SOLUTION: Somebody please help me with this problem... and thank you in advance!! Square ABCD with sides 2 units long is constructed in a semicircle with radius r and diameter PQ. a) D

Question 24525: Somebody please help me with this problem... and thank you in advance!!
Square ABCD with sides 2 units long is constructed in a semicircle with radius r and diameter PQ.
a) Determine the radius of the circle
b) Show that rectangle ABQR is a golden rectangle
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
It would have been helpful to see your diagram of the problem. The first part can be answered but second part is problematical beacause you don't say where point R is located.
1) I can't do diagrams here so we'll have to rely on my verbal skills.
If a square is circumscribed by a semicircle, then one side of the square will rest on the diameter of the semicircle and the two upper corners of the square will just touch the circumference of the semicircle. Now if you bisect the side of the square resting on the diameter, the centre of this side will coincide exactly with the centre of the diameter.
Now draw a line from this point to one of the upper corners of the square to create a right triangle inside the square. You can see that the hypotenuse of this triangle is also the radius of the semicircle.
Using the Pythagorean theorem, you can find the length of the hypotenuse/radius.
The base of the right triangle is half the length of one side of the square, which makes it equal to 1.
The height of the triangle is equal to the length of the side of the square which is 2.
Call the hypotenuse/radius r. So, applying the Pythagorean theorem:

This is the length of the radius.
The ratio of the base to the height of the "golden rectangle" is: .
Your point R must lie outside of the semicicle. If you were to extend a line perpendicular to the diameter from point Q so that it intersects the extension of the top side of the square, the intersection would be point R.
So we have constructed the rectangle ABQR whose dimensions are:
Height is 2 and base is and you can see that the ratio of the base to the height: is that of the "golden rectangle"
"A picture is worth a thousand words"

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