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

Algebra of Limits

If a right an...

Question

If a right angled $△ A B C$ of maximum area is inscribed within a circle of radius $R$ , then

Δ = 2 R^{2}

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

\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} = \frac{\sqrt{2} + 1}{R}

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

r = (\sqrt{2} - 1) R

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

s = (1 + \sqrt{2}) R

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

Open in App

Solution

The correct option is C $\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} = \frac{\sqrt{2} + 1}{R}$
For a right-angled triangle inscribed in a circle of radius $R,$ the length of the hypotenuse is $2 R$

$∴$ the area is maximum when the triangle is isosceles with each side $= \sqrt{2} R$

$∴ s = \frac{1}{2} (2 \sqrt{2} + 2) R = (\sqrt{2} + 1) R$

$∴ △ = \frac{1}{2} \sqrt{2} R . \sqrt{2} R = R^{2} \Rightarrow \frac{1}{r} = \frac{(\sqrt{2} + 1)}{R}$

$∴ \frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} = \frac{s - a}{△} + \frac{s - b}{△} + \frac{s - c}{△} = \frac{s}{△} = \frac{1}{r} = \frac{(\sqrt{2} + 1)}{R}$

Suggest Corrections

Similar questions

Δ

r_{1}, r_{2}, r_{3}

\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} - \frac{1}{r}

A B C

R

△

s

△ A B C

- 1

Δ A B C

a < b < c

r

r_{1}, r_{2}, r_{3}

A, B, C

r < r_{1} < r_{2} < r_{3}

- 2

Δ A B C

r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = \frac{r_{1} r_{2} r_{3}}{r}

A B C

I

r

D, E, F

I

B C, C A

A B

r_{1}, r_{2} r_{3}

\frac{r_{1}}{r - r_{1}}_\frac{r_{2}}{r - r_{2}} + \frac{r_{3}}{r - r_{3}} = \frac{r_{1} r_{2} r_{3}}{(r - r_{1}) (r - r_{2}) (r - r_{3})}

Join BYJU'S Learning Program