In the below figure, $△ A B C$ is equilateral with point O as its circumcentre. Here, area of $△ B O C$ is equal to area of ____.

△ A B C

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\frac{1}{3} △ A B C

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\frac{1}{2} △ A B C

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\frac{1}{4} △ A B C

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Solution

The correct option is B $\frac{1}{3} △ A B C$
Consider $△ A O B, △ A O C a n d △ B O C$
$A B = B C = A C$ (sides of equilateral triangle)
$A O = B O = C O$ (Radii of circumcircle)
$∴ △ A O B ≅ △ A O C ≅ △ B O C$ (SSS congruence)
$∴ A r e a o f △ A O B = A r e a o f △ A O C = A r e a o f △ B O C$
But $A r e a o f △ A O B + A r e a o f △ A O C + A r e a o f △ B O C = A r e a o f △ A B C$
$∴ A r e a o f △ B O C = \frac{1}{3} A r e a o f △ A B C$

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