1956 AMC 12 Problem 27

Attempt Problem 27 of the 1956 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 1956 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

27.

If an angle of a triangle remains unchanged but each of its two including sides is doubled, then the area is multiplied by:
22
33
44
66
more than 66
Answer: C
Concepts:triangle areaincluded anglescaling
Small hint:
Use the area formula 12absin⁡C\frac12ab\sin C for two sides and their included angle
Big hint:
Doubling both side factors multiplies their product by 2⋅22\cdot2
Solution:
If the included sides are aa and b,b, and their unchanged angle is C,C, the original area is 12absin⁡C.\frac12ab\sin C. After both sides are doubled, the area is 12(2a)(2b)sin⁡C=2absin⁡C=4(12absin⁡C). \begin{gathered} \frac12(2a)(2b)\sin C \\ =2ab\sin C\\ =4\left(\frac12ab\sin C\right). \end{gathered}
Thus, the correct answer is C.
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