# The volume of a cylinder is 48.125 cm3, which is formed by rolling a rectangular paper sheet along the length of the paper. If cuboidal box (without any lid i.e., open at the top) is made from the same sheet of paper by cutting out the square of side 0.5 cm from each of the four corners of the paper sheet, then what is the volume of this box

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1. We do have a way for this question.Only way is to think out of the box. Let me tell you how it goes:
Volume of cylinder= (pi) (r)(r)(h)= 48.125
(r)(r)h = 15.3125
15.3125 is not a perfect square. To make it perfect square, h should be 5cm.
Now (r)(r)=15.3125/5 = 3.0625 which makes r = 1.75cm
Therefore, rectangular sheet has length of 2(pi)(r)=11cm and width (h) of 5cm.
After cutting a square of 0.5cm from each corner, the dimension of the cuboid becomes:
Therefore, Volume of cuboid= 10 x 4 x 0.5 = 20 cubic cm.
Thank you for reading it!
Have a nice day Reiny!
Jay Bankoti

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2. Volume of a cylinder = pi * r^2 * h

As per the question,
pi * r^2 * h = 48.125
r^2 * h = 48125/10000 * 7/22
r^2 * h = (49 * 5) / 16
r^2 * h = (7^2 * 5) / 4^2
r^2 = (7^2 * 5) / 4^2 * h

To make r^2 perfectly get converted into a value with a power of 1, we must cancel out the extra 5 that is making the RHS unable to be completely squared. We can so this by the extra 'h' variable that we have. So the height of the cylinder or the breadth of the paper is 5 cm.

r^2 = 7^2 / 4^2
r = 7/4

Since the length of the paper would be 2 * pi * r
l = 2 * 22/7 * 7/ 4 = 11 cm

If we draw 4 o.4 squares at the 4 corners of the page, we may observe the fact that when we fold the remains inward, it forms a cuboid without a top, with a hight of 0.5 cm. The breadth of the page is 5cm, but because 0.5cm squares are at both ends, 1 cm is reduced to calculate the real breadth of the cuboid. In the same way, the length of the page will also be reduced by 1 to calculate the length of the cuboid.

So (for the cuboid),
length = 11-1 = 10cm
breadth = 5-1 = 4cm
height = 0.5cm
Therefore the volume of the cuboid (box) = 10 * 4 * 0.5 = 20cm^3

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3. let the width of the sheet of paper be x
let its length be y

so the cylinder made from it would have a height of y
the width would be its circumference
2πr = x
r = x/(2π)

volume = πr^2 h = π(x^2/(4π^2) y = x^2 y/(4π)
48.125 = x^2 y/(4π)
192.5π = x^2y
y = 192.5/x^2

now to the box:
length = (y-1)
width = (x-1)
height = .5
volume = .5(x-1)(y-1) = .5(x-1)(192.5/x^2 - 1)
= (1/2)(192.5/x - x - 192.5/x^2 + 1)

getting nowhere .....

not enough information

in the first part, there is no unique solution for x and y
which would change the shape of the sheet of paper, which would change the volume of the resulting box

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