Question

one side of a triangle is 2 cm shorter than the base, $x$. the other side is 1 longer than the base. what lengths of the base will allow the perimeter of the triangle to be at least 11 cm?

Explanation:

Step1: Express side - lengths

The base is \(x\) cm. One side is \((x - 2)\) cm and the other side is \((x+1)\) cm.

Step2: Set up perimeter inequality

The perimeter \(P=x+(x - 2)+(x + 1)\). We want \(P\geq11\), so \(x+(x - 2)+(x + 1)\geq11\).

Step3: Simplify the left - hand side

Combining like terms, \(x+(x - 2)+(x + 1)=x+x - 2+x + 1=3x-1\). So, \(3x-1\geq11\).

Step4: Solve the inequality for \(x\)

Add 1 to both sides: \(3x-1 + 1\geq11 + 1\), which gives \(3x\geq12\). Then divide both sides by 3: \(\frac{3x}{3}\geq\frac{12}{3}\), so \(x\geq4\).

Answer:

The lengths of the base \(x\) should satisfy \(x\geq4\) cm.