Two Isoceles Triangles

Two noncongruent triangles have the same area but different perimeters. Triangle A has sides lengths of 8, 8 and 6 cm. Triangle B has two sides with length of 8 cm. Find the length of the third side of Triangle B. Round your answer to the nearest hundredth of a cm.
Source: mathcontest.olemiss.edu 10/2/2006

SOLUTION

In isosceles \triangle ADE , altitude \overline{AC} is also the perpendicular bisector of base \overline{DE}.
In right \triangle ACD,
AC^2+CD^2=AD^2
AC^2+3^2=8^2
AC^2+9=64
AC^2=64-9=55
AC=\sqrt{55}
Area of \triangle ADE=\frac{1}{2}\times base\times height
=\frac{1}{2}\times DE\times AC
=\frac{1}{2}\times 6\times \sqrt{55}
=3\sqrt{55}


In isosceles \triangle BFG, altitude \overline{BE} is also the perpendicular bisector of base \overline{FG}.
Let FG=2x.
Area of \triangle BFG=\frac{1}{2}\times base\times height
=\frac{1}{2}\times FG\times BE
=\frac{1}{2}\times 2x\times BE
=x\times BE
Since the two triangles have the same area
3\sqrt{55}=x\times BE
BE=\frac{3\sqrt{55}}{x}

In right \triangle BEF,
BE^2+EF^2=BF^2
\left (\frac{3\sqrt{55}}{x}\right )^2+x^2=8^2
\frac{495}{x^2}+x^2=64
495+x^4=64x^2        multiply by x^2
x^4-64x^2+495=0        (1)

Let u=x^2. Then, Eq.(1) has the following quadratic form
u^2-64u+495=0
\left (u-9\right )\left (u-55\right )=0       
factoring the equation
There are two solutions
u=9 and u=55
If u=9, then x^2=9 which implies that x=3. In this case,
FG=2x=6. We reject this solution because the two triangles are noncongruent.
If u=55, then x^2=55 which implies that x=\sqrt{55}. In this case,
FG=2x
=2\sqrt{55}=14.83

Answer: 14.83

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About mvtrinh

Retired high school math teacher.
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