Exercise 23 Page 411

We are asked to calculate the perimeter of the rectangle which we do using the formula P=2l+2w where l is the rectangle's length and w its width. Since the width is parallel to the y-axis we can calculate w using the Ruler Postulate.

w=|y_2-y_1|

SubstituteII

y_2= 3, y_1= 0

w=| 3- 0|

SubTerm

Subtract term

w=|3|

AbsPos

|3|=3

w=3

We found the width to be 3 units, the same as in the presented solution. Let's continue to calculate the rectangle's length. The length, l, is parallel with the x-axis. Again we can use the Ruler Postulate to calculate the distance.

l=|x_2-x_1|

SubstituteII

x_2= 4, x_1= - 1

l=| 4-( - 1)|

SubNeg

a-(- b)=a+b

l=|4+1|

AddTerms

Add terms

l=|5|

AbsPos

|5|=5

l=5

We found the length to be 5 units, which differs from the result in the presented solution. If we now calculate perimeter we get P=2l+2w=2(5)+2(3)=16. Therefore the correct length of the perimeter is 16 units.