1956 AMC 12 第 27 题

先试着解答 1956 AMC 12 第 27 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1956 AMC 12 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

27.

若三角形的一个角保持不变,而该角的两条夹边都加倍,则面积变为原来的:
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
大于 66 倍
more than 66
答案:C
知识点:三角形面积included anglescaling
小提示:
使用由两边及其夹角求面积的公式 12absin⁡C\frac12ab\sin C
Use the area formula 12absin⁡C\frac12ab\sin C for two sides and their included angle
大提示:
两条边都加倍,会使它们的乘积乘以 2⋅22\cdot2
Doubling both side factors multiplies their product by 2⋅22\cdot2
解答:
若两条夹边为 aa 和 bb,保持不变的夹角为 CC,则原面积为 12absin⁡C\frac12ab\sin C。两条边都加倍后,面积为 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}\text{。}
因此,正确答案是 C。
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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