Question

The perimeter of a triangle is 142cm. The medium side is 8cm longer than the shortest side and the longest side is 2cm longer then twice the shortest side. Find the lengths of each side

Answer 1

Let's consider the shortest side of the triangle to be $s$. We can say then that the medium side is $s + 8$, since it is 8cm longer than the shortest side $s$. We can also say that the longest side is $2s + 2$, since it is 2cm longer than twice the shortest side $s$.

The perimeter of a shape is the sum of the lengths of its sides. Using this, we can say that the perimeter of the triangle is:

$s + (s + 8) + (2s + 2)$

By combining like terms, we can simplify the expression:

$4s + 10$

The problem stated that the perimeter was 142 cm, so we can set the expression we just found equal to 142 and solve:

$4s + 10 = 142$

$4s = 132$

$s = 33$

We now know the length of the shortest side, which is 33 cm. By substituting the value we just found for $s$ into our other side lengths, we can find them as well.

• Medium side: $s + 8 \Rightarrow 33 + 8 = 41$
• Longest side: $2s + 2 \Rightarrow 2(33) + 2 = 68$

The side lengths of our triangle are 33 cm, 41 cm, and 68 cm.