AMC8 1997

AMC8 1997 · Q15

AMC8 1997 · Q15. It mainly tests Area & perimeter, Coordinate geometry.

Each side of the large square in the figure is trisected. The corners of an inscribed square are at these trisection points, as shown. The ratio of the area of the inscribed square to the area of the large square is

图中大正方形的每条边被三分等分。内接正方形的顶点位于这些三分点上，如图所示。内接正方形面积与大正方形面积之比为

(A) $\frac{\sqrt{3}}{3}$ $\frac{\sqrt{3}}{3}$

(B) $\frac{5}{9}$ $\frac{5}{9}$

(D) $\frac{\sqrt{5}}{3}$ $\frac{\sqrt{5}}{3}$

(E) $\frac{7}{9}$ $\frac{7}{9}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Taking the side of the large square to be 3 inches gives an area of 9 square inches. Each of the four right triangles has an area of $\frac{1}{2}(2)(1)=1$ sq. inch. The area of the inscribed square is the area of the large square minus the area of the four right triangles, that is, $9-4(1)=5$ sq. inches. The desired ratio is $\frac{5}{9}$.

答案（B）：设大正方形的边长为 3 英寸，则面积为 9 平方英寸。四个直角三角形中每一个的面积为 $\frac{1}{2}(2)(1)=1$ 平方英寸。内接正方形的面积等于大正方形的面积减去四个直角三角形的面积，即 $9-4(1)=5$ 平方英寸。所求比值为 $\frac{5}{9}$。

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