Question

Geometry! I need help! Medal + Fan is promised :)

Answer 1

Circles and segments create the many relationships addressed in the theorems throughout this unit. Use those theorems and the information given for the following diagram to find each indicated measure. Show all your work. Note that the importance of justifications for your answers is equal to or greater than the importance of simply the answers in the earning of points.

Answer 2

b) Find KI if LM =5.

c) Find EP if PL = 6, PF = 4, and OP = 3.



e) Find CJ if CK = 6.

f) Find FH if AF = 10 and AO = 12.

g) Find AE if AF = 10, AL = 16 and AH = 14.

Answer 3

@AccessDenied

Answer 4

@hihi67

Answer 5

Was there one in particular you needed help with?

Answer 6

c-g

Answer 7

For (c): Do you know of a theorem for two intersecting chords and the lengths of each piece? That would be very useful here.

Answer 8

No

Answer 9

We learned all of this a semester ago and this is our review paper.

Answer 10

(if i dont respond I had to do stuff)

Answer 11

That's fine, and the site has been acting all funky too so I won't blame you on being late to reply.

Anyways I recommend reviewing these theorems, they should be valuable in filling out these problems c-g:
http://www.mathopenref.com/chordsintersecting.html
http://www.mathopenref.com/secantsintersecting.html

Answer 12

(c) uses the first theorem. EP is part of EL, along with PL. There is an intersecting chord OF, made of OP and PF
So we can set up this equation: The product of EP and PL equals the product of OP and PF.
EP * PL = OP * PF
And what we do from here is just substitute what information we were given and find the remaining segment length:
c) Find EP if PL = 6, PF = 4, and OP = 3.
EP * 6 = 3 * 4
Solve for EP.

Answer 13

EP*6=12

Answer 14

So do you divide 6 to 12 and get Ep that way?

Answer 15

@AccessDenied

Answer 16

Yep. That is correct.

Answer 17

So Ep is 2?

Answer 18

I agree with that. * Should say EP, both of those are points. But the value is correct.

Answer 19

Okay thanks one moment.

Answer 20

Okay onto E

Answer 21

(e) is sort of a corollary for that second link I posted.
CJ and CK are both tangent lines that intersect at C. We consider a tangent line to be a secant line with the two intersecting poitns being the same, so that this relationship holds:
CJ^2 = CK^2
Which can be simplified down to:
CJ = CK
That makes it easy! We were given CK = 6, so CJ is the same thing!

Answer 22

So CJ is 6? (I want to make sure I am understanding this.)

Answer 23

Yep, CJ = 6 is correct!

Answer 24

Okay ! (sorry the website updated) Onto the next one? @AccessDenied

Answer 25

Yep. (f)
Find FH if AF = 10 and AO = 12.
This is another condition of the intersecting secant theorem (2nd link). But we can't directly find FH through that. But notice that FH is part of AH, and AF is as well. If we can find AH, we just have to subtract AF to get AH - AF = HF.

Can you see how to set up the equation to find AH first? (Consider the tangent AO, AH and AF here)

Answer 26

Ummmm
12(AO)-10(AF)= AH?

Answer 27

Not quite... the theorem gives this framework: secant1 * secant2 = tangent^2

Answer 28

Wait so what

Answer 29

In that image, we have something similar to this (removing all the noise around it):

Can you see how this compares to the previous frame (which was for a general situation)?

Answer 30

The setup would then be the same as well: AO ² = AF * AH
Noting that AO is the tangent part, and AF and AH are parts of the secant line.

From here we just plug in our information to find AH.
AF = 10 and AO = 12.

Answer 31

Okay so how do we find AH
Youve lost me.

Answer 32

Do you understand how I created the equation? After that we are just plugging in AF and AO. That leaves only AH to solve for in the equation.

Answer 33

Where are we plugging AF and AO in

Answer 34

I derived the equation to use from this theorem:
Intersecting Secant-Tangent Theorem
http://www.mathopenref.com/secantsintersecting.html
AO² = AF * AH

If ever we have two secants, a secant and a tangent, or two tangent lines intersecting, this theorem holds. We just need to know the length from the intersection point to each point of intersection on the circle.

Answer 35

Okay

Answer 36

from AO^2 = AF * AH
We were told that AF = 10, and AO = 12. It is put directly into the equation, leaving AH to solve for.
(12)^2 = 10 * AH

Answer 37

so does (12)^2 stay that or do you actually square it

Answer 38

You would want to square it. 12^2 = 144. We want to solve for the value of AH, so we are going to simplify everything anyways.
144 = 10 * AH

Answer 39

Okay so now do you divide 144 by 10 or what do you do during this step?

Answer 40

Im honestly 100% sorry if I seem not smart. I am trying really hard.

Answer 41

You are doing fine. And yes, we divide 144 by 10 to solve for AH.
AH = 144 / 10

Answer 42

14.4?

Answer 43

Yep. That is AH.
Now earlier I said that AH contains HF and AF. In fact AH = AF + FH. We want to find FH, and we happen to now know AH and AH. Now we just take the difference.
AH - AF = HF Plugging in AF = 10, AH = 14.4 we just found.
14.4 - 10 = HF

Answer 44

We know AH and AF*

Answer 45

HF = 4.4

Answer 46

Looks good!

Answer 47

Next one!

Answer 48

(After this I have 2 more)

Answer 49

g) Find AE if AF = 10, AL = 16 and AH = 14.
Look for these segments in the diagram. I am thinking we need to implement that secant theorem one last time.

Answer 50

Okay so how do we set up the problem to find AE
AF+AL+AH=AE?

Answer 51

It will look very similar to the last few equations
something * something = something * something

Answer 52

AF*Al=AH*AE?

Answer 53

Much closer! But we multiply the ones that are on the same line:
AF * AH = AL * AE

Answer 54

Oh okay
10*14=16*AE
140=16*AE
140/16=8.75
AE=8.75

Answer 55

Right?

Answer 56

Looks great! :)

Answer 57

Thats my next question. And after that 1 more!

Answer 58

It's been a while since I've seen this theorem:
http://www.mathwarehouse.com/geometry/circle/angles-of-intersecting-chords-theorem.php
Basically, the angle where the intersecting chords meet is equal to the average of the two arc measures.

Answer 59

Or in symbols:

In our case, we have:
m<OPL = 1/2 ( m arc OL + m arc EF )