Circumradius of a quadrangle

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austral

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Hello everyone.

Screenshot_20220418-010713~2.png

I'm trying to solve this exercise, but I'm having the following problems:

-I saw a video where it was said that since BAC = 60°, then BDC = 60° too, but I don't understand why.
-I also easily get confused with the graph, I tried to draw a quadrangle accurately according to the information given (and it took me too long I think), but in the video I mentioned it seems that they drew the same thing but upside-down, and I know the graph should only be used as a reference but all the lines and lengths end up confusing me.
-Then, I don't know how to find the circumradius. Do I need to know a formula? In the video they find the circumradius of the triangle BCD, which is 1, and then say that it is the correct result for 1). But in the exercise they are asking for the circumradius of the quadrangle ABCD, not the triangle BCD. I'm also a bit lost here.

If I manage to understand 1), I think I could do 2) without major difficulties.

Any help would be much appreciated.
 
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austral said:
Hello everyone.

View attachment 32233

I'm trying to solve this exercise, but I'm having the following problems:

-I saw a video where it was said that since BAC = 60°, then BDC = 60° too, but I don't understand why.
-I also easily get confused with the graph, I tried to draw a quadrangle accurately according to the information given (and it took me too long I think), but in the video I mentioned it seems that they drew the same thing but upside-down, and I know the graph should only be used as a reference but all the lines and lengths end up confusing me.
-Then, I don't know how to find the circumradius. Do I need to know a formula? In the video they find the circumradius of the triangle BCD, which is 1, and then say that it is the correct result for 1). But in the exercise they are asking for the circumradius of the quadrangle ABCD, not the triangle BCD. I'm also a bit lost here.

If I manage to understand 1), I think I could do 2) without major difficulties.

Any help would be much appreciated.
Click to expand...
Please "sketch" the rectangle ABCD showing all the pertinent labels. Make the sketch "approximately" accurate.

1650324461135.png
 
austral said:
I saw a video where it was said that since BAC = 60°, then BDC = 60° too, but I don't understand why.
Click to expand...
Do you know the theorem about inscribed angles and arcs?
austral said:
I also easily get confused with the graph, I tried to draw a quadrangle accurately according to the information given (and it took me too long I think), but in the video I mentioned it seems that they drew the same thing but upside-down, and I know the graph should only be used as a reference but all the lines and lengths end up confusing me.
Click to expand...
Accuracy isn't really needed. Your first sketch, at least, can totally ignore the angles and lengths; it just has to provide a place to label the given facts.

(Once you have more ideas, you can redraw it if you want -- I often go through 2 or 3 sketches (some in my head) before I clearly see everything.)
austral said:
I guess this is approximately accurate...

View attachment 32237
Click to expand...
That makes a reasonable start. I would have drawn A, B, C, and D on a circle, so you can see inscribed angles that share the same arc (hint, hint).

And once you see that, in fact, D is also 60 degrees, you can redraw this a little more accurately, using an important fact about triangle ACD.
 
Dr.Peterson said:
Do you know the theorem about inscribed angles and arcs?
Click to expand...
I had never heard of the theorem before, but I searched it up on Google and I found this image:
inscribed-angle-theorem.png
So I drew the quadrangle into a circle following what you said here:
Dr.Peterson said:
I would have drawn A, B, C, and D on a circle, so you can see inscribed angles that share the same arc (hint, hint).
Click to expand...
And here it is:
IMG_20220418_213731324.jpg
I guess the arc from B to C is the one that makes angles BAC and BDC equal?

I wonder what the important fact about triangle ACD is...

Also, thanks for the tip about sketching - I always take so much time thinking about the "correct way" to draw whatever I'm being asked to, before even starting to do it, so I don't get "confused" later due to the sketch being misleading, but I should probably redraw it as the exercise becomes clearer. I'm supposed to take no longer than 5~10 minutes to solve this, so I better stop overthinking the drawing part.
 
austral said:
I guess the arc from B to C is the one that makes angles BAC and BDC equal?
Click to expand...
Yes.
austral said:
I wonder what the important fact about triangle ACD is...
Click to expand...
Label angles and lengths in your drawing, and you should see it!
austral said:
Now that I think about it, is it maybe that the arc from B to D makes the angle BAD 90°, meaning that the angle CAD is 30°?
Click to expand...
No. Don't say "maybe"; have a good reason. (These are false.)
 
In your (first) sketch (above) you appear to have marked two lines: the line CD and the line (or part of the line) BD as being equal (in length) but that cannot possibly be true! (Do you understand why?)
(I now see this has been fixed; the site made me wait an hour before posting this message! ?)

