Question

A large right triangle is going to be a part

Answer 1

The given triangle in the question is a right-angled triangle. In order to get the length of each side, we will apply the Pythagoras theorem.

The Pythagoras theorem is,

`\text{Hypotenuse}^2=Opposite^2+Adjacent^2`

Where,

`\begin{gathered} \text{Hypotenuse}=5 \\ \text{Opposite}=2x-2 \\ \text{Adjacent}=x \end{gathered}`

Therefore,

`5^2=(2x-2)^2+x^2`

Let us expand the above

`\begin{gathered} 25=2x(2x-2)-2(2x-2)+x^2 \\ 25=4x^2-4x-4x+4+x^2 \\ 25=5x^2-8x+4 \end{gathered}`

Switch sides

`5x^2-8x+4=25`

Subtract 25 from both sides

`5x^2-8x+4-25=25-25`

Simplify

`5x^2-8x-21=0`

Solve with the quadratic formula

`x_(1,\: 2)=(-\left(-8\right)\pm√(\left(-8\right)^2-4\cdot\:5\left(-21\right)))/(2\cdot\:5)`

Thus

`\sqrt[]{(-8)^2-4\cdot\: 5(-21)}=22`

Therefore,

`x_(1,\: 2)=(-\left(-8\right)\pm\:22)/(2\cdot\:5)`

Separate the solutions

`x_1=(-\left(-8\right)+22)/(2\cdot\:5),\: x_2=(-\left(-8\right)-22)/(2\cdot\:5)`

Hence,

`\begin{gathered} x=(-(-8)+22)/(2\cdot\: 5)=3 \\ x=(-(-8)-22)/(2\cdot\: 5)=-(7)/(5) \end{gathered}`

The solutions to the quadratic equations are

`x=3,\: x=-(7)/(5)`

Therefore, from the above result, the length of a triangle can never be negative.

Hence, x = 3

Let us now solve the length of the remaining side

`\begin{gathered} 2x-2=2(3)-2=6-2=4 \\ \therefore2x-2=4 \end{gathered}`

Therefore, the length of each leg is

`3ft,4ft,5ft`