Quadrilateral Missing Length

In quadrilateral $ABCD$ , it is given that $AD = 8$ , $DC = 12$ , $CB = 10$ , and $\angle A = \angle B = 60^\circ$ . Find the length of $AB$ . Generalize.

#Geometry #Quadrilaterals

Easy Math Editor

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Comments

image link: http://s29.postimg.org/n3gl6sjl3/Untitled.png

Let $AD$ and $BC$ intersect at $E.$ Notice that $\triangle EAB$ is equilateral. If $AB=x,$ we have $CE = x-10$ and $DE= x-8$ . By Cosine rule on $\triangle ECD$ , $12^2 = (x-8)^2 + (x-10)^2 - 2 \cdot \dfrac{1}{2} \cdot (x-8)(x-10) .$ Solving yields $x= 9+\sqrt{141}$ .

The generalization attempt should work out similarly-- denote $AD \cap BC = X,$ use sine rule to find the ratios $\dfrac{CX}{XB}$ and $\dfrac{DX}{AX},$ and then use cosine rule on $\triangle XAB.$ I'm too lazy to carry out the details at the moment.

Sreejato Bhattacharya - 6 years, 8 months ago

Correct; nice solution!

Ahaan Rungta - 6 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$