Also, are you aware that: \(\displaystyle Tan (60°)=\sqrt{3}\) ?

That fact may be of use to you too. ?
 
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austral said:
Now that I think about it, is it maybe that the arc from B to D makes the angle BAD 90°, meaning that the angle CAD is 30°?
Click to expand...
As Dr. Peterson points out, you can't just say "maybe"! The line BD is a Diameter of your circle, so the arc from B to D is a Semi-Circle. What do you know about angles within a semi-circle? (Google that too, if you need to find out.)
 
Okay, it's really late now (1:51am where I live) so I'm going to carefully read each reply tomorrow and see if I can finally understand this. Thank you both very much for your help ?
 
austral said:
Okay, it's really late now (1:51am where I live) so I'm going to carefully read each reply tomorrow and see if I can finally understand this. Thank you both very much for your help ?
Click to expand...
I trust you've had a good night's sleep. ?

I have now looked in more detail at the problem pictured in your original post and I can now see why you marked those lines as 'equal' in your first attempt to sketch the quadrilateral.

Unfortunately, you have chosen to name the vertices (corners) in your sketches in a way that makes it impossible to construct your figure accurately! I could just say move them all round clockwise so that A replaces B, etc. and start again to try and fit all the stated conditions into your diagram but I think it would be more useful and informative, for someone at your level (if I have correctly assessed the level you are at), if I give you a much better way to proceed, so....

Start on a clean sheet and draw a vertical line 6 cm long. Mark the point at the top of your line as A and the point at the bottom as C; also mark a point in the middle of the line (ie: 3 cm down from A and 3 cm up from C), let's label this point M.

Now take a pair of compasses and draw a circle whose circumference passes through both A & C by placing the pin of your compasses at M. I trust you do have a pair of compasses. If not, then you will just have to do your best to draw the circle "freehand". (Alternatively, you could trace around a round object and then draw the vertical line inside that circle marking it A at the top, C at the bottom and M in the middle.)

The question states that \(\displaystyle 2AB=AC \Rightarrow AB=\frac{1}{2}AC\) but you have drawn \(\displaystyle AC=6 cm \Rightarrow AB=3 cm\) so now take your ruler and draw a line 3 cm long from A out to the circumference of your circle to place point B. (If you've gone down the route of tracing the circle round an object then you will need to measure the length AM and use that instead of 3 cm)

You should now be able to complete your sketch (very) accurately and find the answers you need after having done so. (Ignore the actual lengths (in cm) of the lines you have drawn and just mark everything on your sketch in accordance with the conditions set out in your original problem.)

PS: My earlier hint that \(\displaystyle Tan(60°)=\sqrt{3}\) was, perhaps, a bit hasty.
It would be more useful for you to know that \(\displaystyle Tan(30°)=\frac{1}{\sqrt{3}}\) but it might also be useful for you to study the diagram (below) and fill in the missing values then, perhaps make a copy of it in your workbook(s); it's always useful to have that information 'to hand' or, better still, to have it thoroughly memorized!

CENTER]

​
If you click on the diagram it will open up in a new frame and at the top right hand side you should find a little "Download" icon that will allow you to save a copy of it to your device.
 
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austral said:
Okay, it's really late now (1:51am where I live) so I'm going to carefully read each reply tomorrow and see if I can finally understand this. Thank you both very much for your help ?
Click to expand...
Hi again, Austral,

I need to make a quick alteration to my last post!

When I was typing it up I just copied & pasted the "LaTex" expression I had composed for \(\displaystyle Tan(60°)=\sqrt{3}\) but after I changed the degrees (60→30) and what followed, in my haste it slipped my mind that I actually wanted to suggest a different trigonometric ratio to you (other than the Tangent)

So please ignore the line where it says: \(\displaystyle Tan(30°)=\frac{1}{\sqrt{3}}\) because, although that is true, it is not a piece of information that you need!
What I should have said is: \(\displaystyle Cos(30°)=\frac{\sqrt{3}}{2}\)

That is the piece of information on the worksheet that should be of considerable use to you!

However, if you go through the worksheet I gave you (diagram above), filling in all the missing answers, it should become immediately obvious to you what in there is of most use in solving your problem.
 
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Wow, thank you for taking the time to write such long and detailed answers... I know I said I was going to read everything and try to solve the exercise today, but I arrived home late and apart from maths I need to study Chemistry and History... I'm busy :(
I'll see if I can follow the steps you gave me in the replies today but in 3-4 hours I should already be sleeping again ?
I'm writing this message because I can't just see your replies and not answer anything. It would appear that I'm ignoring you... I don't want that to happen.
Again, thanks for your help. I've never received an answer online that was this complete.
 
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