A geometry problem by Ronak Singha

Geometry Level pending

In the figure below points A, B, C and D are on a circle. Point O is the intersection of chords AC and BD. The area of triangle BOC is 15; the lenght of AO is 10 and the length of OB is 5. What is the area of triangle AOD?


The answer is 60.

Ronak Singha by Ronak Singha , 27, India

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2 solutions

Nguyen Thanh Long
Jul 23, 2015

According to the theory for the relation between two chords that intersect in a circle that we have: O A × O C = O B × O D → O A O B = O C O D = 2 OA \times OC = OB \times OD \rightarrow \frac{OA}{OB}=\frac{OC}{OD}=2 [ A O D ] = 4 × [ B O C ] = 60 [AOD] = 4 \times [BOC] = \boxed{60}

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As △ D O A ∼ △ C O B \bigtriangleup DOA \sim \bigtriangleup COB , the proportionality constant is 2 2 , therefore the area is 15 ⋅ 2 2 = 60 15\cdot 2^2 = 60

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