Chapter 9, Problem 14RP

To determine

To find:The perimeter of square base JKMO.

Answer to Problem 14RP

The perimeter is $P=120$ .

Explanation of Solution

Given information:The following values have been given

  $PR=20$

  $PS=25$

Calculation:We know that in the right triangle PRS , if we apply Pythagoras, we get

  $R{S}^2+P{R}^2=P{S}^2$

Hence, we substitute the values of PR and RS, and we get

  $R{S}^2+{20}^2={25}^2\to RS=15$

Now, we know that each side of the square will be

  $a=2RS=2(15)=30$

Hence, the perimeter will be

  $P=4a=4(30)=120$