Width of rectangle: 50/10 = 5 cm
0.5
The width of the rectangle is 13 meters. The perimeter of the rectangle is 50 meters, and its length is 12 meters. The 50 meters is 2 times the length plus 2 times the width. With a length of 12 meters, twice that is 24 meters. That leaves 50 meters - 24 meters for twice the width. And 50 - 24 = 26, which is twice the width. The 26 meters divided by 2 = 13 meters, which is the width of the rectangle.
108 feet is the perimeter of a rectangle with a length of 50 feet and a width of 4 feet.
Let the width be x: 2(length+width) = perimeter 2(20+x) = 50 40+2x = 50 2x = 50 -40 2x = 10 x = 5 centimeters Check: 5+5+20+20 = 50 centimeters
From the statement of the problem, if w is the width, the area is 2w2 , the product of the width and the length, which is stated to be twice the width. Since 2w2 must be less than 50, w2 < 25, and the width must be less than 5 meters.
Width would be 14 cm
length 18 width 7
If "50 inches" is the perimeter all you can say is that length plus width = 25 inches.
To answer the question as written: If the length of a rectangle is given as being 50 feet, then the length of the rectangle is 50 feet. To answer the question that is really being asked, if the width of a rectangle is 12.4 feet and the length of 50 feet, the area can be found by multiplying the length by the width. Therefore, the are of the rectangle would be 620 square feet. If however, the question was to determine the perimeter, you would use the formula 2l+2w=p so you would add 12.4 +12.4+50+50, for a perimeter of 124.8 feet.
It is: 10 times 5 = 50 square in
The perimeter is found by multiplying the width by the length. So divide 50 by 12 and you get 4.166 width.