I will love you, if you can solve this geometry problem....

[slipperyslope]

My class got a take home quiz problem. We are allowed to use any resource we want and each other. Out of 24 people, we can't solve this problem. We spent about 2 hours trying to figure it out. Please help us solve it.

Here is a link to a drawing I made.
Problem

Please be gentle to the web server

Jim

 

[The Stigenator]

easy..

 

[boyRacer]

AX=3...

just kidding... i dunno.

 

[richardycc]

8 times cos of 60degree?

 

[The Stigenator]

here are some basic tips to get you started.

Draw a line from Point "A" to the other side of teh cirle. THe like will form a 90 degree angle to the tangent. If my calculations are right it should fall at "R". This assuming everything is upto scale and the angle at "A" is 90 degrees.

What ever the case may be the line will pass thru the center of the circle.
We know PR = 9 and we know Angle A is 90 degrees. Looking at the larger triangle.. we can see it forms a Right angled triangle with the angles at 90, 60 and 30.

We know the Distance between AP = PX therefore their angles are the same too..That means the angle at P and angle at X are the same. We already know P's angle (from larger triangle)
so we can say what X's angle is.

Once we have that .. put in the formula.. and you will find you distance AX..

simple..

 

[The Stigenator]

<< 8 times cos of 60degree? >>



are you nuts.. its not 8times cos60.. check your calculations again..

plus dont give the answer away.. let him figure out how to get the distance.. its easy .. the forumla is in teh book if you look.

 

[gopunk]

good guy -

how do you prove it falls at R?

i can't find a way to do that, and saying that "it looks like it should" is not acceptable in a proof.

 

[GoodToGo]

<< THe like will form a 90 degree angle to the tangent. If my calculations are right it should fall at "R". >>



It will only 90 degrees if it passes through the center of the circle and we dont know whether it does or does not, so ur assumption is wrong.

 

[The Stigenator]

just draw it.. and it will fit in form.

Draw a circle of Z radius.
Draw a chord of 8 cm accross. Extend it by 1 cm and label P. From Point P draw a Tangent and mark A. Draw your 90 degree and it should go right thru the center.

Edit:

Measure the distance between AP and mark point on PR for point X.. and then Draw line thru point X for point B...

I am normally not wrong about geometery problems...

 

[gopunk]

Draw a circle of Z radius.
Draw a chord of 8 cm accross. Extend it by 1 cm and label P. From Point P draw a Tangent and mark A. Draw your 90 degree and it should go right thru the center.

no i'm sorry, you may be right, but that is not a proof.

 

[GoodToGo]

How do u prove that triangles are similar?

 

[FrontlineWarrior]

42

 

[slipperyslope]

Realize that I drew that in paint myself. I don't think it will go through the center of the circle.

Jim

 

[McPhreak]

<< How do u prove that triangles are similar? >>



That would be the SAS, SSS, and one other one, I forget (SSA?) theorems

 

[GoodToGo]

<<

<< How do u prove that triangles are similar? >>



That would be the SAS, SSS, and one other one, I forget (SSA?) theorems >>



How do u exactly say that triangles are similar? I mean do we say that ratio of 2 sides equal to ratio of other 2 sides or wat?

 

[The Stigenator]

<< How do u prove that triangles are similar? >>



Its a theorm.. if any two angles in a triangle are the same as a that of a second triangle then the triangles are similar triangles. Their sides maybe be reduced or increased, but they are similar triangles.

Remember a triangle is 180 degrees, and an angle is 90 degrees.

There is quite a few like this.. go read your Math 101 book.. its in there.

 

[urameatball]

my teacher always called it A-S-S.

but I won't be solving your problem because you don't love me NOW!!

I want to be loved NOW NOW NOW!

 

[The Stigenator]

<< How do u prove that triangles are similar? >>



Its a theorm.. if any two angles in a triangle are the same as a that of a second triangle then the triangles are similar triangles. Their sides maybe be reduced or increased, but they are similar triangles.

Remember a triangle is 180 degrees, and an angle is 90 degrees.

There is quite a few like this.. go read your Math 101 book.. its in there.

 

[EvilYoda]

hehe, I second whoever said "42".......apparently it slipped by everyone else. ;-)

I'd help, but I'm too busy studying for my diff eq exam tomorrow, sorries, hope someone helps you out.

 

[McPhreak]

<<

<<

<< How do u prove that triangles are similar? >>



That would be the SAS, SSS, and one other one, I forget (SSA?) theorems >>



How do u exactly say that triangles are similar? I mean do we say that ratio of 2 sides equal to ratio of other 2 sides or wat? >>



If the ratio of 2 sides equal the ratio of 2 sides of a different triangle, they are only similar if the angle between the sides on both triangles are equal. Or, you can compare the ratio of 3 sides of both triangles to prove they are similar.

 

[The Stigenator]

you got to have a mimum of this:



SSS
AAA
ASS
SAS
<one more>
Where A=angle S=side.

 

[slipperyslope]

<< here are some basic tips to get you started.

Draw a line from Point "A" to the other side of teh cirle. THe like will form a 90 degree angle to the tangent. If my calculations are right it should fall at "R". This assuming everything is upto scale and the angle at "A" is 90 degrees.

What ever the case may be the line will pass thru the center of the circle.
We know PR = 9 and we know Angle A is 90 degrees. Looking at the larger triangle.. we can see it forms a Right angled triangle with the angles at 90, 60 and 30.

We know the Distance between AP = PX therefore their angles are the same too..That means the angle at P and angle at X are the same. We already know P's angle (from larger triangle)
so we can say what X's angle is.

Once we have that .. put in the formula.. and you will find you distance AX..

simple.. >>



Nothing in that picture is to scale really. I did the best I could. On my version the line RP is a little higher so it is obvious that line RA would not be the diameter. This is college geometry not HS geometry and you are thinking this is easier than it is. I know the basics of geometry. If what you said was true then this would be a simple problem.

I appreciate the help though.

Jim

 

[jchu14]

The length of AX is 4

There are 2 theorems that you need to use.

1)AP^2=PQ*PR (found this in a textbook, still looking for proof, but it's right)

2)Triangle AXQ and triangle RXB are similar. (AAA-The angles share the same arc on the circle)

If you need the problem worked out, just ask, i'll check back later.

 

[morkinva]

I get 4.5

 

[jchu14]

AP^2=PQ * PR (1)
PR=QR+PQ=8+1=9
AP=SQR(PQ*PR)=SQR(1*9)=3
AP=PX=XB=3

QX=PX-PQ=3-1=2
RX=QR-QX=8-2=6
AX:XQ=RX:XB (2)
AX=(XQ*RX)/XB=(2*6)/3=4 .......ANS

(1) Theorem 9-11 (2) Theorem 9-12
both on page 362 / chapter 9
Geometry By: Ray c. Jurgensen, Richard G. Brown, John W. Jurgensen
Houghton Mifflin Company. Boston
ISBN 0395-46146-4