Given,
The measure of base of triangle is 20 cm.
The measure of altitude of the triangle is 12 cm.
The expression of area of the triangle is,
[tex]A=\frac{1}{2}\times base\times\text{altitude}[/tex]
Substituting the values in the expression then,
[tex]\begin{gathered} A=\frac{1}{2}\times20\times\text{1}2 \\ A=10\times\text{1}2 \\ A=120cm^2 \end{gathered}[/tex]
The area of the triangle is 120 cm^2.
Consider,
The third side of the triangle is 2x cm.
The semiperimeter of the triangle is,
[tex]S=\frac{20+12+2x}{2}=16+x[/tex]
Area of triangle,
[tex]\begin{gathered} A=\sqrt[]{(16+x)(16+x-12)(16+x-20)(16+x-x)} \\ A=\sqrt[]{(16+x)(x+4)(x-4)16} \\ A=\sqrt[]{(256+16x)(x^2-16)} \\ A=\sqrt[]{256x^2+16x^3-256x-4096} \end{gathered}[/tex]
Equation both area then,
[tex]\begin{gathered} 120=\sqrt[]{16x^3+256x^2-256x-4096} \\ 14400=16x^3+256x^2-256x-4096 \\ \frac{18496}{16}=x^3+16x^2-16x \\ x^3+16x^2-16x-1156=0 \end{gathered}[/tex]
After solving this expression, the value of x is 7.38.
The third side of the triangle is,
[tex]2x=2\times7.38=[/tex]
