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 **\(\frac{800} {\sqrt{3}}\)**

## Comments

Sort by:

TopNewestThis was a 2013 secodary level olympiad problem in Khulna – Rifath Rahman · 1 year, 9 months ago

Log in to reply

– Partho Kunda · 1 year, 9 months ago

You're right.....Good memory of course.......Log in to reply

– Rifath Rahman · 1 year, 9 months ago

You qualified that time?Log in to reply

– Partho Kunda · 1 year, 9 months ago

Yeah! 1st Runners up.....I qualified in the regional for 4 times.....Log in to reply

– Rifath Rahman · 1 year, 9 months ago

This will be my 1st timeLog in to reply

\(\dfrac{10\sqrt{6}}{3}\)? – Krishna Sharma · 1 year, 9 months ago

Log in to reply

– Partho Kunda · 1 year, 9 months ago

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

\(\frac { 20 }{ \sqrt { 3 } }\) 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 ) – Homo Sapiens · 1 year, 9 months ago

Log in to reply

– Partho Kunda · 1 year, 9 months ago

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

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 ) – Homo Sapiens · 1 year, 9 months ago

Log in to reply

– Partho Kunda · 1 year, 9 months ago

why AP+PB=AS+BQLog in to reply

– Homo Sapiens · 1 year, 9 months ago

If two tangents are drawn to a circle from an external point, the distances from that point to the points of contact are equal.Log in to reply

– Partho Kunda · 1 year, 9 months ago

oh! I forgot it.....shit.......and thanks......Log in to reply