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How do I show that the golden ratio in a quadradratic golden rectangle is [(1+square...

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r10 | Student, Grade 9 | Honors

Posted March 3, 2010 at 5:23 PM via web

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How do I show that the golden ratio in a quadradratic golden rectangle is [(1+square root of 5)/2]?

The golden rectangle is the rectangle defiened by the following statement:

The golden rectangle can be divided into a square and a smaller rectangle by a line which is parallel to its shorter sides, and the smaller rectangle is similar to the original rectangle.

Thus, if ABCD is the golden rectangle, ADXY is a square and BCXY is similar to ABCD, (i.e., BCXY is a reduction of ABCD).

The ratio of AB/AD for the golden rectangle is called the golden ratio.

show that the golden ratio is [(1+square root of 5)/2]

(Hint: let AB = x units and BC = 1 unit.)

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neela | High School Teacher | Valedictorian

Posted March 3, 2010 at 5:46 PM (Answer #1)

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Let ABCD be the golden rectangle.

In the rectangle ABCD, let AB =CD = x and AD = BC =1 unit.

Now we will divide CD at Xand ABat Y such that DX = AD= XY =AY = 1

Then BCXY is a similar golden Rectangle, by data.

Therefore, XY/DC = x by the definition (given in the data).

But XY = AD = 1 and DC =CD-DX= x-1, bu construction.

Therefore, XY/DC = x, This implies 1/(x-1) = x .Or

1 = (x-1)x . Or,

0 = x^2 - x -1 . Or

1*x^2 - 1*x - 1 = 0.............(1). This is a quadratic equation . A standard quadratic equation in the form ax^2+bx+c = 0 has the soltion for x given by:

c = { -b +or- (b^2-4ac)^(1/2)}/(2a).

Here in (1), a=b=1 and c=-1. So,

x = {-(-1) +Or- [ (-1)^2 + 4*1]^(1/2)}/(2*1).

= [1+5^(1/2)]/2 is the positive value of the soltion.

So the golden ratio is 1:x = 1 : [1+5^(1/2)]/2.

Or 1 : 1.618033989.

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