+1641 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.
+15000 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!
\(\pi r^2=25\pi \)
\(r=5\)
\(\overline{OM}^2=3^2+5^2-2\cdot 3\cdot 5\cdot cos\ \alpha=7^2+5^2-2\cdot 7\cdot 5\cdot cos\ \alpha\\ \overline{OM}^2=34-30\cdot cos\ \alpha=74-70\cdot cos\ \alpha \)
\((70-30)\cdot cos\ \alpha=74-34\)
\(cos\ \alpha=\frac{40}{40}=1\) \(\alpha = zero!\)
\(\overline{OM}^2=34-30\cdot cos\ \alpha=34-30\cdot 1=4\)
\(\overline{OM}=2\)
!
\(\)
+118687 Thanks asinus,
Once you know that r=5, another way to proceed is this:
\(\overline{PM}*\overline{MR}=\overline{SM}*\overline{MQ}\\ 7*3=(r+\overline{OM})*(r-\overline{OM)}\\ 21=r^2-\overline{OM}^2\\ 21=25-\overline{OM}^2\\ \overline{OM}=2 \)
+1641 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) = r2 - (OM)2
+15000 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
!
That diagram of yours is not correct. If the diameter is 10, then the cord PR cannot be 10.
+1641 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!
+118687 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 :))
+15000 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.
\(\pi r^2=25\pi \)
\(r=5\)
\(\overline{OM}^2=3^2+5^2-2\cdot 3\cdot 5\cdot cos\ \alpha=6^2+5^2-2\cdot 6\cdot 5\cdot cos\ \alpha\\ \overline{OM}^2=34-30\cdot cos\ \alpha=61-60\cdot cos\ \alpha \)
\((60-30)\cdot cos\ \alpha=61-34\)
\(cos\ \alpha=\frac{27}{30}=0.9\) \(\alpha = 25.8^0\)
\(\overline{OM}^2=34-30\cdot 0.9=34-27=7\\ \overline{OM}=\sqrt{7}\)
\(\overline{OM}=2.646\)
!
