Calculate the radius of the circle, given 14cm chord is midway the radius

[chijioke]

A chord of length 14 cm is mid-way the radius of the circle. Calculate the radius of the circle.
If I must solve this problem, I think need help in interpreting the phrase is mid-way the radius of the circle.
Below is the drawing I managed to deduce from the words of the problem

But I can tell you that I have not been able to do anything meaningful with drawing. I think I need help here.

 

[skeeter]

Pythagoras ...

[math]r^2 - \left(\dfrac{r}{2}\right)^2 = 7^2[/math]

 

[Dr.Peterson]

A chord of length 14 cm is mid-way the radius of the circle. Calculate the radius of the circle.
If I must solve this problem, I think need help in interpreting the phrase is mid-way the radius of the circle.
Below is the drawing I managed to deduce from the words of the problem
View attachment 34986
But I can tell you that I have not been able to do anything meaningful with drawing. I think I need help here.

To me, at least in my dialect, "a chord is mid-way the radius of the circle" seems ungrammatical. But I would assume it means that it intersects the perpendicular radius at its midpoint, which agrees with your drawing, as far as you got.

 

[Harry_the_cat]

I think it means that the height of your triangle you have drawn is half the radius. (Agreeing with Post #2)

 

[chijioke]

To me at this point, I think the problem is useless and has no solution.

 

[Dr.Peterson]

To me at this point, I think the problem is useless and has no solution.

Why would you say that? It has a reasonable interpretation, and can be solved. Three of us have interpreted it the same way.

 

[Harry_the_cat]

Skeeter has shown in Post #2 that you can use Pythagoras where the hypotenuse of your right-angled triangle is r, the vertical height is r/2 and the base is 7. Can you solve that equation to find r? Give it a go and show us what you got.

 

[chijioke]

Skeeter has shown in Post #2 that you can use Pythagoras where the hypotenuse of your right-angled triangle is r, the vertical height is r/2 and the base is 7. Can you solve that equation to find r? Give it a go and show us what you got.

I solved using the relationship given below.

Pythagoras ...

[math]r^2 - \left(\dfrac{r}{2}\right)^2 = 7^2[/math]

I got r as 14 cm. That is why I said the problem is useless and not worth wasting time on.

 

[Dr.Peterson]

I solved using the relationship given below.

I got r as 14 cm. That is why I said the problem is useless and not worth wasting time on.

Did you check whether [imath]14^2 - \left(\dfrac{14}{2}\right)^2 = 7^2[/imath] works??

This is the wrong answer. Try again; and this time show us your work.

 

[chijioke]

Did you check whether [imath]14^2 - \left(\dfrac{14}{2}\right)^2 = 7^2[/imath] works??

This is the wrong answer. Try again; and this time show us your work.

[math]r^2 - \left(\dfrac{r}{2}\right)^2 = 7^2[/math][math]4 r ^ { 2 } - r ^ { 2 } = 7 ^ { 2 } × 4[/math][math]r^2(4-1)=49×4[/math][math]r^2(3)=49×4[/math][math]r^2=\frac{49×4}{3}[/math][math]=\sqrt{\frac{49×4}{3}}[/math][math]r=\frac{14\sqrt{3}}{3}[/math][math]\therefore~ r=8.1cm[/math]Hope it is correct now.

Pythagoras ...

[math]r^2 - \left(\dfrac{r}{2}\right)^2 = 7^2[/math]

So how did you get this?

 

[Dr.Peterson]

[math]r^2 - \left(\dfrac{r}{2}\right)^2 = 7^2[/math]...
[math]r=\frac{14\sqrt{3}}{3}[/math][math]\therefore~ r=8.1cm[/math]Hope it is correct now.

I hope you checked your answer!

If [imath]r=8.1[/imath], then [imath]r^2-\left(\frac{r}{2}\right)^2=8.1^2-\left(\frac{8.1}{2}\right)^2=8.1^2-4.05^2=65.61-16.4025=49.2075[/imath]. That is just a little off due to rounding. If you used a more precise value, such as 8.083, you would get closer to 49.

So how did you get this?

Please try to apply the Pythagorean Theorem to the figure. You have been told that this is where it came from.

 

[chijioke]

To me, at least in my dialect, "a chord is mid-way the radius of the circle" seems ungrammatical.

So please what is the correct grammar?

But I would assume it means that it intersects the perpendicular radius at its midpoint, which agrees with your drawing, as far as you got.

Why would you assume that?

 

[chijioke]

I hope you checked your answer!

If [imath]r=8.1[/imath], then [imath]r^2-\left(\frac{r}{2}\right)^2=8.1^2-\left(\frac{8.1}{2}\right)^2=8.1^2-4.05^2=65.61-16.4025=49.2075[/imath]. That is just a little off due to rounding. If you used a more precise value, such as 8.083, you would get closer to 49.


Please try to apply the Pythagorean Theorem to the figure. You have been told that this is where it came from.

Yes, I just applied it with the interpretation given in post #4. I understand now.

 

[chijioke]

I hope you checked your answer!

Yes I did- from book answer page. The solutions given is 8.1 cm.

If you used a more precise value, such as 8.083, you would get closer to 49.

But that is what I just got. Have you checked to see that [math]r=\frac{14\sqrt{3}}{3}[/math]is the same as 8.083 if not for approximation which is done to the nearest decimal place. So what are you saying exactly?

 

[Dr.Peterson]

So please what is the correct grammar?

Perhaps change "a chord is mid-way the radius of the circle" to "a chord is mid-way along the radius of the circle", or, better, "a chord's midpoint is at the midpoint of a radius of the circle". The word "midway" is not a preposition, but an adverb.

Yes I did [check the answer] - from book answer page. The solutions given is 8.1 cm.

No, you don't check the answer to a problem by looking in the book; books can be wrong, and you don't have one available when you take a test (or solve a real life problem).

You check by putting your answer into the statement of the problem. Haven't you been taught that?

And that is what I showed you:

If [imath]r=8.1[/imath], then [imath]r^2-\left(\frac{r}{2}\right)^2=8.1^2-\left(\frac{8.1}{2}\right)^2=8.1^2-4.05^2=65.61-16.4025=49.2075[/imath]. That is just a little off due to rounding. If you used a more precise value, such as 8.083, you would get closer to 49.

That is what I meant by checking!

In general, you should do this check using the number you actually calculated, before rounding; the rounded value is only for the final answer you show.

A fuller check would be to put the numbers into the picture and apply the Pythagorean theorem:

​

That confirms that the formula is appropriate.

 

[chijioke]

Perhaps change "a chord is mid-way the radius of the circle" to "a chord is mid-way along the radius of the circle", or, better, "a chord's midpoint is at the midpoint of a radius of the circle". The word "midway" is not a preposition, but an adverb.

That is a good one. I just pray l assimilate this.

No, you don't check the answer to a problem by looking in the book; books can be wrong, and you don't have one available when you take a test (or solve a real life problem).

Very true.

You check by putting your answer into the statement of the problem. Haven't you been taught that?

I have been thought.

And that is what I showed you:

That is what I meant by checking!

In general, you should do this check using the number you actually calculated, before rounding; the rounded value is only for the final answer you show.

A fuller check would be to put the numbers into the picture and apply the Pythagorean theorem:

View attachment 35073​

That confirms that the formula is appropriate.

Yes, the formula is appropriate.
I but at least my final rounding is correct. Isn't it?

 

[Dr.Peterson]

but at least my final rounding is correct. Isn't it?

Yes, if that's what they told you to do (which is suggested by the answer they gave).