Relation between Exradius ,semiperimeter and circumradius can be given by ( r 1 is the radius of the circle opposite the angle A) Options: A . &nbsp r 1 = 2 Δ s − a , r 1 = 4 R c o s A 2 s i n B 2 s i n C 2 B . &nbsp r 1 = Δ s − a , r 1 = 4 R s i n A 2 c o s B 2 c o s C 2 C . &nbsp r 1 = 2 Δ ( s − a ) ( s − b ) , r 1 = 4 R c o s A 2 s i n B 2 s i n C 2 D . &nbsp r 1 = Δ ( s − a ) ( s − b ) ( s − c ) , r 1 = 4 R c o s A 2 s i n B 2 s i n C 2

Answer: Option B : B Relation between Exradius and semiperimeter can be given by, r 1 = Δ s − a where r 1 is the radius of the circle opposite the angle A. and it's helpful to remember the identity r 1 = 4 R s i n A 2 c o s B 2 c o s C 2

Relation between Exradius ,semiperimeter and circumradius can be given by (r1

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Question -

Relation between Exradius ,semiperimeter and circumradius can be given by
( $r_{1}$ is the radius of the circle opposite the angle A)

Options:

A .

r_{1} = \frac{2 Δ}{s - a}, r_{1} = 4 R c o s \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2}

B .

r_{1} = \frac{Δ}{s - a}, r_{1} = 4 R s i n \frac{A}{2} c o s \frac{B}{2} c o s \frac{C}{2}

C .

r_{1} = \frac{2 Δ}{(s - a) (s - b)}, r_{1} = 4 R c o s \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2}

D .

r_{1} = \frac{Δ (s - a)}{(s - b) (s - c)}, r_{1} = 4 R c o s \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2}

Answer: Option B
:
B
Relation between Exradius and semiperimeter can be given by,

r_{1} = \frac{Δ}{s - a}

where

r_{1}

is the radius of the circle opposite the angle A. and it's helpful to remember the identity

r_{1} = 4 R s i n \frac{A}{2} c o s \frac{B}{2} c o s \frac{C}{2}

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Discussion Forum : Solution Of Triangles

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