The lengths of the legs of an isosceles right triangle are x and √(6x+16) a. Find the length of the hypotenuse, in terms of x, in simplest radical form b. Find the length of all…

Question

The lengths of the legs of an isosceles right triangle are x and
`√(6x+16)`

a. Find the length of the hypotenuse, in terms of x, in simplest radical form
b. Find the length of all three sides of the triangle, in simplest radical form.

Answer 1

Check the picture below. Let's recall is an isosceles, so it has twin sides and is a right-triangle.

`x^2+(√(6x+16))^2\implies √(x^2+6x+16)\qquad \impliedby hypotenuse \\\\[-0.35em] ~\dotfill\\\\ x~~ = ~~√(6x+16)\implies \stackrel{\textit{squaring both sides}}{x^2=6x+16}\implies x^2-6x-16=0 \\\\\\ (x-8)(x+2)=0\implies x= \begin{cases} 8 ~~ \textit{\LARGE \checkmark}\\\\ -2 ~~ \bigotimes \end{cases} \\\\[-0.35em] ~\dotfill`

`\underset{\textit{length of each side}}{\stackrel{√((8)^2 + 6(8) + 16)}{{\Large \begin{array}{llll} 8√(2) \end{array}}}\hspace{5em}\stackrel{x}{\text{\Large 8}}\hspace{5em}\stackrel{√(6(8) - 16)}{\text{\Large 8}}}`

now, let's notice, the negative value is a legitimate value, however for this case is invalid since the lengths of the triangle can't be negative.