SOLUTION: An isosceles riangle has a perimeter of 50 cm. The sum of the length of the base and the ehight of the triangle is 31 cm. Find the area of the triangle. I set it up for 2a

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Question 120790This question is from textbook Algebra 2 and Trigonometry
: An isosceles riangle has a perimeter of 50 cm. The sum of the length of the base and the ehight of the triangle is 31 cm. Find the area of the triangle.

I set it up for 2a + b = 50
and b + h = 31

I have a lot of triangle formulas at my disposal, such as the area of a triangle = 1/2 bh, or Heron's formula for the area of a triangle. I know that we can also use the special formula for the area of an isosceles triangle. The quadratic formula of a squared + b squared = c squared too. But I guess I keep getting stuck with the fact that I've got 3 unknowns rather than 2, re: a, b, and h. In the other problems in the book, I generally have one linear equation, which I solve for a, then I substitute a into the quadratic equation. This one has me really stuck. All help very much appreciated. This question is from textbook Algebra 2 and Trigonometry

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Yes, but h is a function of a and b.
h, a, and b/2 form a right triangle where a is the hypotenuse.
a%5E2=%28b%2F2%292%2Bh%5E2
h%5E2=+a%5E2-%28b%2F2%292
From above, you also know that
h=31-b
h%5E2=+a%5E2-%28b%2F2%292
%2831-b%29%5E2=+a%5E2-%28b%2F2%292
2a%2Bb=50
b=50-2a
%2831-50%2B2a%29%5E2=+a%5E2-%2825-a%29%5E2 Now you have one quadratic equation in a to solve.
%282a-19%29%5E2=+a%5E2-%2825-a%29%5E2
4a%5E2-76a%2B361=a%5E2-%28625-50a%2Ba%5E2%29
4a%5E2-76a%2B361=a%5E2-625%2B50a-a%5E2%29
4a%5E2-76a%2B361=-625%2B50a%29
4a%5E2-126a%2B986=0%29
Use the quadratic formula.
a=%28-%28-126%29%2B-sqrt%28%28-126%29%5E2-4%2A4%2A986+%29%29%2F%282%2A4%29
a=%28126%2B-sqrt%2815876-15776%29%29%2F%288%29
a=%28126%2B-sqrt%28100%29%29%2F%288%29
a=%28126%2B-10%29%2F%288%29
There are two solutions.
a%5B1%5D=%28126%2B10%29%2F%288%29
a%5B1%5D=%28136%29%2F%288%29
a%5B1%5D=17
a%5B2%5D=%28126-10%29%2F%288%29
a%5B2%5D=%28116%29%2F%288%29
a%5B2%5D=14.5
Placing the values of "a" in the original equations, you get,
a[1]=17
b[1]=16
h[1]=15
Area[1]=120
a[2]=14.5
b[2]=21
h[2]=10
Area[2]=105