You mean the that bit of whirrling uncertainty....
How do I find the radius of the circle? (An isosceles triangle ABC has its vertices on a circle....)
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I don't. Ok please. Could you just mention the similar triangles and I will make equation and find r?
Dr.Peterson
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What other triangle(s) have the same angle as angle MBA?
And please at least try to think systematically, rather than saying you can't. That's what mathematics is all about.
lev888
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We can. But I think you should be more interested in filling gaps in your knowledge/skills than in getting the right answer as quickly as possible. If you don't see all triangles that share the given angle it's a problem. There are no shortcuts in math. If you don't understand something you will not be able to proceed to more complex topics.
[math]r=\frac{13\times 6.5}{12} \approx 7cm[/math]This is my first time using geogebra to create and post a sketch in this forum. I think am going to create a new thread sometime later concerning some things I need to learn in creating geometrical shapes as its relates to maths. So where should such thread go? I mean which forum would be appropriate to fix such thread.
lev888
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Geometry and Trig forum sounds like a good match for posts about geometrical shapes.
I mean asking about using Geogebra in creating some shapes.
Dr.Peterson
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I'd still call it Geometry, since we don't have a Technology forum! And geometry helps a lot in using GeoGebra.
Actually, I usually don't even pay attention to where a question is; it doesn't matter much.
Dr.Peterson
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Yes, that looks right -- though it would help if you told us what triangles you used, and show the proportion itself.
Here is my own GeoGebra drawing, showing that your answer is correct:
I used triangle BMA and BNO though I forgot to show the center point O when I was doing the construction.
jonah2.0
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Beer induced algebraic alternative solution follows.
r^2=5^2+[(LengthofPO)-(Length of PM)]^2
r^2=5^2+[r-(2*r-12)]^2
What do I need to think in order to find PM. I no that BP = 2r
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.
r^2=5^2+(Length of MO)^2
r^2=5^2+[(LengthofPO)-(Length of PM)]^2
r^2=5^2+[r-(2*r-12)]^2
Dr.Peterson
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Yes, that's what my hints were intended to lead you to; so the proportion is [math]\frac{BO}{BN}=\frac{BA}{BM}\\\\\frac{r}{6.5}=\frac{13}{12}\\\\r=\frac{13\cdot6.5}{12}\approx7.0417[/math]
Two triangles are similar if the angles of one are respectively equal to the angle of the other.
Let's take a look at the triangle once again.
There are four similar Δ, -----, ΔPBO, ΔBON, ΔONA, and ΔPOC. All these Δs, are ~ to the two bigger congruent Δs, ΔBOC and ΔBOM. There are similar because all Δs have right angles.
Let say, I pick just two pairs of the Δs and compare them. We are comparing ΔPBO and ΔBOC.
Since [math]∠BPO = ∠OMC = 90 \degree[/math] Their corresponding sides are r and 13 cm.
My question now is, making reference to ΔPBO and ΔBOC, which other two angles are equal and why?
Let's take a look at the triangle once again.
There are four similar Δ, -----, ΔPBO, ΔBON, ΔONA, and ΔPOC. All these Δs, are ~ to the two bigger congruent Δs, ΔBOC and ΔBOM. There are similar because all Δs have right angles.
Let say, I pick just two pairs of the Δs and compare them. We are comparing ΔPBO and ΔBOC.
Since [math]∠BPO = ∠OMC = 90 \degree[/math] Their corresponding sides are r and 13 cm.
My question now is, making reference to ΔPBO and ΔBOC, which other two angles are equal and why?
Dr.Peterson
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If we label according to corresponding parts, the triangles you refer to are ΔBOP, ΔBON, ΔAON, and ΔCOP, which are in fact congruent; these are similar to the larger triangles ΔBCM and ΔBAM (and others), not ΔBOC and ΔBOM, which are not right triangles (in fact, BOM isn't a triangle at all). What are you thinking??
Please correct your triangles and ask again. Did you mean ΔPBO and ΔMBC, perhaps?
I made mistake in my earlier post. This is what I intend to post.
There are four similar Δ, -----, ΔPBO, ΔBON, ΔONA, and ΔPOC. All these Δs, are ~ to the two bigger congruent Δs, ΔBMC and ΔBMA. There are similar because all the Δs have right angles.
Let say, I pick just two pairs of the Δs and compare them. We are comparing ΔPBO and ΔMCB
Since
[math]∠BPO=∠OMC= 90 \degree[/math]Their corresponding sides are r and 13 cm.
My question now is, making reference to ΔPBO and MCB, which other two angles are equal and why?
I am asking so that we can establish similarity between the
Dr.Peterson
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These are all similar -- in fact, congruent -- but, again, it is clearer if you name them using corresponding parts, as ΔPBO, ΔNBO, ΔNAO, ΔPCO, and then ΔMBC and ΔMBA.
Also, it is not enough to say that they are similar because they all have right angles; not all right triangles are similar. Do you see that the other angles I've marked as corresponding are all congruent? Or is that also part of your question?
In my terms, these are ΔPBO and ΔMBC; you're right about which angles are right angles (at P and M), but it would be clearer to name them ∠BPO and ∠BMC, for consistency. The sides you identify as corresponding are BO = r and BC = 13 cm; I suppose you say they are corresponding because they are the hypotenuse of each triangle?
I was thinking that you were claiming these are similar, since you talked about corresponding sides; how can you not already know which angles correspond? That is how you know the triangles are similar, and how I've been naming them.
Look at the triangles you refer to. Do you see that both have a vertex at B, and the angles there, ∠PBO and ∠MBC, are the same angle? That's why they correspond. (Angles at the same vertex in different triangles do not always correspond.)
Then, clearly, the other pair of vertices, O and C, also correspond, and are congruent. (What is that angle equal to?)
So, the first thing to do is to identify congruent angles, and so identify them as corresponding. Sometimes this is obvious, because they are the same angle; other times, it takes more thought (perhaps using parallel lines, or the fact that two angles are each half of congruent angles, and so on).