AMC10 2014 B

AMC10 2014 B · Q21

AMC10 2014 B · Q21. It mainly tests Pythagorean theorem, Area & perimeter.

Trapezoid ABCD has parallel sides $\overline{AB}$ of length 33 and $\overline{CD}$ of length 21. The other two sides are of lengths 10 and 14. The angles at A and B are acute. What is the length of the shorter diagonal of ABCD?

梯形ABCD有平行边 $\overline{AB}$ 长33和 $\overline{CD}$ 长21。另外两条边长分别为10和14。A和B处的角度是锐角。ABCD的较短对角线的长度是多少？

(A) $10\sqrt{6}$ $10\sqrt{6}$

(B) 25 25

(D) $18\sqrt{2}$ $18\sqrt{2}$

(E) 26 26

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Assume without loss of generality that $DA=10$ and $BC=14$. Let $M$ and $N$ be the feet of the perpendicular segments to $\overline{AB}$ from $D$ and $C$, respectively. The four points $A,M,N,B$ appear on $\overline{AB}$ in that order. Let $x=DM=CN$, $y=AM$, and $z=NB$. Then $x^2+y^2=10^2=100$, $x^2+z^2=14^2=196$, and $y+21+z=33$. Therefore $z=12-y$, and it follows that $\sqrt{196-x^2}=12-\sqrt{100-x^2}$. Squaring and simplifying gives $24\sqrt{100-x^2}=48$, so $x^2=96$ and $y=\sqrt{100-96}=2$. The square of the length of the shorter diagonal, $\overline{AC}$, is $(y+21)^2+x^2=23^2+96=625$, so $AC=25$.

答案（B）：不失一般性，设 $DA=10$ 且 $BC=14$。令 $M$ 与 $N$ 分别为从 $D$、$C$ 向 $\overline{AB}$ 作垂线的垂足。四点 $A,M,N,B$ 按此顺序位于 $\overline{AB}$ 上。设 $x=DM=CN$，$y=AM$，$z=NB$。则 $x^2+y^2=10^2=100$，$x^2+z^2=14^2=196$，且 $y+21+z=33$。因此 $z=12-y$，从而有 $\sqrt{196-x^2}=12-\sqrt{100-x^2}$。两边平方并化简得 $24\sqrt{100-x^2}=48$，故 $x^2=96$ 且 $y=\sqrt{100-96}=2$。较短对角线 $\overline{AC}$ 的长度平方为 $(y+21)^2+x^2=23^2+96=625$，因此 $AC=25$。

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