# 2006 iTest Problems/Problem U2

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The following problem is from the Ultimate Question of the 2006 iTest, where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted.

## Problem

Points $A$ and $B$ lie on a circle centered at $O$ such that $\angle AOB$ is right. Points $C$ and $D$ lie on radii $OA$ and $OB$ respectively such that $AC = 7$, $CD = 5$, and $BD = 6$. Determine the area of quadrilateral $ACDB$. $[asy] draw(circle((0,0),10)); draw((0,10)--(0,0)--(10,0)--(0,10)); draw((0,3)--(4,0)); label("O",(0,0),SW); label("C",(0,3),W); label("A",(0,10),N); label("D",(4,0),S); label("B",(10,0),E); [/asy]$

## Solution

Let $x$ be the length of $CO$. The radius of the circle is $7+x$, so the length of $DO$ is $x+1$. By the Pythagorean Theorem, \begin{align*} x^2 + (x+1)^2 &= 25 \\ x^2 + x^2 + 2x + 1 &= 25 \\ 2x^2 + 2x - 24 &= 0 \\ x^2 + x - 12 &= 0 \\ (x+4)(x-3) &= 0 \end{align*} Since lengths must be positive, $x = 3$. The area of $ABCD$ equals the area of $AOB$ minus the area of $COD$, so $[ABCD] = \tfrac12 \cdot 10^2 - \tfrac12 \cdot 3 \cdot 4 = \boxed{44}$.