Question

one leg of a right triangle has a length of 7 m. the other sides have lengths that are consecutive integers. find these lengths. the other leg is m.

Explanation:

Step1: Let the other leg and hypotenuse

Let the other leg be \(x\) m. Since the other sides are consecutive integers, the hypotenuse is \((x + 1)\) m.

Step2: Apply the Pythagorean theorem

By the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) (where \(a = 7\), \(b=x\), \(c=x + 1\)), we have \(7^{2}+x^{2}=(x + 1)^{2}\).
Expand \((x + 1)^{2}\) using the formula \((a + b)^{2}=a^{2}+2ab+b^{2}\), so \(49+x^{2}=x^{2}+2x + 1\).

Step3: Solve the equation for \(x\)

Subtract \(x^{2}\) from both sides of the equation \(49+x^{2}-x^{2}=x^{2}+2x + 1-x^{2}\).
We get \(49=2x + 1\).
Subtract 1 from both sides: \(49-1=2x+1 - 1\), so \(48 = 2x\).
Divide both sides by 2: \(x=\frac{48}{2}=24\).

Answer:

The other leg is \(24\) m.