You visited us 1 times! Enjoying our articles? Unlock Full Access!

Sine Rule

In a △ ABC,...

Question

In a $△ A B C,$ let $∠ C = \frac{π}{2} .$ If r and R are the inradius and the circumradius, respectively, of the triangle then $2 (r + R)$ is equal to

a + b

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

b + c

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

c + a

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

a + b + c

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

Open in App

Solution

The correct option is A $a + b$
In $Δ A B C$
$∠ C = \frac{π}{2}$
Therefore, $a^{2} + b^{2} = c^{2}$
$\Rightarrow 2 a b = {(a + b)}^{2} - c^{2} = (a + b + c) (a + b - c) = 2 s (a + b - c)$ ....(1)

Now, $y = 2 (r + R) = \frac{2 △}{s} + c sin C = \frac{a b sin C}{s} + c sin C$
$\Rightarrow y = a + b - c + c = a + b$ [ from (1) ]

Suggest Corrections

Similar questions

△ A B C

∠ C = \frac{π}{2}

r

R

2 (r + R)

△ A B C

∠ C = \frac{π}{2}

r

R

2 (r + R)

In a triangle ABC, Let , $∠ C = \frac{π}{2}$ , if r is the inradius and R is the circumradius of the triangle ABC, then 2(r+R) equals

In a $Δ$ ABC, let $∠$ C = $π / 2$ .If r is the inradius and R is the circumradius of the triangle, then 2 (r +R) is equal to

A B C

∠ C = \frac{π}{2}

r

R

A B C

2 (r + R)

Join BYJU'S Learning Program