Question

Each side of the large square in the figure is trisected (divided into three equal parts). 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

$\dfrac{\sqrt{3}}{3}$

$\dfrac{5}{9}$

$\dfrac{2}{3}$

$\dfrac{\sqrt{5}}{3}$

$\dfrac{7}{9}$

Solution:

Let $3x$ be the side length of the large square. Then we can find the side length of the inner square via $\sqrt{(2x)^2 + x^2} = \sqrt{5x^2} = x\sqrt{5}$ from the Pythagorean Theorem.

The area of the larger square is $(3x)^2 = 9x^2$ and that of the inner square is $(x\sqrt{5})^2 = 5x^2.$

The ratio of the areas is then $\dfrac{5}{9}.$

Thus, B is the correct answer.

1997 AMC 8 Problem 15

Solution:

Problem 15 in Other Years