Chapter 7.1, Problem 17PSB

a

To determine

To calculate : The Perimeter of JKMO.

Answer to Problem 17PSB

The perimeter of JKMO is $P=40$

Explanation of Solution

Given information : The following information has been given

EFGH is a rectangle

  $FH=20$

J,K,M and O are midpoints

Formula used : Pythagoras theorem states that sides of a right triangle can be calculated as

  $a^2+b^2=c^2$

Calculation : We know that JKMO are the midpoints of the rectangle. Thus, we know that

  $JK=\frac{1}{2}FH$

Now, we know that diagonals of a rectangle are equal due to Pythagoras theorem mentioned above. Thus, we can say that

  $JK=\frac{1}{2}FH=KM$

Now, substituting the value of FH, we get

  $Perimeter=4(JK)=4(10)$

Thus, the value of perimeter is $P=40$

b

To determine

The most descriptive name for JKMO is to be given

Answer to Problem 17PSB

The most descriptive name for JKMO is Rhombus

Explanation of Solution

Given information : The following information has been given

EFGH is a rectangle

  $FH=20$

J,K,M and O are midpoints

As we know by Pythagoras theorem in the first part that all sides of JKMO are equal, we can say that the name Rhombus best suits any quadrilateral with four equal sides, when opposite sides are known to be parallel.