I need to know how to solve this problem. Can anyone help me?

In trapezium **ABCD**, **ADIIBC, AD<BC**,unparallel sides are equal.A circle with centre **O** is inscribed in the trapezium. **OAD** is equilateral. Find the radius of the circle if the area of the trapezium is **\(\frac{800} {\sqrt{3}}\)**

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## Comments

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TopNewestThis was a 2013 secodary level olympiad problem in Khulna

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You're right.....Good memory of course.......

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You qualified that time?

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\(\frac { 20 }{ \sqrt { 3 } }\) should be the answer. (by the way, I wrote up a solution and God knows where I made the mistake in LATEXing. Everything started looking like Hebrew :P )

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I really don't know. Do u know how to solve it?

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drop perpendiculars from A & D to BC and call the feet M and N respectively. Let r be the radius and AD = a Say, the circle touches AB . BC , CD , DA at P, Q, R, S respectively, Here AM = 2r and AS = MQ = a/2

Now, OS is the height of equilateral triangle AOD , so OS =r = (root3)a/2 and, AB = AP + PB = AS + BQ = AS + BM +MQ = a + BM [ as AS = MQ = a/2 ] now use Pythagoras in ABM and you'll get a relation between BM and r

so, area of ABCD = ABM + DNC + AMND = 2ABM + AMND [since AMB and DNC are congruent ] Solve the equation and you'll get the answer

(Hope you'll understand..... I messed up again trying to LATEX :P )

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\(\dfrac{10\sqrt{6}}{3}\)?

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I really don't know. Do u know how to solve it?

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