how to find dimensions of a rectangle?

Question

i'm really bad at math and especially geometry... any help appreciated, thanks:

The length of a rectangle is 5 centimeters (cm) less than 4 times the width. If the perimeter is 140 cm, find the dimensions of the rectangle.

Answer 1

The information in the problem gives:
L = 4W - 5
P = 2L + 2W = 140

Substituting the first equation into the second yields:
2(4W - 5) + 2W = 140
8W - 10 +2W = 140
10W - 10 = 140
10W = 150
W = 15

Then, L = 4(15) - 5 = 60 - 5 = 55

Answer 2

Let the length of the rectangle be l cm and
its width be w cm.
Then l = 4w -10.
Now perimeter p = 2(l+w) =140 or putting the expression for l in this:
2(4w -10 +w)=140 or 5w -10 =70 or
w=16 cm, then, l = 54 cm
Answer: l = 54 and w = 16 cm.
Verification: l=4w-10 satisfied.
Perimeter p = 2(54+16) = 140 OK

Answer 3

Let width = w. Let length = 4w - 5. Your equation should read: 2(w + 4w - 6) = 140. Now combine the like terms within the parentheses, giving you 2(5w - 5)= 140. Now distribute the 2:
10w -10 = 140. Add the 10 to both sides: 10w = 150. Divide by 10: w = 15. Now, plug in the value for w into the equation for length (4w - 5) to find the length. You should get 55 for the length.

Answer 4

length = 4*width -5
perimeter = 2( length +width)

140 =2( 5width -5) 5 width = 75 width=15 cm length= 55 cm

Answer 5

In the same style given by technique of bedbye above, the rectangle that provides the perfect section is the only closest to a sq.. for this reason, the length are 7.5 x 7.5, or a suitable sq.. in case you recognize differential calculus, then you definately can clean up this project this way: enable L = length & W = width Perimeter, P = 2L + 2W = 30 2W = 30 - 2L W = 15 - L section, A = LW A = L * (15 - L) = 15L - L^2 To get the utmost section, differentiate the above equation with appreciate to L, and equate to 0. dA/dL = 15 - 2L = 0 2L = 15 L = 7.5 for this reason, W = 15 - L = 15 - 7.5 W = 7.5 for this reason, the length of the rectangle are 7.5 feet in length and seven.5 feet in width (a suitable sq.).

Answer 6

width = w
length = 4w-5

2w + 2l = P // Perimeter Formula

2w+2(4w-5)=140 // substitute known value and solve
2w + 8w -10 = 140
10w = 150
w = 15
L=55

2(15)+2(55)=140 // Always Check Answer
30 + 110 = 140
140 = 140 // True so both [W & L] are correct

Answer 7

L = 4W - 5
2(L + W) = 140
(4W - 5 + W) = 70
5W = 75
W = 15
L = 4*15 - 5 = 55

2(15 + 55) = 2*70 = 140

Answer 8

First we recall the area of a rectangle = base height. Suppose one side of the rectangle is x then the other side should be (100 - 2x)/2 Therefore, the area function A(x) = x[(100-2x)/2] . The maximum occurs at x = -b/(2a) = 25 feet and A(25)=625 square feet. The dimension of this rectangle is 25 by 25, which is a square.

Answer 9

Where x represents the width:

2[(4x - 5) + x)] = 140

2(5x -5) = 140

10x - 10 = 140

10x = 150

x = 15

The width is 15 cm., the length is 55 cm.

Answer 10

140 divided by 20= x
x times 20= 140