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In a quadrilateral $ABCD, BC =10, CD = 14$ and $AD = 12$, it is also known that $\angle A=\angle B=60^\circ$. Now, if $AB = a + \sqrt{b}, \text{ where } a,b \in \mathbb{N}, $ how could we find $a+b$?

I am getting 210, is this correct?

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What are $a$ and $b$? Merely positive numbers that are defined as $AB=a+\sqrt b$? – user21436 Mar 3 '12 at 13:15
@Kannappan: It is not mentioned in the problem, I am guessing that it should be positive. – Quixotic Mar 3 '12 at 13:19
up vote 2 down vote accepted

I will assume that $A$, $B$, $C$, $D$ go counterclockwise in that order. A diagram should go with this solution, but you will have to supply it.

Drop perpendiculars fom $D$ to $P$ on $AB$, from $C$ to $Q$ on $AB$. Easily we have by trigonometry that $AP=6$ and $BQ=5$. The only thing we need to do is to find $PQ$.

Drop a perpendicular from $C$ to $R$ on $DP$. Note that by trigonometry $DP=12\frac{\sqrt{3}}{2}=6\sqrt{3}$. Similarly, $CQ=5\sqrt{3}$, and therefore $DR=\sqrt{3}$.

It follows (suggestion by @Foool, replacing a more awkward calculation) that by the Pythagorean Theorem, $CR=\sqrt{193}$. But $CR =QP$. Thus $AB=11+\sqrt{193}$, and the desired sum is $204$, if we need $a$ and $b$ to be positive integers.

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