Chapter 11.6, Problem 6P

To determine

To find: the length of each side of the right triangle.

Answer to Problem 6P

  $6feet,8feet,10feet$ .

Explanation of Solution

Given:

The length of the right triangle is consecutive even integers in feet.

Calculation:

According to the question, length of the right triangle is consecutive even integers in feet.

Let the first side be x.

Therefore, second and third side will be $x+2andx+4$ .

  

Here, from right triangle ABC, AC is the hypotenuse. Therefore,

By using Pythagoras theorem, $A{C}^2=A{B}^2+B{C}^2$ Thus,

  $\begin{array}{l}A{C}^2=A{B}^2+B{C}^2\\ {\left(x+4\right)}^2={\left(x+2\right)}^2+x^2\left\{\text{substitutes the given value}\right\}\\ \left\{{\left(a+b\right)}^2=a^2+2ab+b^2\right\}\\ x^2+8x+16=x^2+4x+4+x^2\\ \therefore 8x+16=4x+4+x^2\left\{\text{remove }{x}^2\text{from both the sides}\right\}\\ \Rightarrow x^2+4x+4-8x-16=0\left\{\text{equate one side to 0}\right\}\\ x^2-4x-12=0\left\{\text{simplify}\right\}\end{array}$

Now, solve the quadratic equation, so that to find the length of the sides of the triangle.

  $\begin{array}{l}{x}^2-4x-12=0\left\{\text{taking square root on both the sides}\right\}\\ x^2+2x-6x-12=0\{\text{simplify}}\\ x\left(x+2\right)-6\left(x+2\right)=0\{\text{simplify}}\\ \left(x-6\right)\left(x+2\right)=0\{\text{taking common factor}}\end{array}$

Either $\left(x-6\right)=0$ or $\left(x+2\right)=0$ .

When $\left(x+2\right)=0\Rightarrow x=-2\left\{\text{negative length neglect}\right\}$ .

When $\left(x-6\right)=0\Rightarrow x=6$ .

Therefore, the sides of the triangle are

  $\begin{array}{l}x=6\\ \text{therefore},x+2=8\text{and}x+4=10\end{array}$

Hence, the length of each side of the right triangle is $6feet,8feet,10feet$ .