In a triangle ABC,AB = AC = 37. Let D be a po | vedclass

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(C) Let $E$ be the foot of the perpendicular from $A$ to $BC$. Since $	riangle ABC$ is isosceles with $AB = AC$,$E$ is the midpoint of $BC$. Let $BE

Vedclass · Accepted Answer

$40$

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(C) Let $E$ be the foot of the perpendicular from $A$ to $BC$. Since $	riangle ABC$ is isosceles with $AB = AC$,$E$ is the midpoint of $BC$. Let $BE = EC = x$. Then $BD = x - DE = 7$ and $CD = x + DE$. In $	riangle ADE$,$AE^2 = AD^2 - DE^2 = 33^2 - DE^2$. In $	riangle ABE$,$AE^2 = AB^2 - BE^2 = 37^2 - x^2$. Equating the two expressions for $AE^2$: $33^2 - DE^2 = 37^2 - x^2$ $x^2 - DE^2 = 37^2 - 33^2$ $(x - DE)(x + DE) = (37 - 33)(37 + 33)$ Since $BD = x - DE = 7$,we have: $7 \cdot CD = 4 \cdot 70$ $7 \cdot CD = 280$ $CD = 40$.

In a $\triangle ABC$,$AB = AC = 37$. Let $D$ be a point on $BC$ such that $BD = 7$ and $AD = 33$. The length of $CD$ is:

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