AMC 8 2002 Problem 20

AMC8 2002 年第 20 题

三角形★★★☆☆

The area of triangle

XYZ

is 8 square inches. Points

A

and

B

are midpoints of congruent segments

\overline{XY}

and

\overline{XZ}

. Altitude

\overline{XC}

bisects

\overline{YZ}

. The area (in square inches) of the shaded region is

三角形

XYZ

的面积是 8 平方英寸。点

A

和点

B

分别是两条相等线段

\overline{XY}

和

\overline{XZ}

的中点，高

\overline{XC}

平分

\overline{YZ}

。那么阴影部分的面积是多少平方英寸？

正确答案 · Correct Answer

解析 · Solution

Solution 1

We know the area of triangle

XYZ

is 8 square inches. The area of a triangle can also be represented as

\displaystyle \frac{bh}{2}

or in this problem

\displaystyle \frac{XC\cdot YZ}{2}

. By solving, we have

\displaystyle \frac{XC\cdot YZ}{2} = 8,

XC\cdot YZ = 16.

With SAS congruence, triangles

XCY

and

XCZ

are congruent. Hence, triangle

\displaystyle XCY = \frac{8}{2} = 4

. (Let's say point

D

is the intersection between line segments

XC

and

AB

.) We can find the area of the trapezoid

ADCY

by subtracting the area of triangle

XAD

from 4. We find the area of triangle

XAD

by the

\displaystyle \frac{bh}{2}

formula-

\displaystyle \frac{XD\cdot AD}{2} = \frac{\frac{XC}{2} \cdot AD}{2}

AD

\displaystyle \frac{1}{4}

YZ

from solution 1. The area of

XAD

\displaystyle \frac{\frac{XC}{2} \cdot \frac{YZ}{4}}{2} = \frac{16}{16} = 1

. Therefore, the area of the shaded area- trapezoid

ADCY

has area

4-1 = \text{(D)3}

. - sarah07

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