2022 AMC 10A Problem 10

Problem 10 of 25EasierAlgebra Geometry

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?

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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)

18.

Thus, E is the correct answer.