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Standard XII

Mathematics

Algebra of Limits

ABC is an iso...

Question

ABC is an isosceles triangle inscribed in a circle of radius r. If $A B = A C$ and h is the altitude from A to BC. If the triangle ABC has perimeter $P$ and area $Δ$ then $lim h \to 0 512 r \frac{Δ}{P^{3}}$ is equal to

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Solution

We have $B C = 2 B D, A D = h$ and $O D = h - r$ .

$∴ B C = 2 \sqrt{r^{2} - (h - r)^{2}} = 2 \sqrt{2 h r - h^{2}}$

$\Rightarrow A B = \sqrt{2 h r - h^{2} + h^{2}} = \sqrt{2 h r}$

so that $P = 2 A B + B C$

$= 2 [\sqrt{2 h r - h^{2}} + \sqrt{2 h r}]$

Also the area of $Δ A B C$ is
$Δ = B D \cdot A D = h \sqrt{2 h r - h^{2}}$

$∴ \frac{Δ}{P^{3}} = \frac{h \sqrt{2 h r - h^{2}}}{8 (\sqrt{2 h r - h^{2}} + \sqrt{2 h r})^{3}}$

$= \frac{h \sqrt{2 h r - h}}{8 (\sqrt{2 h r - h} + \sqrt{2 r})^{3}}$

$\Rightarrow lim h \to 0 512 r \frac{Δ}{P^{3}} = 512 r \frac{\sqrt{2 r}}{8 (2 \sqrt{2 r})^{3}} = 4$

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ABC is an isosceles triangle inscribed in a circle of radius r. If AB=AC and h is the altitude from A to BC. If the triangle ABC has perimeter P and area Δ then limh→0512rΔP3 is equal to

We have BC=2BD,AD=h and OD=h−r.∴BC=2√r2−(h−r)2=2√2hr−h2⇒AB=√2hr−h2+h2=√2hrso that P=2AB+BC=2[√2hr−h2+√2hr]Also the area of ΔABC isΔ=BD⋅AD=h√2hr−h2∴ΔP3=h√2hr−h28(√2hr−h2+√2hr)3=h√2hr−h8(√2hr−h+√2r)3⇒limh→0512rΔP3=512r√2r8(2√2r)3=4

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