You've done well so far.So how can I obtain the radius of the circle?
Glad to see you've drawn a sketch this time. ??View attachment 36266
View attachment 36267
i. Let the height [math]\overline{BM}~\text{be x}[/math][math]x=\sqrt{13^2-5^2}=12 cm[/math][math]\therefore[/math] the height of the triangle [math]\overline{BM}=12cm[/math]ii. So how can I obtain the radius of the circle?
Thank you sir.You've done well so far.
Now add the center and some radii to the figure:
As a gift, I've also added another midpoint, N. which I used to locate the center. (Do you know how to do that?)
Do you see any similar triangles? Use those to write an equation you can solve for r.
I love this your sketch. Which application did you use to make that happen?
No. But I would love to know?As a gift, I've also added another midpoint, N. which I used to locate the center. (Do you know how to do that?)
Yes, I am seeing triangle AOM and MOC as similar. And triangle AON and ONB as similar. - their corresponding sides are in the same ratio. But I am finding it difficult using it to write an equation for r.Do you see any similar triangles? Use those to write an equation you can solve for r.
Thank you sir.Glad to see you've drawn a sketch this time. ??
I will take my time to go through them.
Triangles in those pairs are equal, not just similar. But you can find a more useful of similar but unequal triangles in that sketch.Yes, I am seeing triangle AOM and MOC as similar. And triangle AON and ONB as similar.
In other words they are congruent.Triangles in those pairs are equal, not just similar.
I can't for now. Could you please help?But you can find a more useful of similar but unequal triangles in that sketch.
By way, is the value for the radius I got correct?Good luck.
You would not be right since you didn't prove that O is the midpoint of the segment MB.Or would I be right to say that from triangle AOM, OM = 6cm since MB = 12 cm.
Then which triangles are similar but not congruent in the sketchYou would not be right since you didn't prove that O is the midpoint of the segment MB.
You are saying you haven't found a pair yet. What do you plan to do in order to find them? You can do more than ask others to do your work ...I can't [find a pair of similar but not congruent triangles] for now. Could you please help?
I've probably mentioned it before: GeoGebra.I love this your sketch. Which application did you use to make that happen?
Please read answer #3. That's what he was helping you with.No. But I would love to know [how to find the center of a circle]?
I can't still find triangles similar without being equal. All I keep seeing is a pair of similar and equal triangles.You are saying you haven't found a pair yet. What do you plan to do in order to find them? You can do more than ask others to do your work ...
You might think about triangles that have the same angles. Maybe pick an angle and mark all other angles that are congruent to it. If that doesn't suggest a pair of triangles to consider, repeat with another angle. Once you find two triangles that share two angles, you've got it.
Give it a try, and show us what you did.
I've probably mentioned it before: GeoGebra.
Please read answer #3. That's what he was helping you with.
Please list ALL triangles that contain the angle ABM. Are there triangles among them where the other 2 angles are equal?I can't still find triangles similar without being equal. All I keep seeing is a pair of similar and equal triangles.
View attachment 36266
View attachment 36267
i. Let the height [math]\overline{BM}~\text{be x}[/math][math]x=\sqrt{13^2-5^2}=12 cm[/math][math]\therefore[/math] the height of the triangle [math]\overline{BM}=12cm[/math]ii. So how can I obtain the radius of the circle?
I can't still find triangles similar without being equal. All I keep seeing is a pair of similar and equal triangles.
@chijioke, note that this new suggestion is a different method than my own hint (presumably chosen as a way of using what you do see, rather than what you do not); it is not uncommon that there are multiple methods, and what one person sees as most natural may be harder for someone else to see.You only need to find an expression for the Length of MO in terms of r and you're done. You can easily do that by first finding an expression for the Length of PM (also in terms of r). Good luck.
As an aside, I've actually found that a bit of a buzz (not too heavy) makes Quantum Mechanics easier to understand.(No beer is needed.)
What do I need to think in order to find PM. I no that BP = 2rBeer influenced vision follows.
View attachment 36283
Focus your attention for a moment on the black right triangle CMO.
Clearly, r^2=5^2+(Length of MO)^2.
You only need to find an expression for the Length of MO in terms of r and you're done. You can easily do that by first finding an expression for the Length of PM (also in terms of r). Good luck.
Have you tried the advice in post #12?What do I need to think in order to find PM. I no that BP = 2r
The two triangles are triangle ABM and ABO. The two equal angles are OBA and OAB.Please list ALL triangles that contain the angle ABM. Are there triangles among them where the other 2 angles are equal?
That's not all. Do you have a systematic approach to finding all triangles with the given angle?The two triangles are triangle ABM and ABO. The two equal angles are OBA and OAB.
Yes.What do I need to think in order to find PM. I no that BP = 2r
i. Let the height [math]\overline{BM}~\text{be x}[/math][math]x=\sqrt{13^2-5^2}=12 cm[/math][math]\therefore[/math] the height of the triangle [math]\overline{BM}=12cm[/math]...