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# geometry problem

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A quarter-circle (R) is inscribed in a square.  A circle (r) is drawn.  Find the ratio of r to R. Sep 25, 2020

#1
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(√2r2 - r) : r = (√2r2 + r) : R Sep 26, 2020
#2
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The diagonal of the square is     $$\sqrt{2R^2}$$ You need to check my working for stupid errors.

Sep 26, 2020
#3
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Shame on me I made a boo-boo but I'm not the only one. jugoslav  Sep 26, 2020
edited by jugoslav  Sep 26, 2020
edited by jugoslav  Sep 26, 2020
edited by jugoslav  Sep 27, 2020
edited by jugoslav  Sep 27, 2020
edited by jugoslav  Sep 27, 2020
#4
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ok jugoslav,

I will mark just one of your errors

DB = sqrt(2 * R)             sqrt(2 * R2)   wrong

$$DB=\sqrt{R^2+R^2}\\ DB=\sqrt{2R^2}\\ DB=\sqrt2 * \sqrt{R^2}\\ DB=\sqrt{2} *R$$

Your answer does not give a ratio at all. We already know that the ratio is r:R    that is given in the question.

Yours does not simplify to that so it is not correct.

PLUS the whole point is to put R in terms of    r

Note: it is really good that you are questioning my solution.

That shows everyone here that you are actively learning.

Melody  Sep 27, 2020
edited by Melody  Sep 27, 2020
#5
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1 : 3 - 2√2               ( 1 : -0.17157287525381 )

The correct answer is:      1 : 3 + 2√2         or       1 : 5.828427124746190666

jugoslav  Sep 27, 2020
#6
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Maybe so, I already said my answer needed to be checked for stupid errors.

Everyone can see that your second post answer #3, which was completely wrong,  has been edited in the past half hour, (after I posted mine.)

Stop trying to be so dishonest.

Melody  Sep 27, 2020
edited by Melody  Sep 27, 2020
#8
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$$\displaystyle r = (3-2\sqrt{2})R.$$

If

$$\displaystyle r = (3+2\sqrt{2})R$$

then r would be greater than R which is nonsense.

Guest Sep 27, 2020
#9
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Yes that is true.

All answers here must be checked for careless errors.

Melody  Sep 28, 2020