2022 AMC 10A Problems/Problem 10

Problem

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 centimeters, as shown below. What is the area of the original index card?

$\textbf{(A) }14 \qquad \textbf{(B) }10 \sqrt{2}$ $\qquad \textbf{(C) }16 \qquad \textbf{(D) }12 \sqrt{2}$ $\qquad \textbf{(E) }18$

Solution 1(Simple coordinates and basic algebra)

$[asy] size(8cm); draw((0,0)--(6,0)); draw((0,0)--(0,3)); draw((0,3)--(6,3)); draw((6,0)--(6,3)); draw((0,1)--(1,1)); draw((1,1)--(1,0)); draw((5,2)--(6,2)); draw((5,2)--(5,3)); label("$A$",(0,0),W); label("$D$",(6,3),E); label("$E$",(0,1),W); label("$F$",(1,0),S); [/asy]$

We will use coordinates here. Label the bottom left corner of the larger rectangle(without the square cut out) as $A(0,0)$ and the top right as $D(w,l)$ , where $w$ is the width of the rectangle and $l$ is the length. Now we have vertices $E(0,1)$ , $F(1,0)$ , $G(w-1,l)$ , and $H(w,l-1)$ as vertices of the irregular octagon created by cutting out the squares. Label $I(1,1)$ and $J(w-1, l-1)$ as the two closest vertices formed by the squares. The distance between the two closest vertices of the squares is thus $IJ$ = ( $4$ $\sqrt{2})^2.$ Substituting, we get

$(IJ)^2 = (w-2)^2 + (l-2)^2 = (4\sqrt{2})^2 = 32.$

Using the fact that the diagonal of the rectangle = $8$ , we get

$w^2+l^2 = 64$ .

Subtracting $(2)$ from $(1)$ and simplifying, we get $w+l$ = $10$ . Squaring yields $w^2 + 2wl + l^2$ = $100$ and thus area of the original rectangle = $wl$ = $18$ = $\boxed{\textbf{(E) } 18}.$