$In the given figure, Δ A B C is equilateral with point O as its circumcentre. Here, area of Δ B O C = \frac{1}{3} area of Δ A B C$ .

True

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False

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Solution

The correct option is A True
$Concider Δ A O B, Δ A O C and Δ B O C . A B = B C = A C (Sides of equilateral triangle) A O = B O = C O (Radii of circumcircle) ∴ Δ A O B ≅ Δ B O C ≅ Δ C O B (SSS congruence) . ∴ Area of Δ A O B = Area of Δ B O C = Area of Δ A O C . But Area of Δ A O B + Area of Δ B O C + Area of Δ A O C = Area of Δ A B C . ∴ Area of Δ B O C = \frac{1}{3} Area of Δ A B C$
$Hence, the given statement is true.$

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Similar questions

In the given figure, O is the circumcentre of an equilateral triangle ABC. Here, area of $Δ B O C = \frac{1}{3} area of Δ A B C$ .

In the given figure, Δ A B C is equilateral with point O as its circumcentre. Here, area of Δ B O C = \frac{1}{3} area of Δ A B C

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Area of the given quadrilateral can be written as Area of ΔABC +Area of ΔACD

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In the given figure, ΔABC is equilateral with point O as its circumcentre. Here, area of ΔBOC=13 area of ΔABC.

$In the given figure, Δ A B C is equilateral with point O as its circumcentre. Here, area of Δ B O C = \frac{1}{3} area of Δ A B C$ .