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Geometry Level 2

Rhombus OABC is drawn inside a circle whose centre is O in such a way that vertices A,B and C of rhombus are on circle. If area of rhombus is 32\sqrt{3} then radius of circle is


The answer is 8.

Aditya Todkar by Aditya Todkar , 24, India

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1 solution

Aayush Patni
Dec 1, 2014

O is center of circle. Let radius of circle = r . Therefore OA=AB=BC=OC=r ( in a rhombus all sides are equal. Also one of the diagonals= r. Let OB=r Let the diagonals OB and AC intersect at D. In triangle OAD ,

By Pythagoras theorem

OA^2= OD^2+AD^2

r^2= (r/2)^2+ AD^2 ( OD=1/2 OB)

AD= (3/4)^1/2 r therefore AC=2AD=(3)^1/2 r

Area of rhombus= 1/2 * OB*AC

32 (3)^1/2= 1/2 * r (3)^1/2 *r

r=8

Sorry for the presentation.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

It would be good if you could change the wording and use 32 3 32\sqrt3 , that is put together the following symbols: \ ( 32 \ sqrt 3 \ ), so it would look right. I first thought that it was supposed to mean 32 3 \frac{32}{\sqrt3} and it took me quite a while to find something that would give me an integer answer. Thank you.

Marta Reece - 3 years, 11 months ago

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