Length 17 cm. Width 10 cm.
Anything greater than 27.713cm (4 x √48).
width:10,length;17 17*2+10*2=54
5.7 is greater.
25 millimeters is only 2.5 centimeters; so the 4 centimeters is greater.
6x4 has a perimeter of 2*(6+4) = 2*10 = 20 cm 6.5x2 has a perimeter of 2*(6.5+2) = 2*8.5 = 17 cm So the first has the larger perimeter.
The maximum length is 25 cm.
Length 17 cm. Width 10 cm.
Anything greater than 27.713cm (4 x √48).
width:10,length;17 17*2+10*2=54
5.7 is greater.
Of course, a rectangle can have a greater perimeter and a greater area. Simply double all the sides: the perimeter is doubled and the area is quadrupled - both bigger than they were.
25 millimeters is only 2.5 centimeters; so the 4 centimeters is greater.
Yes.
Rectangular perimeter = 2(Length + Width) Width = a Length = a+5 Then, 84 = 2(a+a+5) 42 = 2a+5 (42-5)/2 = a = 18.5m = Width Length = a+5 = 23.5m
yes it will have a greater area
Let's set up an inequality to represent all possible values of the width (w) of the rectangle given the information provided. The length of the rectangle is three times its width, so the length (L) can be expressed as L = 3w. The perimeter (P) of a rectangle is given by the formula: P = 2(L + w). The perimeter is greater than 64 centimeters, so we have P > 64. Now, substitute the expression for L from step 1 into the perimeter formula from step 2: P = 2(3w + w) Simplify the expression inside the parentheses: P = 2(4w) P = 8w Now, we have the perimeter in terms of the width: P = 8w. We already know that P > 64, so we can write the inequality: 8w > 64 To isolate w, divide both sides of the inequality by 8: w > 64 / 8 w > 8 So, the inequality representing all possible values of the width (w) is: w > 8 This means that the width of the rectangle must be greater than 8 centimeters for the perimeter to be greater than 64 centimeters.