How do I find the radius of the circle? (An isosceles triangle ABC has its vertices on a circle....)

[chijioke]

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?

 

[Dr.Peterson]

So how can I obtain the radius of the circle?

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.

 

[The Highlander]

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?

Glad to see you've drawn a sketch this time. ??

Read this or this and then this.

Hope that helps. ?

 

[chijioke]

You've done well so far.

Thank you sir.

Now add the center and some radii to the figure:

View attachment 36271​

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.

You've done well so far.

Now add the center and some radii to the figure:

View attachment 36271​

I love this your sketch. Which application did you use to make that happen?

As a gift, I've also added another midpoint, N. which I used to locate the center. (Do you know how to do that?)

No. But I would love to know?

Do you see any similar triangles? Use those to write an equation you can solve for r.

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.
Or would I be right to say that from triangle AOM, OM = 6cm since MB = 12 cm. So that [math]r=\sqrt{5^2+6^2} \approx 8 cm[/math]

 

[chijioke]

Glad to see you've drawn a sketch this time. ??

Thank you sir.

Read this or this and then this.

Hope that helps. ?

I will take my time to go through them.

 

[blamocur]

Yes, I am seeing triangle AOM and MOC as similar. And triangle AON and ONB as similar.

Triangles in those pairs are equal, not just similar. But you can find a more useful of similar but unequal triangles in that sketch.
Good luck.

 

[chijioke]

Triangles in those pairs are equal, not just similar.

In other words they are congruent.

But you can find a more useful of similar but unequal triangles in that sketch.

I can't for now. Could you please help?

Good luck.

By way, is the value for the radius I got correct?

 

[lev888]

Or would I be right to say that from triangle AOM, OM = 6cm since MB = 12 cm.

You would not be right since you didn't prove that O is the midpoint of the segment MB.

 

[chijioke]

You would not be right since you didn't prove that O is the midpoint of the segment MB.

Then which triangles are similar but not congruent in the sketch

 

[Dr.Peterson]

I can't [find a pair of similar but not congruent triangles] for now. Could you please help?

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 love this your sketch. Which application did you use to make that happen?

I've probably mentioned it before: GeoGebra.

No. But I would love to know [how to find the center of a circle]?

Please read answer #3. That's what he was helping you with.

 

[chijioke]

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.

I can't still find triangles similar without being equal. All I keep seeing is a pair of similar and equal triangles.

 

[lev888]

I can't still find triangles similar without being equal. All I keep seeing is a pair of similar and equal triangles.

Please list ALL triangles that contain the angle ABM. Are there triangles among them where the other 2 angles are equal?

 

[jonah2.0]

Beer influenced vision follows.

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?

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.

 

[Dr.Peterson]

I can't still find triangles similar without being equal. All I keep seeing is a pair of similar and equal triangles.

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.

@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.

What you need to do is to stop focusing on one expectation, and let your mind open up to other possibilities. Try anything that comes to mind beyond what you have seen yet, and see what happens.

We can all get stuck in a problem, and getting unstuck is an important skill to learn.

(No beer is needed.)

 

[topsquark]

(No beer is needed.)

As an aside, I've actually found that a bit of a buzz (not too heavy) makes Quantum Mechanics easier to understand.

Seriously!

-Dan

 

[chijioke]

Beer 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.

What do I need to think in order to find PM. I no that BP = 2r

 

[blamocur]

What do I need to think in order to find PM. I no that BP = 2r

Have you tried the advice in post #12?

 

[chijioke]

Please list ALL triangles that contain the angle ABM. Are there triangles among them where the other 2 angles are equal?

The two triangles are triangle ABM and ABO. The two equal angles are OBA and OAB.

 

[lev888]

The two triangles are triangle ABM and ABO. The two equal angles are OBA and OAB.

That's not all. Do you have a systematic approach to finding all triangles with the given angle?

 

[jonah2.0]

Beer induced reaction follows.

What do I need to think in order to find PM. I no that BP = 2r

Yes.
Now you just need to review the following:

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]...