Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
        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 ABC has perimeter p and area a then lim(h tending to 0) 512ra/p^3 equals to?
7 years ago

Jitender Singh
IIT Delhi
158 Points
										Ans: 4Area of triangle:$a = h\sqrt{r^{2}- (h-r)^{2}}$Perimeter of triangle:$p = 2\sqrt{r^{2}- (h-r)^{2}}+2\sqrt{2hr}$$L = \lim_{h\rightarrow 0}\frac{512.r.a}{p^{3}}$$L = \lim_{h\rightarrow 0}\frac{512.r.h\sqrt{r^{2}-(h-r)^{2}}}{(2\sqrt{r^{2}- (h-r)^{2}}+2\sqrt{2hr})^{3}}$$L = \lim_{h\rightarrow 0}\frac{512.r.h\sqrt{2hr-h^{2}}}{(2\sqrt{2hr-h^{2}}+2\sqrt{2hr})^{3}}$$L = \lim_{h\rightarrow 0}\frac{512.r\sqrt{2r-h}}{(2\sqrt{2r-h}+2\sqrt{2r})^{3}}$$L = \frac{512r}{8.8(\sqrt{2r})^{2}} = 4$Thanks & RegardsJitender SinghIIT DelhiaskIITians Faculty

3 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Differential Calculus

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details