Find the area of the rectangle whose length is 2 more that twice the width and the perimeter is 22 Find the area of the rectangle whose length is 2 more that twice the width and the perimeter is 22
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calendarEducator since 2009
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First, we have to solve for the length and the width. We can do this by setting up the following equations.
l = 2w + 2
2l + 2w = 22
2 (2w +2) + 2w = 22
4w + 4 + 2w = 22
6w + 4 = 22
6w = 18
w = 3
If w = 3, then the length of the rectangle is 2 more than twice that. That means that length = 2*3 + 2 or l = 6 + 2 or l = 8
So the width is 3 and the length is 8. To get the area, we multiply these and we get 24.
So the area is 24.
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calendarEducator since 2008
write3,662 answers
starTop subjects are Math, Science, and Social Sciences
Area of the rectangle is:
a = length * width = L*w
But we know that:
L = 2+ 2w
Also:
2L + 2W = 22
Now substitute with L value:
==> 2L + 2W = 22
==> 2(2+2w) + 2w = 22
==> 4 + 4w + 2w = 22
==> 6w = 18
==> w = 3
==> L = 2+ 2*3 = 8
==> a = L*w = 3*8 = 24
==> the area of the rectangle is 24 square unit.
To calculate the area of the rectangle, we'll have to know the values of the lengths of the sides of the rectangle.
Let's note the length as a and the width as b.
From enunciation we know that:
a = 2 + 2b (1)
From enunciation we also know that the perimeter is 22 units.
The formula for perimeter of a rectangle is:
P = 2(a+b)
22 = 2(a+b)
We'll divide by 2:
a+b = 11 (2)
To calculate a and b we have to solve the system of equations (1) and (2).
We'll substitute (1) in (2):
2 + 2b + b = 11
3b = 11-2
3b = 9
We'll divide by 3:
b = 3
We'll substitute b in (2):
a+3 = 11
a = 11-3
a = 8
Now, we'll calculate the area of the rectangle:
A = a*b
A = 8*3
A = 24 square units
Let:
Width of rectangle = w = x
Then:
length of rectangle = l = 2x + 2
Perimeter of rectangle = 2(w + l) = 2(x + 2x + 2)= 22
Therefore:
2x + 4x + 4 = 22
6x = 22 - 6 = 18
Therefore:
x = 18/6 = 3
Width = x = 3
Length = 2x + 2
= 2*3 + 2
= 6 + 2 = 8
Area of rectangle = w*l = 3*8 = 24
To fibd the area , A .
Area = length*breadth = lb.
Given is the realation: l = 2 more than width or l=2+2b. And Perimeter p = 22.
Algebraically, P = 2(l+b) = 2[(2+2b)+b] = 2(2+3b) =4+6b which should equal to 22 or
4+6b = 22.
6b = 22-4 = 18
6b/6 = 18/6 =3
b = 3.
So by give realtion l = 2+2b , l = 2+2*3 = 8.
So l= 8 and b = 3. Therefore area A = lb = 8*3 =24
Let the length of the rectangle be L and let the width be W.
As the length is 2 more than twice the width: L=2+2W
The perimeter is 22, therefore 2(L+W)=22
=>2(2+2W+W)=22
=>2+3W=11
=>3W=9
=>W=9/3=3
As L=2+2W=2+2*3=8
Therefore the area of the rectangle which is L*W=8*3=24
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