The length of a rectangle is 5 cm more than 2 times its width. If the area of the rectangle is 76 cm^2,?

Question

find the dimensions of the rectangle to the nearest thousandth.

Please help

Answer 1

length =g, width = w
if g =5 + 2W then area= (5+2w)*w =76 sq cm
2w^2 +5w -76 =0
w= (-5 +/- sqrt(25 + 4*2*76)) / 2*2 ;
w= (-5 +/- sqrt(25+ 608)) /4
w = (-5 +/- 25.16)/4
w = 20.16/4 (length positive value)
w = 5.04 cm (width)
g = 15.08 cm (length)

Answer 2

Hi,

x = width
2x + 5 = length

Since area = length x width:
x(2x + 5) = 76
2x² + 5x = 76
2x² + 5x - 76 = 0

This does not factor, so use the quadratic formula to solve:
................._______
x = [-b ± √(b² - 4ac)/(2a)

.................__________
x = [-5 ± √(5² - 4(2)(-76)/(2(2))

.................________
x = [-5 ± √(25 + 608)/(4)

................___
x = -5 ± √633
......---------------
...........4

x can't be negative, so eliminate that answer.

(-5 + √633)/4 = 5.040 cm = x = width

2x + 5 = 2(5.040) + 5 = 15.080 cm = length

The width and length of the rectangle are 5.040 and 15.080 respectively.

I hope that helps!! :-)

Answer 3

Let width be W
length = 2W + 5
W.(2W + 5) = 76
2W² + 5W - 76 = 0
W = [ - 5 ± √(25 + 608) ] / 4
W = [- 5 ± √(633) ] / 4
W = [ - 5 ± 25.159 ] / 4
W = 20.159 / 4 (taking +ve value)
W = 5.04 cm
L = 15.08 cm

Answer 4

l = 2w + 5
w = w

A = lw

76 = w(2w + 5)
76 = 2w^2 + 5w
2w^2 + 5w - 76 = 0

w = (-b ± sqrt(b^2 - 4ac))/(2a)

w = (-5 ± sqrt(5^2 - 4(2)(-76)))/(2(2))
w = (-5 ± sqrt(25 + 608))/4
w = (-5 ± sqrt(633))/4

l = 2((-5 ± sqrt(633))/4) + 5
l = ((-5 ± sqrt(633))/2) + 5
l = (-5 ± sqrt(633) + 10)/2
l = (5 ± sqrt(633))/2

since you can't have negative length or width

w = (-5 + sqrt(633))/4
l = (5 + sqrt(633))/2

ANS :
width = about 5.040 cm
length = about 15.080 cm

Answer 5

l=2w+5
A=76=lw=w(2w+5)
76=2w²+5w subtract 76 from each side
2w²+5w-76=0
w=(-5±√(25+4*2*76))/4
w=(-5±√633)/4 since w>0
w=(-5+√633)/4
w=5.040cm
l=2w+5=2*5.04+5=15.080cm

Answer 6

width = 5.0398728 ~= 5.040
length = 15.0797456 ~= 15.080