Equal Triangles

Two different triangles have the same area but different perimeters. Triangle A has side lengths of 10,10 and 6 cm. Triangle B has two sides with length of 10 cm. Find the length of the third side of triangle B. Round your answer to the nearest tenth of a cm.
Source: mathcontest.olemiss.edu 9/9/2013

SOLUTION

Since triangle A is isosceles, altitude h_A is the perpendicular bisector of the base.
h_A^2+3^2=10^2
h_A^2=91
h_A=\sqrt{91}
Area of triangle A
\frac{1}{2}\times 6\times \sqrt{91}=3\sqrt{91}

Since triangle B is isosceles, altitude h_B is the perpendicular bisector of the base.
h_B^2+x^2=10^2\qquad \left (1\right )
Area of triangle B = area of triangle A
\frac{1}{2}\times 2x\times h_B=3\sqrt{91}
xh_B=3\sqrt{91}
x^2h_B^2=819
h_B^2=\frac{819}{x^2}
Substitute the value of h_B^2 into Eq. \left (1\right )
\frac{819}{x^2}+x^2=10^2
819+x^4=100x^2
x^4-100x^2+819=0
The equation is quadratic in x^2
x^2=\frac{100\pm\sqrt{6724}}{2}
=\frac{100\pm 82}{2}
First solution
x^2=\frac{100+82}{2}=91
x=\sqrt{91}
Length of third side
2x=2\sqrt{91}=19.1 cm

Second solution
x^2=\frac{100-82}{2}=9
x=3
Length of third side
2x=2\times 3=6 cm which is the length of the base of triangle A.

Answer: 19.1 cm

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About mvtrinh

Retired high school math teacher.
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