AMC8 2024

AMC8 2024 · Q3

AMC8 2024 · Q3. It mainly tests Area & perimeter, Polygons.

Four squares of side length $4, 7, 9,$ and $10$ are arranged in increasing size order so that their left edges and bottom edges align. The squares alternate in color white-gray-white-gray, respectively, as shown in the figure. What is the area of the visible gray region in square units?

四个边长分别为 $4、7、9$ 和 $10$ 的正方形按从小到大的顺序排列，使得它们的左边缘和底边缘对齐。这些正方形颜色交替为白-灰-白-灰，如图所示。可见灰色区域的面积是多少平方单位？

(A) \ 42 \ 42

(B) \ 45 \ 45

(D) \ 50 \ 50

(E) \ 52 \ 52

Answer

Correct choice: (E)

正确答案：（E）

Solution

We work inwards. The area of the outer shaded square is the area of the whole square minus the area of the second largest square. The area of the inner shaded region is the area of the third largest square minus the area of the smallest square. The sum of these areas is \[10^2 - 9^2 + 7^2 - 4^2 = 100 - 81 + 49 - 16 = 19 + 33 = \boxed{\textbf{(E)}\ 52}\] This problem appears multiple times in various math competitions including the AMC and MATHCOUNTS.

我们向内计算。最外层阴影正方形的面积是整个正方形的面积减去第二大正方形的面积。最内层阴影区域的面积是第三大正方形的面积减去最小正方形的面积。这些面积之和为 \[10^2 - 9^2 + 7^2 - 4^2 = 100 - 81 + 49 - 16 = 19 + 33 = \boxed{\textbf{(E)}\ 52}\] 这道题出现在各种数学竞赛中，包括 AMC 和 MATHCOUNTS。

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