A geometry problem by nishant kumar

Geometry Level 3

What is the area of triangle with side lengths a a , b b , and c c which is inscribed in a circle of radius R R ?

π R 2 a b c 2 \frac {\pi R^2 abc}2 a b c 2 R \frac{abc}{2R} a b c 4 R \frac{abc}{4R} π R 2 − a b c \pi R^2-abc

Nishant Kumar by nishant kumar , 21, India

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2 solutions

David Vreken
Jun 8, 2020

By the law of sines , a sin ⁡ A = b sin ⁡ B = c sin ⁡ C = d = 2 R \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d = 2R , so sin ⁡ C = c 2 R \sin C = \frac{c}{2R} .

The area of a triangle is therefore T = 1 2 a b sin ⁡ C = 1 2 a b ( c 2 R ) = a b c 4 R T = \frac{1}{2} ab \sin C = \frac{1}{2} ab (\frac{c}{2R}) = \boxed{\frac{abc}{4R}} .

1 Helpful 2 Interesting 1 Brilliant 0 Confused

Thanks, Sir!

Vinayak Srivastava - 1 year ago

Some more proofs . The last one is the one mentioned by @David Vreken

Mahdi Raza - 1 year ago

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Wow, nice proof! Actually I don't understand trignometry so your proof helps!

Vinayak Srivastava - 1 year ago

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Yeah, simple construction

Mahdi Raza - 1 year ago
Zakir Husain
Jun 8, 2020

Circumradius' formula: R = a b c 4 A R=\frac{abc}{4A} A = a b c 4 R \boxed{A=\frac{abc}{4R}} Note:

-In the formula above A A is the area of triangle

-If you need proof for the formula just ask in the comments

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I did the same.

Marvin Kalngan - 1 year ago

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