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I have a quadrilateral containing two opposite right-angles and two known angles, and two equal known sides, thus:-

enter image description here

I am trying to calculate the area, which appears to be defined, as all the angles and two sides are known. But I can't figure out how to get it. None of the formulae listed here seem to apply.

I've also tried dividing it into two triangles, and also extending the short side to make a larger right triangle, but haven't been able to get anywhere; I keep introducing more ancillary elements than I can find equations for.

Could someone give me a hint?

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Draw the diagonal bisecting(check it ) angle $a$ , prove two triangles formed are congruent and then compute area of individual triangle and sum them.

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  • $\begingroup$ I see. The two triangles are mirror-images because they have the diagonal and edge d the same length and both have a right angle in the same place, so the short side upper left is the same length as the long side, etc...thanks for the tip. $\endgroup$ – Brian Hooper Sep 28 '12 at 7:43
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Hint: Extending sounds good. Extend the "top" side leftward. Then extend the long diagonal side labelled $d$ in the northwest direction. You get a big triangle and a little triangle inside it and similar to it.

The area and hypotenuse of the big triangle are easy, using trig functions of $a$. From this you can find the hypotenuse of the little triangle. That tells you the scaling factor that needs to be applied to the big triangle to get the little one. Then subtract.

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