This is a little harder than a typical SAT question, and obviously formatted with a bit more flair and color than the College Board would use, but it deals with the same concepts you’ll need to master to kick the SAT where it hurts most, so have a go at it:

I looked out my window the other day and realized that the shadow that my apartment building was casting was a perfect square. If the distance from the edge of the shadow to the top of the building (the dotted line on the diagram) was 100 feet, and my building (which has a square base) is 7 times as tall as it is wide, what was the area of the shadow?

I’ll refrain from answering this myself just yet. Put your answer in the comments, and check back in a few days to see if you were right!

**Update:** This post had LOTS of views and very few folks attempting an answer, so either people were finding this post by mistake when they Googled for something else, people are shy, or this question was harder than I thought. Answer and explanation below the cut.

Basically, this problem requires three insights to solve:

- If the shadow is a perfect square, then it has to be exactly as long as it is wide. That means it has to be as long as the building is wide. Once you’ve got that figured out…
- This is a right triangle problem. The dotted line in the diagram is the hypotenuse of the triangle formed by the height of the building and the length of the shadow (which is the same as the width of the building).
- The area of the shadow will be the width of the building squared.

*x*and get to work:

*x*^{2} + (7*x*)^{2} = 100^{2}

50*x*^{2} = 10000

*x*^{2} = 200

**The area of the shadow is 200 ft**. Nice job to the one kid who got it right, and all the others who got it right and didn’t have the confidence to post it. Maybe next time I’ll offer a prize or something.

^{2}