First slide

Introduction to limits

Question

$A B C i s a n i s o s c e l e s t r i a n g l e i n s c r i b e d i n a c i r l e o f r a d i u s r . I f A B = A C a n d h i s t h e a l t i t u d e f r o m A t o B C, t h e n t h e t r i a n g l e$ ABC has perimeter P = $2 [\sqrt{(2 h r - h^{2})} + \sqrt{2 h r}]$ and area A then $\lim_{h \to 0} \frac{A}{P^{3}}$

Moderate

A

$\frac{1}{r}$
B

$\frac{1}{64 r}$
C

$\frac{1}{128 r}$
D

$\frac{1}{2 r}$

Solution

In $∆$ ABC, AB = AC AD $⊥$ BC (D is mid point of BC) Let r = radius of circumcircle $\Rightarrow$ OA =OB =OC =r Now BD = $\sqrt{B O^{2} - O D^{2}} = \sqrt{r^{2} - {(h - r)}^{2}} = \sqrt{2 r - h^{2}} \Rightarrow B C = 2 \sqrt{2 r h - h^{2}}$ $∴ A r e a o f △ A B C = \frac{1}{2} \times B C \times A D = h \sqrt{2 r h - h^{2}}$ Also $\lim_{h \to 0} \frac{A}{p^{3}} = \frac{h \sqrt{2 r h - h^{2}}}{8 {(\sqrt{2 r h - h^{2}} + \sqrt{2 h r})}^{3}}$ $= \lim_{h \to 0} \frac{h^{3 / 2} \sqrt{2 r - h}}{8 h^{3 / 2} {(\sqrt{2 r - h} + \sqrt{2 r})}^{3}}$ $= \lim_{h \to 0} \frac{\sqrt{2 r - h}}{8 {(\sqrt{2 r - h} + \sqrt{2 r})}^{3}}$ = $\frac{\sqrt{2 r}}{8 {(\sqrt{2 r} + \sqrt{2 r})}^{3}} = \frac{\sqrt{2 r}}{8.8.2 r . \sqrt{2 r}} = \frac{1}{128 r}$

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