1967 AHSME Problems/Problem 32

In quadrilateral $ABCD$ with diagonals $AC$ and $BD$ , intersecting at $O$ , $BO=4$ , $OD = 6$ , $AO=8$ , $OC=3$ , and $AB=6$ . The length of $AD$ is: $\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$

$\sqrt{166}$

After drawing the diagram, we see that we actually have a lot of lengths to work with. Considering triangle ABD, we know values of $AB, BD(BD = BO + OD)$ , but we want to find the value of AD. We can apply stewart's theorem now, letting $m = 4, n = 6, AD = X, AB = 6$ , and we have $10 \cdot 6 \cdot 4 + 8 \cdot 8 \cdot 10 = x^2 + 36 \cdot 6$ , and we see that $x = \sqrt{166}$ , $\boxed{E \sqrt{166}}$

1967 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 31	Followed by Problem 33
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1967 AHSME Problems/Problem 32

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