Question

question 19 of 22. the perimeter of a square box is 112 inches; if the diagonal of the square box is 2k\sqrt{2}, where k is a constant, then what is the value of the constant k?

Explanation:

Step1: Find the side - length of the square

The perimeter of a square is given by $P = 4s$, where $P$ is the perimeter and $s$ is the side - length. Given $P=112$ inches, then $s=\frac{P}{4}$. So, $s = \frac{112}{4}=28$ inches.

Step2: Use the Pythagorean theorem for the diagonal of a square

In a square, if the side - length is $s$, and the diagonal is $d$, by the Pythagorean theorem $d^{2}=s^{2}+s^{2}=2s^{2}$. So, $d = s\sqrt{2}$. We know $d = 2k\sqrt{2}$. Since $d=s\sqrt{2}$ and $s = 28$ inches, then $2k\sqrt{2}=28\sqrt{2}$.

Step3: Solve for $k$

Divide both sides of the equation $2k\sqrt{2}=28\sqrt{2}$ by $2\sqrt{2}$. We get $k=\frac{28\sqrt{2}}{2\sqrt{2}} = 14$.

Answer:

14