A piece of string was wrapped around a cylindrical pillar to find the circumference of the circular cross section of the pillar. The length of the string is then measured with a ruler that has 1cm as the smallest devision. The measurement is recorded as 113cm.

What are the lower and upper bounds of the area of the circular cross section of the pillar, correct to the nearest square centimetre?

The problem describes that the circumference of the pillar has an error margin of 1 cm, meaning that its value can be as low as 112cm (lower bound) or as high as 114cm (upper bound).

To find the lower and upper bounds of the cross section's area, we identify the radius from each bound and plug them into the circle area formula.

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Finding R and the Bounds' Areas

Using the circumference equation we can equate it to each bound's value and rearrange for their r values.

Lower bound:

112 = 2πr

(112/2π) = r

Upper bound:

114 = 2πr'

(114/2π) = r'.

(the apostrophe on the upper bound's r is so we can differentiate both bounds' radius values)

Now we plug them into the area formula.

Lower bound:

r²π = (112/2π)²π ≈ 998

Upper bound:

(r')²π = (114/2π)²π ≈ 1034.