AMC12 2008 B

AMC12 2008 B · Q18

AMC12 2008 B · Q18. It mainly tests Fractions, Triangles (properties).

A pyramid has a square base $ABCD$ and vertex $E$. The area of square $ABCD$ is $196$, and the areas of $\triangle ABE$ and $\triangle CDE$ are $105$ and $91$, respectively. What is the volume of the pyramid?

一个棱锥的底面是正方形 $ABCD$，顶点为 $E$。正方形 $ABCD$ 的面积为 $196$，并且 $\triangle ABE$ 与 $\triangle CDE$ 的面积分别为 $105$ 和 $91$。该棱锥的体积是多少？

(A) 392 392

(B) 196\sqrt{6} 196\sqrt{6}

(D) 392\sqrt{3} 392\sqrt{3}

(E) 784 784

Answer

Correct choice: (E)

正确答案：（E）

Solution

Let $h$ be the height of the pyramid and $a$ be the distance from $h$ to $CD$. The side length of the base is $14$. The heights of $\triangle ABE$ and $\triangle CDE$ are $2\cdot105\div14=15$ and $2\cdot91\div14=13$, respectively. Consider a side view of the pyramid from $\triangle BCE$. We have a systems of equations through the Pythagorean Theorem: $13^2-(14-a)^2=h^2 \\ 15^2-a^2=h^2$ Setting them equal to each other and simplifying gives $-27+28a=225 \implies a=9$. Therefore, $h=\sqrt{15^2-9^2}=12$, and the volume of the pyramid is $\frac{bh}{3}=\frac{12\cdot 196}{3}=\boxed{784 \Rightarrow E}$.

设 $h$ 为棱锥的高，设 $a$ 为从高的垂足到 $CD$ 的距离。底面边长为 $14$。$\triangle ABE$ 和 $\triangle CDE$ 的高分别为 $2\cdot105\div14=15$ 和 $2\cdot91\div14=13$。从 $\triangle BCE$ 的侧视图考虑，用勾股定理得到方程组： $13^2-(14-a)^2=h^2 \\ 15^2-a^2=h^2$ 令两式相等并化简得 $-27+28a=225 \implies a=9$。因此 $h=\sqrt{15^2-9^2}=12$，棱锥体积为 $\frac{bh}{3}=\frac{12\cdot 196}{3}=\boxed{784 \Rightarrow E}$。

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