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What is the width of a rectangle with length 14 cm and area 161 cm2?
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∙ 13y ago
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∙ 13y ago
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What is the width of a rectangle with length 14 cm and area 161cm?
It is: 161/14 = a width of 11.5 cm
What is the width of a rectangle with lenght 14cm and area 161cm?
To find the answer you would do 161 divided by 14 because width plus length equals area. You just do the problem backwards. the answer ends up being width=11.5cm
What is the square root of 161 in radical form?
23
Given the arithmetic sequence -11-16-21 what is S30?
-161.
A multiple of 23?
23, 46, 69, 92, 115, 138, 161, 184, 207, 230
What is the width of a rectangle with length 14 cm and area 161 cm2 with the steps please?
area=length*width width=area/length w=161/14 w=11.5
What is the width of a rectangle with length 14 cm in an area of 161 cm²?
Its width is: 161/14 = 11.5 cm
What is the width of a rectangle with length 14 cm and area 161cm2?
Width = 161/14 = 11.5 cm
What is the width of a rectangle with length 14 cm and area 161cm?
It is: 161/14 = a width of 11.5 cm
What is the width of a rectangle with the length of 14cm and area 161cm squared?
Width = 161/14 = 11.5 cm
What is the width of a rectangle with length 14 cm and area 161 cm2 show all the work?
Its width is 11.5 cm.Knowing that, it will be easy for you to reproduce the work.
What is the width of a rectangle with lenght 14cm and area 161cm?
To find the answer you would do 161 divided by 14 because width plus length equals area. You just do the problem backwards. the answer ends up being width=11.5cm
What are the rectangles that have the same perimeter as area?
Perimeter is a length, and a length cannot be the same as an area.Ignoring the units, all rectangles that have the same numerical perimeter as area are those that satisfy:2 x (length + width) = length x widthwhich can be rearranged to give:width = 2 x length/(length - 2)Meaning that given any length over 2, a width can be found to give a rectangle that meets the requirement that its numerical perimeter is the same as its numerical area. (For a length greater than 0 and less than 2, the width would be negative and not possible; similarly for a length less than 0, the length is negative and not possible. When the length is 2, the width is undefined and so not possible. When the length is 0 the width is 0 and it is not a rectangle.)At some stage as the length increases, the length will equal the width and as the length continues to increase the rectangle then given will match the previous rectangles with the length and widths swapped. This occurs when:length = 2 x length/(length - 2)⇒ length x (length - 4) = 0⇒ length = 4.So, as long as the length is greater than or equal 4 it will be the longer side - the length, by convention, is the longer side. Thus all rectangles satisfy:width = 2 x length/(length - 2)with length ≥ 4, will have the numerical value of their perimeter the same as the numerical value of their area.For example:4 cm x 4 cm: perimeter = 16 cm, area = 16 cm25 cm x 31/3 cm: perimeter = 162/3 cm, area = 162/3 cm26 cm x 3 cm: perimeter = 18 cm, area = 18 cm241/2 cm x 33/5 cm: perimeter = 161/5 cm, area = 161/5 cm2etcNote: the first example is a square which is a rectangle with all the sides the same length.
The perimeter of a rectangular painting is 322 centimeters If the width of the painting is 68 centimeters what is its length?
Do-it-in-your-head method: Length plus width is half the perimeter ie 161 cm, of which 68 is width so the length is 93 cm.
If the width of a rectangle is 2 less than twice its length and the area of the rectangle is 161 square cm what is the length of the diagonal?
Suppose the length is L cmThen the width is 2L - 2 cmTherefore, the area is L*W = L*(2L - 2) = 2L^2 - 2L sq cm.Then 2L^2 - 2L = 161or 2L^2 - 2L - 161 = 0therefore L = [1+sqrt(323)]/2 [the other root of the quadratic equation is negative and so cannot be a length].and then W = sqrt(323) - 1By Pythagoras,L^2 = 1/4 + 1/4*323 + sqrt(323)/4 = 81 + sqrt(323)/2W^2 = 323 + 1 - 2*sqrt(323) = 324 - 2*sqrt(323)and then D^2 = L^2 + W^2 = 405 - 3/2*sqrt(323) = 378.0417 approxand so D = 19.4433 cm, approx.
What would be the length and width of a shape if the area is 161cm?
Sorry, but this cannot be calculated. The shape could be long and skinny, like 1cm x 161 cm, or it could be a wider shape like, say 7 cm x 23 cm.
What is the length when 4173281 the volume?
161
