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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

ABC

is an isosceles triangle inscribed in a circle of radius

r

. If

AB=AC

and

h

is the altitude from

A

BC

and

P

be the perimeter of

ABC

then

\displaystyle\lim_{h\to 0}\dfrac{\Delta}{P^{3}}

is equal to (where

\Delta

is the area of the triangle)

\dfrac{1}{32r}

\dfrac{1}{64r}

\dfrac{1}{128r}

correct

DNone

Solution

Step 1: For an isosceles triangle inscribed in a circle of radius

r

with altitude

h

from

A

: using the chord-length formula, half-base is

\sqrt{h(2r-h)}

, so

BC=2\sqrt{h(2r-h)}=\sqrt{4h(2r-h)}=\sqrt{8rh-4h^{2}}

. Step 2:

AB=AC=\sqrt{h^{2}+\dfrac{BC^{2}}{4}}=\sqrt{h^{2}+(2rh-h^{2})}=\sqrt{2rh}

. Step 3: Perimeter

P=2\sqrt{2rh}+\sqrt{8rh-4h^{2}}=2\sqrt{2rh}+2\sqrt{2rh-h^{2}}

P=2\left[\sqrt{2rh}+\sqrt{2rh-h^{2}}\right]

Step 4: Area

\Delta=\dfrac{1}{2}\cdot BC\cdot h=\dfrac{1}{2}\cdot 2\sqrt{2rh-h^{2}}\cdot h=h\sqrt{2rh-h^{2}}

. Step 5: As

h\to 0

\sqrt{2rh-h^{2}}\sim\sqrt{2rh}

and

\sqrt{2rh}+\sqrt{2rh-h^{2}}\sim 2\sqrt{2rh}

, so

P\sim 4\sqrt{2rh}

. Step 6: Compute the limit:

\lim_{h\to 0}\dfrac{h\sqrt{2rh-h^{2}}}{P^{3}}=\lim_{h\to 0}\dfrac{h\sqrt{2rh}}{(4\sqrt{2rh})^{3}}=\lim_{h\to 0}\dfrac{h\sqrt{2rh}}{64\cdot(2rh)\sqrt{2rh}}

=\lim_{h\to 0}\dfrac{h}{64\cdot 2rh}=\dfrac{1}{128r}

Correct answer: (3)

\dfrac{1}{128r}

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