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.

jugoslav Sep 15, 2020

#1**+2 **

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!**

\(\pi r^2=25\pi \)

\(r=5\)

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

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

\(cos\ \alpha=\frac{40}{40}=1\) \(\alpha = zero!\)

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

\(\overline{OM}=2\)

!

\(\)

asinus Sep 15, 2020

#2**+2 **

Thanks asinus,

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

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

Melody Sep 15, 2020

#6**0 **

Hello, Melody! Could you please delete this post. When the question is not correct, then the answers do not make much sense. Thanks!

**Melody, look at the 2nd row of your answer:**

** (r + OM) * (r - OM) = r ^{2} - (OM)^{2} **

jugoslav
Sep 15, 2020

#10**+1 **

**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

#3**+2 **

That diagram of yours is not correct. If the diameter is 10, then the cord PR cannot be 10.

Guest Sep 15, 2020

#5**+3 **

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

#7**0 **

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?

I want to see not just your answer but how you have got your answer :))

Melody
Sep 15, 2020

#9**+3 **

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.

\(\pi r^2=25\pi \)

\(r=5\)

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

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

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

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

\(\overline{OM}=2.646\)

!

asinus Sep 15, 2020