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

True

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

False

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
True

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

Suggest Corrections

Similar questions

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

△ A B C

△ B O 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

The area of an equilateral triangle with side 12cm is $36 \sqrt{3} c m^{2}$

The area of an equilateral triangle with side 12cm cannot be found.

Join BYJU'S Learning Program