Question

Use the Pythagorean Theorem to find the lengths of the

sides of the triangle.
26
2x-14

Answer 1

`\begin{array}{llll} \textit{using the pythagorean theorem} \\\\ c^2=a^2+o^2 \end{array} \qquad \begin{cases} c=\stackrel{hypotenuse}{26}\\ a=\stackrel{adjacent}{2x}\\ o=\stackrel{opposite}{2x-14} \end{cases} \\\\\\ (26)^2= (2x)^2 + (2x-14)^2\implies 676 = (4x^2)+(4x^2-56x+14^2) \\\\\\ 676=4x^2+4x^2-56x+14^2\implies 676=8x^2-56x+196 \\\\\\ 0=8x^2-56x-480\implies 0=8(x^2-7x-60) \\\\\\ 0=x^2-7x-60\implies 0=(x-12)(x+5)\implies x= \begin{cases} ~~ 12 ~~ \checkmark\\ -5 ~~ \bigotimes \end{cases}`

now, -5 is a valid value for "x", however in this case we can't use it, because that makes one of our sides negative and all sides must be a positive value.

`\stackrel{ 2(12) }{\text{\LARGE 24}}\hspace{5em}\stackrel{ 2(12)-14 }{\text{\LARGE 10}}\hspace{5em}\text{\LARGE 26}`