Question

the perimeter of a rectangle is 50 cm the length of the rectangle is 5m less than 4 times its breadth find the width of the rectangle

Answer 1

Answer:

Assuming L is length and W is width.

Assuming L is length and W is width."The perimeter of rectangle is 50 meters" 2L + 2W = 50

Assuming L is length and W is width."The perimeter of rectangle is 50 meters" 2L + 2W = 50"The length of the rectangle is 8 meters less than 4 times the width" L = 4W - 8

Assuming L is length and W is width."The perimeter of rectangle is 50 meters" 2L + 2W = 50"The length of the rectangle is 8 meters less than 4 times the width" L = 4W - 8Now, plug in the last equation for the L values in the perimeter formula to result in an equation with only one variable: 2 (4W - 8) + 2W = 50. Solve for W:

Assuming L is length and W is width."The perimeter of rectangle is 50 meters" 2L + 2W = 50"The length of the rectangle is 8 meters less than 4 times the width" L = 4W - 8Now, plug in the last equation for the L values in the perimeter formula to result in an equation with only one variable: 2 (4W - 8) + 2W = 50. Solve for W: 2 (4W - 8) + 2W = 50

Assuming L is length and W is width."The perimeter of rectangle is 50 meters" 2L + 2W = 50"The length of the rectangle is 8 meters less than 4 times the width" L = 4W - 8Now, plug in the last equation for the L values in the perimeter formula to result in an equation with only one variable: 2 (4W - 8) + 2W = 50. Solve for W: 2 (4W - 8) + 2W = 508W - 16 + 2W = 50

Assuming L is length and W is width."The perimeter of rectangle is 50 meters" 2L + 2W = 50"The length of the rectangle is 8 meters less than 4 times the width" L = 4W - 8Now, plug in the last equation for the L values in the perimeter formula to result in an equation with only one variable: 2 (4W - 8) + 2W = 50. Solve for W: 2 (4W - 8) + 2W = 508W - 16 + 2W = 5010W - 16 = 50

Assuming L is length and W is width."The perimeter of rectangle is 50 meters" 2L + 2W = 50"The length of the rectangle is 8 meters less than 4 times the width" L = 4W - 8Now, plug in the last equation for the L values in the perimeter formula to result in an equation with only one variable: 2 (4W - 8) + 2W = 50. Solve for W: 2 (4W - 8) + 2W = 508W - 16 + 2W = 5010W - 16 = 5010W = 66

Assuming L is length and W is width."The perimeter of rectangle is 50 meters" 2L + 2W = 50"The length of the rectangle is 8 meters less than 4 times the width" L = 4W - 8Now, plug in the last equation for the L values in the perimeter formula to result in an equation with only one variable: 2 (4W - 8) + 2W = 50. Solve for W: 2 (4W - 8) + 2W = 508W - 16 + 2W = 5010W - 16 = 5010W = 66W = 6.6 meters

Assuming L is length and W is width."The perimeter of rectangle is 50 meters" 2L + 2W = 50"The length of the rectangle is 8 meters less than 4 times the width" L = 4W - 8Now, plug in the last equation for the L values in the perimeter formula to result in an equation with only one variable: 2 (4W - 8) + 2W = 50. Solve for W: 2 (4W - 8) + 2W = 508W - 16 + 2W = 5010W - 16 = 5010W = 66W = 6.6 metersL = 4 (6.6) -8

Assuming L is length and W is width."The perimeter of rectangle is 50 meters" 2L + 2W = 50"The length of the rectangle is 8 meters less than 4 times the width" L = 4W - 8Now, plug in the last equation for the L values in the perimeter formula to result in an equation with only one variable: 2 (4W - 8) + 2W = 50. Solve for W: 2 (4W - 8) + 2W = 508W - 16 + 2W = 5010W - 16 = 5010W = 66W = 6.6 metersL = 4 (6.6) -8L = 26.4 - 8

Assuming L is length and W is width."The perimeter of rectangle is 50 meters" 2L + 2W = 50"The length of the rectangle is 8 meters less than 4 times the width" L = 4W - 8Now, plug in the last equation for the L values in the perimeter formula to result in an equation with only one variable: 2 (4W - 8) + 2W = 50. Solve for W: 2 (4W - 8) + 2W = 508W - 16 + 2W = 5010W - 16 = 5010W = 66W = 6.6 metersL = 4 (6.6) -8L = 26.4 - 8L = 18.4 meters

Assuming L is length and W is width."The perimeter of rectangle is 50 meters" 2L + 2W = 50"The length of the rectangle is 8 meters less than 4 times the width" L = 4W - 8Now, plug in the last equation for the L values in the perimeter formula to result in an equation with only one variable: 2 (4W - 8) + 2W = 50. Solve for W: 2 (4W - 8) + 2W = 508W - 16 + 2W = 5010W - 16 = 5010W = 66W = 6.6 metersL = 4 (6.6) -8L = 26.4 - 8L = 18.4 metersThe dimensions of the rectangle are 6.6 meters by 18.4 meters.

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Answer 2

Answer:

breadth = 6 cm

length = 19 cm

Step-by-step explanation:

let the breadth of rectangle be x

length is 4x - 5

perimeter of rectangle = 2(l + b) = 50

2(4x - 5 + x) = 50

5x - 5 = 50/2

5x - 5 = 25

5x = 25 + 5

5x = 30

x = 30/5 = 6

breadth = x = 6

length = 4x - 5 = 4(6) - 5 = 24 - 5 = 19

hope you get your answer