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Question: ABC is an isosceles triangle described in a circle of radius r. If AB = AC and h is the altitude fro...

ABC is an isosceles triangle described in a circle of radius r. If AB = AC and h is the altitude from A to BC then Limh→0Δp3\underset{h \rightarrow 0}{Lim}\frac{\Delta}{p^{3}} equals to [where D is the area and p the is perimeter of the triangle ABC]

A

1/128r

B

1/215r

C

1/128

D

1/215

Answer

1/128

Explanation

Solution

AB = AC and AD = h

BC = 2BD = 2 OB2−OD2\sqrt{OB^{2} - OD^{2}}

= 2r2−(h−r)22\sqrt{r^{2} - (h - r)^{2}} = 22hr−h22\sqrt{2hr - h^{2}}

AB = AD2+BD2\sqrt{AD^{2} + BD^{2}} = h2+2hr−h2\sqrt{h^{2} + 2hr - h^{2}}

= 2hr\sqrt{2hr}

Perimeter p = 2AB + BC

= 22hr−h2\sqrt{2hr - h^{2}} + 2hr\sqrt{2hr}

Area of Triangle ABC D = 1/2 BC. AD

= h2hr−h2\sqrt{2hr - h^{2}}

Limh→0Δp3\underset{h \rightarrow 0}{Lim}\frac{\Delta}{p^{3}} = Limh→0\underset{h \rightarrow 0}{Lim} 2r−h8[2r−h−2r]3\frac{\sqrt{2r - h}}{8\left\lbrack \sqrt{2r - h} - \sqrt{2r} \right\rbrack^{3}} = 1128\frac{1}{128}