ABC is an isosceles triangle inscribed in a circle of radius r- If AB - AC and h is altitude from A to BC- then lim h - 0- p 3 is equal to where is the area and P is the perimeter of the triangle ABC A- 1 - 32 rB- 1 - 64 rC- 1 - 128 rD- 1 - 216 r

The correct option is C 1/128r ∵ AD = h ∴ OD = (h r) Then, BC = 2DC = 2√(r^2 (h r)^2) ∴ BC = 2√(2hr h^2) = 1/2.BC.h = h√((2hr h^2)) = h^3/2√((2r h)) and AB = ...

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Byju's Answer

Standard XII

Mathematics

Factorization Method Form to Remove Indeterminate Form

ABC is an iso...

Question

ABC is an isosceles triangle inscribed in a circle of radius r. If AB=AC and h is altitude from A to BC, then $lim h \to 0 \frac{△}{p^{3}}$ is equal to (where $△$ is the area and P is the perimeter of the triangle ABC)

1/32r

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1/64r

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1/128r

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1/216r

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Solution

The correct option is C
1/128r

$∵$ AD = h

$∴$ OD = (h-r)

Then, BC = 2DC = $2 \sqrt{r^{2} - (h - r)^{2}}$

$∴$ BC = $2 \sqrt{2 h r - h^{2}}$

$△$ = $\frac{1}{2} . B C . h$ = $h \sqrt{(2 h r - h^{2})}$

= $h^{3 / 2} \sqrt{(2 r - h)}$

and AB = AC = $\sqrt{h^{2} + (D C)^{2}}$ = $\sqrt{h^{2} + r^{2} - (h - r)^{2}}$

= $\sqrt{2 h r}$

$∴$ P = AB + BC + CA

= $2 \sqrt{2 h r} + 2 \sqrt{(2 h r - h^{2})}$

= $2 (h)^{1 / 2} (\sqrt{2} r + \sqrt{2 r - h})$

Then, $lim h \to 0 \frac{△}{p^{3}}$ = $lim h \to 0 \frac{h^{3 / 2 \sqrt{(2 r - h)}}}{8 h^{3 / 2} (\sqrt{2 r} + \sqrt{2 r - h})^{3}}$

= $\frac{\sqrt{2 r}}{8 (2 \sqrt{2 r})^{3}}$ = $\frac{1}{128 r}$

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ABC is an isosceles triangle inscribed in a circle of radius r. If AB=AC and h is altitude from A to BC, then limh→0△p3 is equal to (where △ is the area and P is the perimeter of the triangle ABC)

ABC is an isosceles triangle inscribed in a circle of radius r. If AB=AC and h is altitude from A to BC, then $lim h \to 0 \frac{△}{p^{3}}$ is equal to (where $△$ is the area and P is the perimeter of the triangle ABC)