Proportion

[Guest]

A line that has length d is divided into two segments so that the larger segment x is the mean proportional between the whole line and the smaller segment. Determine the proportion for the statement.

[Guest]

d/x = x/(d-x)

x^2+dx-d^2 = 0

Solve for x = 0.618d = larger segment
Smaller segment is (d - 0.618d) = 0.382d

0.681 : 0.382
1.55 : 1

[Guest]

Last post should read

1.62 : 1

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[Guest]

"d/x = x/(d-x)"

Please explain how you obtained the equation and what is meant by the mean proportional.

[Guest]

If A and B are 2 numbers and a number N between them such that A x r = N and N x r = B where r is the common ratio of such numbers in the series. Then the series is A, Ar, Ar^2, Ar^3 etc etc. This is a geometric series or geometric progression and the number N is the geometric mean of the numbers A and B. or sometimes called the mean proportional.
So.... A/N = N/B they are the same ratio. Cross multiply gives N^2 = AB or N = sqroot(AB).

That is all I did when I did this....
The full length is same ratio to the long bit as the long bit is to the short bit.
d/x = x/(d-x)

x^2+dx-d^2 = 0

[Guest]

I understand it now. Thanks.

[Alicelewis11]

he mean proportional is additionally called the geometric mean. One approach to characterize it is to say Gm(a,b) = sqrt(a*b). This says "the geometric mean of an and b is the square foundation of their item." Here is an illustration:
Gm(4,9) = sqrt(4*9) = sqrt(36) = 6
So 6 is the mean relative (geometric mean) of four and nine.
The reason it is known as the mean relative is that the inquiry can
likewise be composed as an extent. We need to discover a number x for 4 and 9 with the goal that the accompanying extent is genuine:

4 / x = x / 9

[jackwilson]

Thanks for sharing it.

[Guest]

You are welcome. I ran across the problem on an advanced algebra exam from 1958.