Question

"One side of a triangle is twice the smallest side. The third side is four feet more than the shortest side. The perimeter is 12 feet. Find the lengths of all three sides (in feet). (Enter your answers as a comma-separated list.)" I have been working on this problem for a while reading all the notes and cant figure it out.

Answer 1

Let a, b, and c be the lengths of the sides (in ft) of the triangle.

Since one side of a triangle is twice the smallest side, the third side is four feet more than the shortest side and the perimeter is 12 feet, then f c is the length of the smallest side, then we can set the following system of equations:

`\begin{gathered} a=2c, \\ b=c+4, \\ a+b+c=12. \end{gathered}`

Substituting the first and second equations in the third one we get:

`2c+c+4+c=12.`

Adding like terms we get:

`4c+4=12.`

Subtracting 4 from the above equation we get:

`\begin{gathered} 4c+4-4=12-4, \\ 4c=8. \end{gathered}`

Dividing the above equation by 4 we get:

`\begin{gathered} (4c)/(4)=(8)/(4), \\ c=2. \end{gathered}`

Finally, substituting c=2 in the first and second equations we get:

`\begin{gathered} a=2\cdot2=4, \\ b=2+4=6. \end{gathered}`

Answer: The lengths of all three sides (in feet) are: 2, 4, and 6.