MyQuestionIcon

1

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

Convexity

lf A1, A2, ...

Question

lf $A_{1}, A_{2}, A_{3}, A$ (in-circle area) are respectively the areas of the ex-circles and in-circle of the $Δ A B C$ , then $\frac{1}{\sqrt{A_{1}}} + \frac{1}{\sqrt{A_{2}}} + \frac{1}{\sqrt{A_{3}}} =$

A

\frac{2}{\sqrt{A}}

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

B

\frac{1}{\sqrt{A}}

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

C

\frac{1}{2 \sqrt{A}}

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

D

\sqrt{A}

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

Open in App

Solution

The correct option is B $\frac{1}{\sqrt{A}}$
$A_{1} = π r_{1}^{2}, A_{2} = π r_{2}^{2}, A_{3} = π r_{3}^{2}, A = π r^{2}$
$= \frac{1}{\sqrt{π}} (\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}})$
$= \frac{1}{\sqrt{π}} (\frac{(s - a) + (s - b) + (s - c)}{Δ})$
$= \frac{1}{\sqrt{π}} (\frac{s}{Δ})$
$= \frac{1}{\sqrt{π} r}$
$= \frac{1}{\sqrt{π r^{2}}}$
$= \frac{1}{\sqrt{A}}$ .

Suggest Corrections

thumbs-up

1

Similar questions

A, A_{1}, A_{2}, A_{3}

be the area of the in-circle and ex-circles, then

\frac{1}{\sqrt{A_{1}}} + \frac{1}{\sqrt{A_{2}}} + \frac{1}{\sqrt{A_{3}}}

A, A_{1}, A_{2}, A_{3}

be the areas of the incircle and exircles, then

\frac{1}{\sqrt{A_{1}}} + \frac{1}{\sqrt{A_{2}}} + \frac{1}{\sqrt{A_{3}}} = \frac{k}{\sqrt{A}}

k =

A_{1}, A_{2},

A_{3}

be the regions on

R^{2}

A_{1} = {(x, y) : x \geq 0, y \geq 0, 2 x + 2 y - x^{2} - y^{2} > 1 > x + y}, A_{2} = {(x, y) : x \geq 0, y \geq 0, x + y > 1 > x^{2} + y^{2}}, A_{3} = {(x, y) : x \geq 0, y \geq 0, x + y > 1 > x^{3} + y^{3}} .

| A_{1} |, | A_{2} |,

| A_{3} |

the areas of the regions

A_{1}, A_{2}

A_{3}

respectively. Then

a_{1}, a_{2}, a_{3}, a_{4}

are the coefficients of

2^{n d}, 3^{r d}, 4^{t h}

5^{t h}

terms in the expansion of

(1 + x)^{n}

respectively, then

\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}}

A, A_{1}, A_{2}, A_{3}

be the area of the incircle and excircles, then

\frac{1}{{\sqrt{A}}_{1}} + \frac{1}{{\sqrt{A}}_{2}} + \frac{1}{{\sqrt{A}}_{3}}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Convexity and Concavity

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon