AMC10 2002 A

AMC10 2002 A · Q25

AMC10 2002 A · Q25. It mainly tests Area & perimeter, Coordinate geometry.

In trapezoid ABCD with bases AB and CD, we have AB = 52, BC = 12, CD = 39, and DA = 5. The area of ABCD is

梯形 ABCD 底边 AB 和 CD，有 AB = 52，BC = 12，CD = 39，DA = 5。ABCD 的面积是

(A) 182 182

(B) 195 195

(D) 234 234

(E) 260 260

Answer

Correct choice: (C)

正确答案：（C）

Solution

(C) First drop perpendiculars from $D$ and $C$ to $\overline{AB}$. Let $E$ and $F$ be the feet of the perpendiculars to $\overline{AB}$ from $D$ and $C$, respectively, and let $h=DE=CF,\quad x=AE,\quad \text{and}\quad y=FB.$ Then $25=h^2+x^2,\quad 144=h^2+y^2,\quad \text{and}\quad 13=x+y.$ So \[ 144=h^2+y^2=h^2+(13-x)^2=h^2+x^2+169-26x=25+169-26x, \] which gives $x=50/26=25/13$, and \[ h=\sqrt{5^2-\left(\frac{25}{13}\right)^2}=5\sqrt{1-\frac{25}{169}}=5\sqrt{\frac{144}{169}}=\frac{60}{13}. \] Hence \[ \text{Area}(ABCD)=\frac12(39+52)\cdot\frac{60}{13}=210. \]

（C）先从 $D$ 和 $C$ 向 $\overline{AB}$ 作垂线。设 $E$ 和 $F$ 分别为从 $D$ 与 $C$ 到 $\overline{AB}$ 的垂足，并令 $h=DE=CF,\quad x=AE,\quad \text{以及}\quad y=FB.$ 则 $25=h^2+x^2,\quad 144=h^2+y^2,\quad \text{且}\quad 13=x+y.$ 所以 \[ 144=h^2+y^2=h^2+(13-x)^2=h^2+x^2+169-26x=25+169-26x, \] 从而得到 $x=50/26=25/13$，并且 \[ h=\sqrt{5^2-\left(\frac{25}{13}\right)^2}=5\sqrt{1-\frac{25}{169}}=5\sqrt{\frac{144}{169}}=\frac{60}{13}. \] 因此 \[ \text{Area}(ABCD)=\frac12(39+52)\cdot\frac{60}{13}=210. \]

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