Question

Daniel finds a rectangular index card and measures its diagonal to be $8$ centimeters. Daniel then cuts out equal squares of side $1$ cm at two opposite corners of the index card and measures the distance between the two closest vertices of these squares to be $4 \sqrt{2}$ centimeters, as shown below. What is the area of the original index card?

$14$

$10 \sqrt{2}$

$16$

$12 \sqrt{2}$

$18$

Solution:

We can label $a$ and $b$ as the width and height as in the diagram. Then we get that $a^2 + b^2 = 64$ and $(a - 2)^2 + (b - 2)^2 = 32.$

The latter expression simplifies to $a^2 + b^2 - 4a - 4b + 4 + 4 = 32,$ which is the same as $72 - 4(a + b) = 32.$ From this we get $a + b = 10.$

Squaring this, we get $a^2 + b^2 + 2ab = 100,$ which gets us that $2ab = 36,$ which means that the area $(ab)$ is $18.$

Thus, E is the correct answer.

2022 AMC 10A Problem 10

Solution:

Problem 10 in Other Years