Question

A certain triangle has a perimeter of 3084 mi. The shortest side measures 76 mi less than the middle side, and the longest side measures 379 mi more than the middle side. Find the lengths of the three sides.

Answer 1

shortest side: 851

middle side:927

longest side : 1306

Explanation

Step 1

Let

x represents the shortest side

y represents the middle side

z represents the longest side

so, set the equations

a) the perimeter of a rectangle is the sum of the sides, so

$\begin{gathered} \text{Perimeter= side1+side2+side3} \\ replace \\ 3084=x+y+z \\ x+y+z=3084\Rightarrow\text{ Equation(1)} \end{gathered}$

b) The shortest side measures 76 mi less than the middle​ side( in other words, you have to subtract 76 from middle side to get the shortest side)

$x=y-76\Rightarrow equation(2)$

c) and the longest side measures 379 mi more than the middle side,( in other words, you have to add 379 to middle side to obtain the longest side)

$z=y+379\Rightarrow equation(3)$

Step 2

solve the equations

a) now, replace the x an z value sfrom equation (2) and (3) into equation(1)

$\begin{gathered} x+y+z=3084\Rightarrow\text{ Equation(1)} \\ (y-76)+y+(y+379)=3084 \\ -76+3y+379=3084 \\ 303+3y=3084 \\ 3y=3084-303 \\ \\ 3y=2781 \\ \text{divide both sides by 3} \\ \frac{3y}{3}=\frac{2781}{3} \\ y=927 \end{gathered}$

now, replace the y value in equatino (2) to find x

b)

$\begin{gathered} x=y-76\Rightarrow equation(2) \\ x=927-76 \\ x=851 \end{gathered}$

c) finally, prelace x and y value in equation (1) to find z

$\begin{gathered} x+y+z=3084\Rightarrow\text{ Equation(1)} \\ 851+927+z=3084 \\ \text{add like terms} \\ 1778+z=3084 \\ \text{subtract 1778 in both sides} \\ 1778+z-1778=3084-1778 \\ z=1306 \end{gathered}$

so

shortest side: 851

middle side:927

longest side : 1306

I hope this helps you