1953 AMC 12 Problem 37

Attempt Problem 37 of the 1953 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1953 AMC 12 solutions, or check the answer key.

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37.

The base of an isosceles triangle is 66 inches and one of the equal sides is 1212 inches. The radius of the circle through the vertices of the triangle is:

7155\dfrac{7\sqrt{15}}5

434\sqrt3

353\sqrt5

636\sqrt3

none of these

Answer: E
Concepts:isosceles trianglecircumradiusHeron formula
Difficulty rating: 1740
Small Hint:

Drop the altitude to split the base into two segments of length 33

Big Hint:

Find the area, then use R=abc4KR=\frac{abc}{4K}

Solution:

The altitude is 12232=315, \sqrt{12^2-3^2}=3\sqrt{15}, so the area is K=12(6)(315)=915.K=\tfrac12(6)(3\sqrt{15})=9\sqrt{15}. Hence the circumradius is R=(12)(12)(6)4(915)=8155. R=\frac{(12)(12)(6)}{4(9\sqrt{15})} =\frac{8\sqrt{15}}5. This value is not listed.

Thus, the correct answer is E.

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Problem 37 in Other Years

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