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AMC10 2001 Test Paper
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Sample Question 24:

In trapezoid \(ABCD\), \(\overline{AB}\) and \(\overline{CD}\) are perpendicular to \(\overline{AD}\), with \(AB+CD=BC\), \(AB<CD\), and \(AD=7\). What is \(AB\cdot CD\)?

\(\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 12.25 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 12.75 \qquad \textbf{(E)}\ 13\)




Answer Keys

Question 24: B


Solutions

Question 24
Step 1: We are given trapezoid \(ABCD\) where \(AB\) and \(CD\) are perpendicular to \(AD\), \(AB < CD\), \(AB + CD = BC\) and \(AD=7\). To find \(AB\cdot CD\), let the length of \(AB = x\) and \(CD = y\). That means \(AB + CD = BC = x + y\) .

Step 2: By the Pythagorean theorem, we can set up the equation \((x+y)^2 = (y-x)^2 + 49\). This simplifies to \(x^2 + 2xy + y^2 = y^2 - 2xy + x^2 + 49\).

Step 3: Cancel out like terms from both sides of the equation. This leaves us with \(4xy = 49\).

Step 4: Finally, dividing both sides by 4, we find that \(xy = 12.25\).

Step 5: Checking answer options, we find \(xy = 12.25\) is answer choice \(\textbf{(B)}\ 12.25\). So, the product of \(AB\) and \(CD\) is \(\textbf{(B)}\ 12.25\).