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`        What is the radius of smallest circular disc large enough to cover every asute angle isoceles triangle of perimeter L`
5 years ago Sumit Majumdar
IIT Delhi
137 Points
```
Dear Student:

To solve the question, first draw an isosceles triangle, base AB, vertex X, so AX = BX and AXB<90°.Let 2θ = angle AXBLet S = AX = BXNow draw the circumscribed circle passing through A, B and X. This is the perimeter of the smallest disc able to cover the triangle.Mark the circle's centre O.The triangle is symmetric about the line OX. From a suitable diagram it should be clear that the total length of the sides is:L = 2(S+Ssin(θ))L = 2S(1+sin(θ)) (equation 1)It the radius of the circle is R, note that OX is a radius and from th ediagram that OXcos(θ) = S/2Rcos(θ) = S/2S = 2Rcos(θ)From equation 1:L = 2(2Rcos(θ))(1+sin(θ))L = 4R[cos(θ)(1+sin(θ))]R = (L/4)/[cos(θ)(1+sin(θ))]Let y = cos(θ)(1+sin(θ))R = (L/4)/yTo get the smallest possible R, we want the largest possible valueof y.We find the maximum of y in the usual way using calculus:y = cos(θ)(1+sin(θ))dy/dθ = 2cos²(θ) - sin(θ) - 1For max/min dy/dθ=0:2cos²(θ) - sin(θ) - 1 = 0Since cos²(θ) = 1 - sin²(θ):2(1-sin²(θ)) - sin(θ) - 1 = 0For neatness, let x = sin(θ)2(1-x²) - x - 1 = 02 - 2x² - x - 1 = 02x² + x - 1 = 0(2x - 1)(x + 1) = 0x = 1/2 or x = -1Find if x=1/2 ro x=-1 gives the maximum of y:y = cos(θ)(1+sin(θ))= (√(1-x²))(1+x)When x = 1/2, y = (√(1-½²))(1+½) = (√¾)(3/2) = ¾√3When x = -1. y = (√(1-1²))(1+1) = 0So the maximum vaue of y is ¾√3So the smallest value of R = (L/4)/ (¾√3) = L/(3√3) = L√3/9

Regards

Sumit

```
5 years ago Sumit Majumdar
IIT Delhi
137 Points
```							Dear student,The following are some of the good colleges:(a) APEX Group of Colleges,Jaipur(b) Malaviya National Institute of Technology, JaipurRegardsSumit
```
5 years ago
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