SOLUTION: find the area of a rectangle whose perimeter is 64 meters and whose length is 4 meters longer than its width the length of rectangle is 21 feet longer than it is wide find the

Question 53900: find the area of a rectangle whose perimeter is 64 meters and whose length is 4 meters longer than its width

the length of rectangle is 21 feet longer than it is wide find the dimensions of the rectangle if its perimter is 90 feet
Found 2 solutions by gsmani_iyer, funmath:
Answer by gsmani_iyer(201) (Show Source): You can put this solution on YOUR website!

First of all, you must know the formula to find the perimeter of a rectangle.
Perimeter of a rectangle = 2(length+width). Now we will come to solve the
problem.

Let the width = x meters.
So the length = x+4 meters. Here we apply the formula.
The perimeter = 2(x+x+4) = 64 meters.
= 2(2x+4) = 64 meters.
= 4x+8 = 64 meters.
So 4x = 64-8 meters. i.e. 56 meters.
x = 56/4 = 14 meters.
So the width of the rectangle = 14 meters and the
the length of the rectangle = 14+4 = 18 meters.
Width = 14 meters; Length = 18 meters. I hope this is clear to you.
Similarly the next problem can also be solved. I guess you can do it now.

gsm

Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
Rectangles are four sided figures where the opposite sides are equal. We find the perimeter of any polygon by adding the sides together. Therefore the formula for the perimeter of a rectangle is:
Peremeter(P)=width(W)+width(W)+length(L)+length(L)
P=W+W+L+L
P=2W+2L
For the first problem:
P=64 m
Let width=x
The length is 4m longer, so let leght=x+4
Plug your values into the formula and solve for x:
64=2x+2(x+4)
64=2x+2x+8
64=4x+8
64-8=4x+8-8
56=4x

14=x
Width:x=14 m
Lenght:x+4=14+4=18m
-------------------------------
For the second one:
P=90
Width=x
Length=x+21
Problem:
90=2(x)+2(x+21)
90=2x+2x+42
90=4x+42
90-42=4x+42-42
48=4x

12=x
Width:x=12 ft.
Length:x+21=12+21=33 ft.
Happy Calculating!!!!
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