# SAT Math Multiple Choice Practice Question 421: Answer and Explanation

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**Question: 421**

**4.**

If and are diameters with lengths of 12, what is the area of the shaded region?

A. 36

B. 30

C. 18

D. 12

E. 9

**Correct Answer:** C

**Explanation:**

**C** Because 135° is one of the middle angles of the circle, the triangles must each have a 45° angle, and are therefore both identical 45°-45°-90° triangles. Remember (or check the front page of the test) that the ratio of the sides in such a triangle is *x*:*x*:*x*. You know that the diameter of the circle is 12, so the hypotenuse (which is equal to the radius of the circle) of each triangle is 6. If 6 is the long side, the other side of each triangle must be . Now you can find the area of one of the triangles and double it. The area of a triangle is *bh*, and both the base and height are equal to . So = 9. You have found the area for one of the triangles, so double it to get 18, or C. Another way to solve this is to use the side of your answer sheet as a ruler and ballpark. If *XZ* = 12, the hypotenuse of each triangle is 6. Now mark off the length of 6 with your homemade ruler and compare that to a side of one of the triangles. You can guesstimate that the side is about 4. Using that approximation, calculate that since the base and height of both triangles is 4, the area of each triangle is × 4 × 4 = 8. The area of both triangles together is 16, which is closest to 18, or C. ETS wants you to do complicated geometry, but all you care about is finding the answer.