questions are for fun. Although they don’t test anything you don’t need to know for the SAT, they’re usually harder than what you’d find on the SAT.

UPDATE: Congrats to al599, who got it first. Solution below the cut.

Ok. So. This is a shaded region question, and as is usually the case with shaded region questions, we’re going to approach it by finding a larger area, and then subtracting easy-to-calculate unshaded parts. Let’s start by finding the area of the whole equilateral triangle DEF. We know the radius of the semicircle is 5, which means the sides of DEF all have a length of 10.

A good thing for you to remember on the SAT is the fact that all the angles in an equilateral triangle measure 60º. This is important here because it means finding the area of an equilateral triangle is as easy as breaking it into two 30º-60º-90º triangles to find the height.

Cool. So the area of DEF is (1/2)(10)(5√3) = 25√3.

Now we need to start chopping up the unshaded parts of DEF so that we can subtract them and be left with only the shaded part.

Note that if you draw segments from B to the points where the semicircle and triangle meet, you create little equilateral triangles in the lower left and right corners. Those equilateral triangles have 1/4 of the area of DEF, so taking two of them away will take away half the area of DEF.

(Take a minute to understand why those are equilateral triangles, and why they’re 1/4 of the triangle. If you’re stuck, ask in the comments.)

So now you’re left with half of the area of DEF, and you still need to subtract the pizza-wedge shape in the center of the semicircle.

To calculate that area, just remember that, because a straight line is 180º, and both gray triangles are equilateral with angles of 60º, the central angle corresponding to that pizza wedge (AKA sector) must also be 60º. We want to find the area of a 60º sector with a radius of 5. A whole circle with radius 5 has an area of 25π, so a 60º sector will have an area of 1/6 of 25π. (For more on how to do this, read this page on circle questions).

And with that, we’re ready to give our final answer!

Comments (17)

Hi, okay… here goes. I’ve never entered a contest on here before, so I don’t know if I’m supposed to provide how I got to my answer, so I’m just going to put the answer.

I got approximately


First I found the area of DEF. I found the height using sqrt(10^2-5^2). Area of DEF = 1/2(sqrt(75)*10) = 43.3. Then I found the area of the semicircle, which is 1/2(25*pi) or 12.5(pi). The hard part would be to get rid of the tiny long semi-ellipses (no idea what they are) at the end of the arcs. So I just drew a line to complete the arc, and I know it would be 60 degree arc because the triangle would be equilateral. So area of the arc is 60/180 * 12.5(pi) = 4.17(pi) or 13.09. Then just find the area of the little triangle, which would be 1/2(sqrt(5^2-2.5^2)*5) = 10.82. Area of arc – area of little triangle = 13.09 – 10.82 = 2.27. Thus, area of semicircle – two arc-end-ellipses (still don’t know what they are called) = 12.5(pi) – 2.27 * 2 = 34.7299. Area of triangle – 34.7299 = shaded area = 43.3 – 34.7299 = 8.5701 or 8.6. Thanks 🙂

Wow I wish I saw this post sooner! Anyway here’s how I did it:
I found the area of every part of the shape included in the figure.
main triangle – (3 small triangles + 1 small arc segment) =
43.301.. – (32.476.. + 2.265..) = 8.560665699

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