SOLUTION: an isosceles triangle has a perimeter of 18 and integer side lengths. what is the smallest possible area and what is the smallest possible angle measure at one of its corners

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Question 1080908: an isosceles triangle has a perimeter of 18 and integer side lengths. what is the smallest possible area and what is the smallest possible angle measure at one of its corners
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
So let the equal side be N.
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From the perimeter you know,
2N%2BB=18
B=18-2N
The height of the triangle would be obtained using the base and N using the Pythagorean theorem,
N%5E2-%28B%2F2%29%5E2=H%5E2
with the condition,
N%5E2%3E=%28B%2F2%29%5E2
N%5E2%3E=%289-N%29%5E2
N%5E2%3E81-18N%2BN%5E2
81-18N%3C0
-18N%3C-81
N%3E81%2F18
N%3E9%2F2
Since N is an integer,
N%3E=5
Also,
2N%2BB=18
B%3E0 so,
2N%3C18
N%3C9
So,
5%3C=N%3C=8
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For the angle the information already given,
sin%28theta%29=H%2FN
sin%28theta%29=sqrt%28N%5E2-%28B%2F2%29%5E2%29%2FN
Make a table of values,
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