Question 704107
A right triangle has a perimeter of 36 cm and an area of 54 cm squared. What are its dimensions?
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Check for integer solutions first.
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Area = a*b/2 = 54
a*b = 108
a & b = 9 and 12
{{{c = sqrt(9^2 + 12^2) = 15}}}
9 + 12 + 15 = 36
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The hard way:
{{{a + b + sqrt(a^2 + b^2) = 36}}}
ab/2 = 54
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ab = 108
b = 108/a
Sub for b
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{{{a + 108/a + sqrt(a^2 + (108/a)^2) = 36}}}
{{{a + 108/a + sqrt(a^2 + 11664/a^2) = 36a}}}
{{{a^2 + 108 + sqrt(a^4 + 11664) = 36a}}}
{{{a^2  - 36a + 108 = -sqrt(a^4 + 11664)}}}
{{{a^4 - 72a^3 + 1512a^2 - 7776a + 11664 = a^4 + 11664}}}
{{{72a^3 - 1512a^2 + 7776a = 0}}}
a = 0 (Ignore)
{{{a^2 - 21a + 108 = 0}}}
(a - 9)*(a - 12) = 0
a = 9, b = 12
or vice versa