Circles

Agnishom · 2013-12-30 01:50:03

1. If an isosceles triangle PQR, in which PQ=PR=12 cm, is inscribed in a circle of radius 18 cm, find the area of the triangle

Nehushtan · 2013-12-30 02:26:22

Agnishom · 2013-12-30 02:41:29

Could you please post the workings?

P.S.: That is correct

atran · 2013-12-30 05:27:58

The problem becomes easy when you draw three radii inside the triangle; each radius connected to one point of the triangle.

Agnishom · 2013-12-30 05:28:49

Thanks for helping. I've drawn the radii. What next?

atran · 2013-12-30 05:29:14

Use the cosine rule first, and then the sine rule.

Agnishom · 2013-12-30 05:43:28

I am allowed to use the pythagoras theorem which is a stripped version of the cosine rule

atran · 2013-12-30 05:56:38

Aren't you allowed to use the cosine rule?

Agnishom · 2013-12-30 05:58:06

possibly not; but I would want to see a solution with it anyway

Nehushtan · 2013-12-30 06:32:20

atran wrote:

The problem becomes easy when you draw three radii inside the triangle; each radius connected to one point of the triangle.

You cant. The sides of the triangle are shorter than the radius of the circle.

Agnishom wrote:

I would want to see a solution with it anyway

atran · 2013-12-30 06:39:03

Nehushtan wrote:

atran wrote:
The problem becomes easy when you draw three radii inside the triangle; each radius connected to one point of the triangle.
You cant. The sides of the triangle are shorter than the radius of the circle.

You're right.

Last edited by atran (2013-12-30 06:39:16)

bobbym · 2013-12-30 06:50:16

Hi;

atran · 2013-12-30 07:10:36

Here is my solution. I hope I haven't made mistakes. If I have, please correct me.
Here is the link to the image:

http://imgur.com/JdjngAY

Look at the triangle PCR; you know all the sides, so you can use the cosine rule to find the angle (a). After a little work you have:

Now you need (x). To find it, you need to use the sine rule. Applying the rule you get:

Next you need to find the height of the original triangle (call it h):

So the area is equal to:

anonimnystefy · 2013-12-30 07:18:49

And sin(arccos(x))=sqrt(1-x^2).

Agnishom · 2013-12-30 14:41:13

Is there no way that the sum can be done using no coordinate geometry or trigonometry?

Nehushtan · 2013-12-30 14:59:37

Agnishom wrote:

Is there no way that the sum can be done using no coordinate geometry or trigonometry?

You certainly can do it that way

In Atrans diagram, produce PC to the other end of the diameter at S. Also let D be the midpoint of QR.

Then SQP is a right angle so

Also the triangles SQP and QDP are similar. Thus

from which you can proceed to calculate PD and hence the area of triangle PQR.

Last edited by Nehushtan (2013-12-30 15:02:16)

Agnishom · 2013-12-30 15:17:43

Thanks, this is what I was looking for

Math Is Fun Forum

#1 2013-12-30 01:50:03

Circles

#2 2013-12-30 02:26:22

Re: Circles

#3 2013-12-30 02:41:29

Re: Circles

#4 2013-12-30 05:27:58

Re: Circles

#5 2013-12-30 05:28:49

Re: Circles

#6 2013-12-30 05:29:14

Re: Circles

#7 2013-12-30 05:43:28

Re: Circles

#8 2013-12-30 05:56:38

Re: Circles

#9 2013-12-30 05:58:06

Re: Circles

#10 2013-12-30 06:32:20

Re: Circles

#11 2013-12-30 06:39:03

Re: Circles

#12 2013-12-30 06:50:16

Re: Circles

#13 2013-12-30 07:10:36

Re: Circles

#14 2013-12-30 07:18:49

Re: Circles

#15 2013-12-30 14:41:13

Re: Circles

#16 2013-12-30 14:59:37

Re: Circles

#17 2013-12-30 15:17:43

Re: Circles

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