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# Geometry

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Circle O has an area of 25pi. Line SQ is a diameter of a circle. If PM equals 7 and RM equals 3 find OM. Sep 15, 2020

#1
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Circle O has an area of 25pi. Line SQ is a diameter of a circle. If PM equals 7 and RM equals 3 find OM.

Kreis O hat eine Fläche von 25 pi. Die Linie SQ ist ein Durchmesser eines Kreises. Wenn PM gleich 7 und RM gleich 3 ist, finde OM.

Hello jugoslav!

$$\pi r^2=25\pi$$

$$r=5$$

$$\overline{OM}^2=3^2+5^2-2\cdot 3\cdot 5\cdot cos\ \alpha=7^2+5^2-2\cdot 7\cdot 5\cdot cos\ \alpha\\ \overline{OM}^2=34-30\cdot cos\ \alpha=74-70\cdot cos\ \alpha$$

$$(70-30)\cdot cos\ \alpha=74-34$$

$$cos\ \alpha=\frac{40}{40}=1$$                       $$\alpha = zero!$$

$$\overline{OM}^2=34-30\cdot cos\ \alpha=34-30\cdot 1=4$$

$$\overline{OM}=2$$ !



Sep 15, 2020
edited by asinus  Sep 15, 2020
edited by asinus  Sep 15, 2020
#2
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Thanks asinus,

Once you know that r=5, another way to proceed is this:

$$\overline{PM}*\overline{MR}=\overline{SM}*\overline{MQ}\\ 7*3=(r+\overline{OM})*(r-\overline{OM)}\\ 21=r^2-\overline{OM}^2\\ 21=25-\overline{OM}^2\\ \overline{OM}=2$$

Sep 15, 2020
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Hello, Melody! Could you please delete this post. When the question is not correct, then the answers do not make much sense. Thanks! (r + OM) * (r - OM) = r2 - (OM)2 jugoslav  Sep 15, 2020
edited by jugoslav  Sep 15, 2020
edited by jugoslav  Sep 15, 2020
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Hello Melody!

$$\overline{PM} \times\overline{MR} =\overline{SM} \times \overline{MQ}$$

Is there a law of geometry that says:
If a chord intersects in a circle with the diameter of the circle, the product of the sections of the chord is equal to the product of the sections of the diameter.
Does this law have a name?

I would be grateful for an explanation.
greetings !

asinus  Sep 15, 2020
edited by asinus  Sep 15, 2020
#13
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Yes, this is called: Intersecting chords theorem(Wikipedia has a solid explanation of this theorem)

Since the diameter is the longest chord of a circle, it's included in this theorem.

jugoslav  Sep 15, 2020
edited by jugoslav  Sep 15, 2020
edited by jugoslav  Sep 15, 2020
edited by jugoslav  Sep 15, 2020
#3
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That diagram of yours is not correct. If the diameter is 10, then the cord PR cannot be 10. Sep 15, 2020
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Well something certainly does not make sense. You are right about that. Thanks for noticing and for telling us.

The area cannot be 25pi.     OR         PR cannot be 10

Melody  Sep 15, 2020
edited by Melody  Sep 15, 2020
#5
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You are right! I made a mistake. PM should be 6 NOT 7.

Correction:         Line segment PM = 6 units I apologize to all of you! jugoslav  Sep 15, 2020
edited by jugoslav  Sep 15, 2020
edited by jugoslav  Sep 15, 2020
edited by jugoslav  Sep 15, 2020
edited by jugoslav  Sep 15, 2020
edited by jugoslav  Sep 15, 2020
edited by jugoslav  Sep 15, 2020
#7
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Apology accepted (from me anyway)

However, now that you have the correct distance PM you can show us how it is done.

Have you learned from us?

Melody  Sep 15, 2020
#8
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Ok, fair enough.

r = sqrt( 25pi / pi) = 5              PM = 6          RM = 3

I do it this way:           (r + OM) * (r - OM) = PM * RM

r2 - (OM)2 = PM * RM

25 - (OM)2 = 18

(OM)2 = 7

OM = √7 jugoslav  Sep 15, 2020
edited by jugoslav  Sep 15, 2020
edited by jugoslav  Sep 15, 2020
edited by jugoslav  Sep 15, 2020
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Nice work Although I do think you should have put the first row in as well You can give yourself a point  Melody  Sep 15, 2020
edited by Melody  Sep 15, 2020
#9
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Circle O has an area of 25pi. Line SQ is a diameter of a circle. If PM equals 6 and RM equals 3 find OM.

Hello jugoslave!

Here is the solution after the new version.

$$\pi r^2=25\pi$$

$$r=5$$

$$\overline{OM}^2=3^2+5^2-2\cdot 3\cdot 5\cdot cos\ \alpha=6^2+5^2-2\cdot 6\cdot 5\cdot cos\ \alpha\\ \overline{OM}^2=34-30\cdot cos\ \alpha=61-60\cdot cos\ \alpha$$

$$(60-30)\cdot cos\ \alpha=61-34$$

$$cos\ \alpha=\frac{27}{30}=0.9$$                       $$\alpha = 25.8^0$$

$$\overline{OM}^2=34-30\cdot 0.9=34-27=7\\ \overline{OM}=\sqrt{7}$$

$$\overline{OM}=2.646$$ !

Sep 15, 2020
#12
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Good job, asinus! jugoslav  Sep 15, 2020