SOLUTION: Jasmine drew a rectangle with the following properties.
The area is 45 square centimeters.
The length is 5 times the width
what is the perimiter
Question 895247: Jasmine drew a rectangle with the following properties.
The area is 45 square centimeters.
The length is 5 times the width
what is the perimiter Found 3 solutions by nerdybill, josgarithmetic, MathTherapy:Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website! Jasmine drew a rectangle with the following properties.
The area is 45 square centimeters.
The length is 5 times the width
what is the perimiter
.
Let w = width
then
5w = length
.
Area:
w(5w) = 456
5w^2 = 45
w^2 = 9
w = 3 centimeters
.
Length:
5w = 5(3) = 15 centimeters
.
Perimeter:
2(width+length)
2(3+15)
2(18)
36 centimeters Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website! You know this:
If w is width, and L is length,
wL=A, using A for area.
Given: L=5w
p=2w+2L, using p for perimeter.
Can you find w, and L, using these facts as symbolized?
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The question asks for p, or perimeter.
You can adjust this as ----which should be easier.
Finish using the given value for A=45, and evaluate p. Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website! Jasmine drew a rectangle with the following properties.
The area is 45 square centimeters.
The length is 5 times the width
what is the perimiter
Let width be W
Then length = 5W
Since area = 45 sq cm, then: W(5W) = 45
Width, or , or W = 3
Therefore, length = 5(3), or 15 cm
Perimeter: 2W + 2L, or 2(3) + 2(15), or 6 + 30, or cm