P=2l+2w The answer is 60ft.
The perimeter of a rectangle with length of 10 miles and width of 5 miles is 30 miles.
The length is 20 units and the width is 10 units
One rectangle that has a perimeter of 30 has a length of 5 and a width of 10.
I think you mean the width is 6. The answer depends entirely on what the length of the rectangle is. For example, if it is a square, so that the length and width are both 6, the answer is 24. I the length is 10, the perimeter is 32. In general, just add the width (6) to the length and double the sum to get the perimeter.
2 sides have length 10 and the other 2 sides have width 8 perimeter is sum of sides = 10+10+8+8 = 36
The perimeter of a rectangle with length of 10 miles and width of 5 miles is 30 miles.
The length is 20 units and the width is 10 units
One rectangle that has a perimeter of 30 has a length of 5 and a width of 10.
length + width = 20/2 = 10 length = 7cm, width = 3cm
The perimeter of a rectangle is given by:P = 2l + 2w, where P is the perimeter of the rectangle, l is the length of the rectangle, and w is the width of the rectangle.P = 2(15) + 2(10) = 50 meters.
I think you mean the width is 6. The answer depends entirely on what the length of the rectangle is. For example, if it is a square, so that the length and width are both 6, the answer is 24. I the length is 10, the perimeter is 32. In general, just add the width (6) to the length and double the sum to get the perimeter.
2 sides have length 10 and the other 2 sides have width 8 perimeter is sum of sides = 10+10+8+8 = 36
What do we know about the perimeter of a rectangle? perimeter = 2 × (length + width) → 2 × (length + width) = 20 in → length + width = 10 in → length = 10 in - width What do we know about the area of a rectangle: area = length × width → length × width = 24.4524 in² But from the perimeter we know the length in terms of the width and can substitute it in: → (10 in - width) × width = 24.4524 in² → 10 in × width - width² = 24.4524 in² → width² - 10 in × width + 24.4524 in² = 0 This is a quadratic which can be solved by using the formula: ax² + bx + c → x = (-b ±√(b² - 4ac)) / (2a) → width = (-10 ±√(10² - 4 × 1 × 24.4524)) / (2 × 1) in → width = -5 ± ½√(100 - 97.8096) in → width = -5 ±½√2.1904 in → width = -5 ± 0.74 in → width = 4.26 in or 5.74 in → length = 10 in - 4.26 in = 5.74 in or 10 in - 5.74 in = 4.26 in (respectively) By convention the width is the shorter length (though it doesn't have to be) making the width 4.26 in and the length 5.74 in. Thus the rectangle is 5.74 in by 4.26 in
multiple the length by 2 then subtract that from the perimeter. then divide that by two. for example length=10 perimeter=30 ... 30-2(10)= 30-20=10 10/2=5 width=5
15 feet
10 inches
20r + 10