mayadavis202
contestada

What is the perimeter of the rectangle shown on the coordinate plane, to the nearest tenth of a unit?

15.3 units
20.4 units
30.6 units
52.0 units

What is the perimeter of the rectangle shown on the coordinate plane to the nearest tenth of a unit 153 units 204 units 306 units 520 units class=

Respuesta :

To calculate the perimeter, consider each side separately. Since this is a rectangle, you have 2 pairs of sides of equal length, so you just have to calculate the length of one of each pair of sides and then double each and add them to find the perimeter.

Top side: If you look carefully, this side is really the hypotenuse of a triangle that is 2 units tall and 10 units long. Using the Pythagorean Theorem, you can calculate the length of the hypotenuse to be 2^2 + 10^2 = c^2 --> c = 10.2

Left side: Just like the top side, this side is the hypotenuse of a triangle that is 5 units tall and 1 unit long. Pythagoras again: 5^2 + 1^2 = c^2 --> c = 5.1

Since you have two of each side that are the same length, double each and add for the final perimeter: 10.2*2 + 5.1*2 = 30.6
ayaahmed16
firstly, we will name the points as the following:
(-6,4)  will be named as A
(4,2) will be named as B
(3,-3) will be named as C 
(-7,-1) will be named as D 
then we will use the distance formula between any 2 points on the graph which equals=[tex] \sqrt({x}1-{x}2)^2 +({y}1-{y}2)^2[/tex]
So,
AB=[tex] \sqrt[/tex](-6-4)^2+(4-2)^2 =2[tex] \sqrt[/tex]26
BC=[tex] \sqrt[/tex](4-3)^2+(2+3)^2 = [tex] \sqrt[/tex]26
CD=[tex] \sqrt[/tex](3=7)^2+(-3+1)^2 = 2[tex] \sqrt[/tex]26
AD=[tex] \sqrt[/tex](-6+7)^2+(4+1)^2 =[tex] \sqrt[/tex]26
Now calculate the sum and it will equal=30.6

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