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The perimeter of rectangle is 50meters.the length of rectangle is eight meters less than four. Times the width. Find the dimensions of the rectangle

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1 Answer

This problem involves more reading comprehension than math! You need to translate the words of the problem into expressions using only numbers and variables:

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

Now, 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

8W - 16 + 2W = 50

10W - 16 = 50

10W = 66

W = 6.6 meters

Plug this value back into the equation for the length of the rectangle: L = 4W - 8

L = 4 (6.6) -8

L = 26.4 - 8

L = 18.4 meters

The dimensions of the rectangle are 6.6 meters by 18.4 meters.