Find the dimensions of a rectangle whose perimeter is 52 and the area is 96

then






with the solution:

If the length is 3 cm and the perimeter is 10 cm, what will the width and area be? You can put this solution on YOUR website!

 

Find the perimeter and area of a rectangle whose length is 5x - 2 and width is x-3.

A rope 60 cm long is made to form a rectangle. if the length is four times its breath, calculate the length and diagonal of the rectangle.

which leads to

the perimeter of a rectangle is 102 inches , and the length of the diagonal is 39 inches. find the dimensions of the rectangle.

1)

P=2(l+w)=102
l+w=51

2) 

l^2+w^2=39^2

using substitution...
(l-51)^2+l^2=1521
l={15, 36}
so the dimensions are 15in x 36in

Perimeter of a rectangle if its length is (4c+7) units and its width is (c-3) units

P=
2(L+W)=
2(4c+7+c-3)=
2(5c+4)=
10c+8
where c > 3

A triangle has 45, 105 and 30 degree angles the smallest side is 8. What is the perimeter?

We can use the Law of Sines: 

 
 
 
 


 

rounded...P=34.77

The length of a rectangle is 5 inches more than double its width. If the perimeter is 70 inches, find the dimensions of the rectangle by using a system of equations.

length of a rectangle is 5 inches more than double its width means:

l=2w+5
the perimeter is 70 inches means:
2l+2w=70
or l+w=35
subbing in for l and simplifying...
(2w+5)+w=35
3w=30
so w=10
then that means
l+10=35
and so l=25
the dimensions must be: 25 inches x 10 inches

The length and width of a rectangle are consecutive even integers. The perimeter is 44 meters. Find the length and width.

2(l+w)=P
2(l+l+2)=44
2l+2=22
2l=20
l=10
length:10, width:12

The length of a football field is 180 feet more than its width. If the perimeter of the field is 1,060 feet, find the length of the field.

P=2(l+w)
l=180+w 

1060=2(180+w+w) 
2w+180=1060/2
2w=530-180
w=175

The length of a football field is 180 feet more than its width. If the perimeter of the field is 1,060 feet, find the length of the field

1)

so

2)

so


adding equations...

Two picture frames have the same area of 96 inches squared but different perimeters. What could the perimeters of the two frames be?

Both pictures must satisfy the equation:

So choosing valid values for any suitable pair of the four dimensions will produce different perimeters. For example:
, 
results in


and so on...

The area of a rectangle is 56 square meters. Find the length and width of the rectangle if it's length is 2 meters greater than it's width.

2(l+w)=56

so our first equation is...
1) l+w=28

comparing length and width...
l=w+2

so our second equation is...
2) l-w=2

system:
1) l+w=28
2) l-w=2
adding equations leads to
2l=30
so
l=15
w=13

The length of a rectangle is 5 inches more than the width. The perimeter is 38 inches. Find the length and width.

1) l=w+5
2) 2(l+w)=38
from 2)
l+w=19

so that
w+5+w=19
2w=14
w=7

l=7+5=12

The perimeter of a rectangular yard is 204 feet. If the width of the yard is 45 feet what is the length of the yard?

1) 2(l+w)=204
2) w=45

l+w=102
l=102-w
l=102-45
l=57 ft

a rectangle has one side equal to (2x+4)cm and a perimeter of (6x+4)cm what is the area of rectangle in terms of x

For convenience let's call the given side the length, so...
 
W are also given the perimeter...6x+4 and we for a rectangle, 
therefore: 

so we have enough information to solve for w: 

 
Now we have enough information to solve for area.

so

length of a dog run is 9 feet more than twice the width and the perimeter is 34 feet, what is the length and width

1) l=2w+9
2) 34=2(l+w) 


from 2)...
 17=l+w
17=(2w+9)+w
17-9=3w
w=8/3 

so 
l=
2*8/3+9=
16/3+27/3=
43/3

the perimeter of a rectangle is 112 cm . the length is 30 cm . calculate the width .

P=2(l+w) 
w=P/2-l
w=112/2-30
w=26 cm

The length of a rectangle is 4cm more than twice its width. The perimeter is 86cm. Find the width and length.

1) l=4+2w
2) 2(l+w)=86 


2) l+w=43 
replacing l...
(4+2w)+w=43
3w=39
w=13cm 



1) becomes
l=4+2*13
l=30cm

THE PERIMETER OF A RECTANGLE IS 24 AND ITS WIDTH IS x WRITE AN EXPRESSION FOR ITS LENGTH.

P = 2(l+w) = 2(l+x) 

24 = 2(l+x) 


12 = l+x 


l = 12-x 


:)