Problem 249 

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John Nolton
Posts: 15

Joined: 10/11/2011
Topic  Problem 249       Flag »  Reply »
You DO NOT need Calculus to solve this problem. Probably the most simplistic equation for the area of an Isosceles triangle is A = 1/2 S^2 sin (2theta) (see problem 249 page 47, July 2012 for Bloch labeling of the triangle). Now lets make it a little easy and call the Apex point "C".  Now the equation will read A= 1/2 S^2 sin C     { equ. #1}
If you remember the graph of the sin function you will know that it is periodic and has a period of 2Pi. Its value (sin) is zero at zero degrees and has its maximum value at 90 Degrees which is 1. By inspection of equ. #1 above you will answer 2 of the 3 questions Bloch is asking. If sin C has a maximum value of +1 then the maximum triangle area is
just equal to    1/2 S^2 .
You have also answered the 1st question at the same time and that is the largest triangle area is when sin C has its maximum value, which is +1 which is 90 degrees (this = angle 2 theta in Bloch drawing).
The last question What is the perimeter of the maximum area triangle is very simple  now that you have a right triangle and should give nobody trouble, I hope.
P=S(2 + sqrt2)

John Nolton
Tombstone, AZ
  Wednesday, July 04, 2012 at 6:06:39 PM

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