You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

Sector OAB is a quarter of a circle of radius 3. A circle is drawn inside this sector, tangent at three points as shown. What is the radius of the inscribed circle? Express your answer in simplest radical form.

I see you have graph paper.

You must be plotting something

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Hi;

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

hi

The sector is ABC with centre B

The smaller circle has centre O with tangent points at D, E and F as shown in the diagram.

For tangents at D and F, the angles at D and F will be 90, so ODBF is a square.

Let radius OD be r.

Use pythag to calculate OB.

Add EO ( = r ) and this is EB ( = 3 )

Hence find r.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

these are all so hard

I see you have graph paper.

You must be plotting something

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

hi

I've got to go out very soon, so I'll just look at Q1 for now.

See diagram. I've put on an extra line XYZ. How long is XY? How long is YZ?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Hi cooljackiec;

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

hi bobbym, I've got methods for 2 and 3, but I'm waiting for some feedback on Q1 from the OP.

Seems we've got all five sorted between us.

I haven't gone as far as answers .... I thought he could do that step.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Hi;

Okay, then this is done.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

Yes. Just one thing missing.

Bob

Offline

**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

yz=xyz-radius of semicircle??

am i supposed to bash with substitution?

I see you have graph paper.

You must be plotting something

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

In the triangle GYZ, YZ is the hypotenuse. and XY is, yes, the radius of the circle.

So that should enable you to get the total length of the line = size of big square.

Bob

Offline

**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

10 + 5sqrt2. derp how did i not see that?!?!?!

how baout 2 and 3?

I see you have graph paper.

You must be plotting something

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

Q1 correct. Well done!

derp how did i not see that?!?!?!

It can happen to us all. I nearly messed up an exam because I hadn't noticed that the length of a radius on one side of a circle was the same size as the radius on the other side of the same circle. doh!

Q2 Call the point where XR and YS cross, point Z.

Triangles ZXY and ZRS are similar. This is because they have a common angle at Z and XY is parallel to SR.

So If you call the height of ZXY k then ZSR has height 4k (because the ratio of sides is 3:12)

So the height (distance) between XY and SR is 5k.

Now look at triangles ZXY and ZPQ. These are similar in the ratio k:6k so now you can calculate PQ.

Q3. I have made a diagram and labelled the points.

PVU is 90 as PV is a tangent

PWS is 90 as WX is the distance you are required to find.

So triangles PVU and PWS are similar.

So you can use the ratio of sides to work out WS, and then WX is easy.

Hope that helps,

Bob

T is the centre of a another circle, diameter PU that goes through V. I thought I needed this circle too, but, it turns out I don't. Interesting though.

*Last edited by bob bundy (2014-08-10 05:27:31)*

Offline

**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

thanks, i got both of them.

Let ABCD be a square, and let M and N be the midpoints of

and , respectively. Find .Triangle ABC has side lengths AB = 9, AC = 10, and BC = 17. Let X be the intersection of the angle bisector of \angle A with side \overline{BC}, and let Y be the foot of the perpendicular from X to side \overline{AC}. Compute the length of \overline{XY}.

Equilateral triangle ABC and a circle with center O are constructed such that \overline{BC} is a chord of the circle and point A is the circumcenter of \triangle BCO in its interior. If the area of circle with center O is 48\pi, then what is the area of triangle ABC?

In a triangle ABC, take point D on \overline{BC} such that DB = 14, DA = 13, DC = 4, and the circumcircles of triangles ADB and ADC have the same radius. Find the area of triangle ABC.

Let

denote the circular region bounded by x^2 + y^2 = 36. The lines x = 4 and y = 3 partition into four regions . Letdenote the area of region If then computeI see you have graph paper.

You must be plotting something

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Hi

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

thanks!!

I see you have graph paper.

You must be plotting something

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Hi;

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

can you do 3?

i appreciate your help

I see you have graph paper.

You must be plotting something

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Hi;

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

final one!

I see you have graph paper.

You must be plotting something

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,211

Hi;

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

Pages: **1**