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When the width is 10Cm and the humus is 35Cm what length would make the perimeter greater than 86Cm?
The only meaning for 'humus' that is in my dictionary is "rich soil".
So, let's just ignore the 'humus' for now.
Given the width of 10cm on a rectangle, there would be two sides that are 10cm in length. That would be 20cm.
So let's subtract those out of 86 to see something about the two ends of the rectangle. 86 - 20 = 66 But 66 is both ends together. Since the two ends of a rectangle are the same length, what number times two equals 66? That would be '33'.
Let's check our work . . . 10 + 33 + 10 + 33 = 86. Yep, we did it right.
But remember, we chose to ignore 'humus' - I can only take a wild guess that a 'humus' of 35cm means that your figure is not a rectangle, at all. To make any calculations involving the 35cm, more information is needed.
What else can I help you with?
The perimeter of a rectangle is 60 cm the width is 5 times greater than the length what is the length?
The length of the length will be 5 and the width will be 25.
Is it sometimes always or never true that the perimeter of a rectangle is greater than the area?
It depends on the length and width... The smaller of the length and the width, the perimeter is greater than the area... But.. The bigger of the length and width, the area is greater than the perimeter. example : length = 5 , width = 2 AREA = 5 x 2 = 10 Perimeter = 2 x ( 5 + 2 ) = 14 example : length = 9 , width = 6 AREA = 9 x 6 = 54 Perimeter = 2 x (9 + 6) = 30 you can see the different.....
What is the width length and perimeter where the area equals 40 cm?
First of all, the area cannot be 40 cm since an area is 2-dimensional whereas the measure given is 1-dimensional. Next, the shape is not defined and so the width, length and perimeter are indeterminate. Even if you assume that the shape is a rectangle, the width, length and perimeter are indeterminate. On the basis that the length is greater than the width, all that you can say is that the length will be greater than or equal to sqrt(40) = 6.324555 (to 6 dp), the width will be 40 divided by that length and the perimeter will be twice their sum. Some examples: Length = 8, width = 5, perimeter = 26 Length = 10, width = 4, perimeter = 28 Length = 40, width = 1, perimeter = 82 Length = 4000, width = 0.01, perimeter = 8000.02 Length = 4,000,000, width = 0.00001, perimeter = 8,000,000.00002 and so on, without limit.
How can you get the width when you have the perimeter and the length?
If the shape is a rectangle, then Perimeter = 2(Length + Width) So Width = Perimeter/2 - Length
The length of a rectangle is 7 centimeters greater than the width The perimeter is 54 centimeters Find the length and the width?
Length 17 cm. Width 10 cm.
The perimeter of a rectangle is 60 cm the width is 5 times greater than the length what is the length?
The length of the length will be 5 and the width will be 25.
Length of a rectange is 5m greater than the width the perimeter is 84m2 what is the length and width?
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
Is it sometimes always or never true that the perimeter of a rectangle is greater than the area?
It depends on the length and width... The smaller of the length and the width, the perimeter is greater than the area... But.. The bigger of the length and width, the area is greater than the perimeter. example : length = 5 , width = 2 AREA = 5 x 2 = 10 Perimeter = 2 x ( 5 + 2 ) = 14 example : length = 9 , width = 6 AREA = 9 x 6 = 54 Perimeter = 2 x (9 + 6) = 30 you can see the different.....
What is the width length and perimeter where the area equals 40 cm?
First of all, the area cannot be 40 cm since an area is 2-dimensional whereas the measure given is 1-dimensional. Next, the shape is not defined and so the width, length and perimeter are indeterminate. Even if you assume that the shape is a rectangle, the width, length and perimeter are indeterminate. On the basis that the length is greater than the width, all that you can say is that the length will be greater than or equal to sqrt(40) = 6.324555 (to 6 dp), the width will be 40 divided by that length and the perimeter will be twice their sum. Some examples: Length = 8, width = 5, perimeter = 26 Length = 10, width = 4, perimeter = 28 Length = 40, width = 1, perimeter = 82 Length = 4000, width = 0.01, perimeter = 8000.02 Length = 4,000,000, width = 0.00001, perimeter = 8,000,000.00002 and so on, without limit.
How can you get the width when you have the perimeter and the length?
If the shape is a rectangle, then Perimeter = 2(Length + Width) So Width = Perimeter/2 - Length
The length of a rectangle is 7 centimeters greater than the width The perimeter is 54 centimeters Find the length and the width?
Length 17 cm. Width 10 cm.
If the perimeter is 48 If the length is twic the width what is the length of rectangle?
perimeter = (2*length) + (2*width) length = 2*width so perimeter = (2*2*width) + 2*width = 6*width perimeter = 48 so you can figure out the width and length
The width of a retangle is fixed at 16cm what lenghts will make the perimeter greater than 78cm?
The perimeter of a rectangle is 2 x length + 2 x width. If the width is 16 then 78 = 2 x length + 2 x 16 2 x length = 78 - 32 = 46 length = 23. For the perimeter to be greater than 78 cm, the length must be greater than 23 cm
How do you find the length of a rectangle if the perimeter is 40 and the width is 2inches?
perimeter = 2 x (length + width) ⇒ length + width = perimeter ÷ 2 ⇒ length = (perimeter ÷ 2) - width So to find the length of the given rectangle, subtract the width from half the perimeter is how to do it. IF you meant "What is the length..." then: length = perimeter ÷ 2 - width = 40 ÷ 2 - 2 in = 20 - 2 in = 18 in
What is one different way to to find the perimeter of a rectangle?
Length + length + width + width = perimeter. Length + width x 2 = perimeter
How do you find width when you have perimeter and length?
width = (perimeter-2*length)/2
How do you get the area if you have the width and the perimeter?
Perimeter = 2 lengths + 2 widths 1/2 perimeter = length + width Length = 1/2 perimeter - width Area = length x width That's (1/2 perimeter - width) x (width) .
