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#1 2014-01-10 19:50:53

thedarktiger
Member
Registered: 2014-01-10
Posts: 83

Difficult geometry prob

Hi,
ABCD is a square. Parallel lines m, n, and p pass through vertices A, B, and C, respectively. The distance between m and n is 12, and the distance between n and p is 17. Find the area of square ABCD.
Im trying to post an image but it wont let im new
roflol
thanks I spent I couple hours on this and have no idea.dunno
-thedarktiger


Good. You can read.

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#2 2014-01-10 21:09:44

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,528

Re: Difficult geometry prob

hi thedarktiger,

Welcome to the forum.

You'll get picture posting rights when you've been a member for a while.

Hopefully the one below is your problem.

Look at the angles I've marked x and y.  They add to 90 and you'll see other angles of the same size in the diagram too.

So show that triangles ABE and BCF have the same angles and one side equal so they are congruent.

Then use Pythagoras to get a side.

Bob

View Image: thedarktiger.gif

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

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#3 2014-01-11 01:53:42

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 90,489

Re: Difficult geometry prob

Hi;


In mathematics, you don't understand things. You just get used to them.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

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#4 2014-01-11 02:19:53

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,528

Re: Difficult geometry prob

hi

Bob


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

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#5 2014-01-11 02:22:29

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 90,489

Re: Difficult geometry prob

Hi Bob;


In mathematics, you don't understand things. You just get used to them.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

Online

#6 2014-01-11 06:16:55

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,528

Re: Difficult geometry prob

hi bobbym,

I did it with:

Bob


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

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#7 2014-01-11 13:50:53

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 90,489

Re: Difficult geometry prob

Hi;


In mathematics, you don't understand things. You just get used to them.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

Online

#8 2014-01-11 17:31:41

thedarktiger
Member
Registered: 2014-01-10
Posts: 83

Re: Difficult geometry prob

Hi,
thanks a ton I got 433 (how to hide?) after you guys
big_smile I was clueless!
EDIT:
im a bob to im robert!

Last edited by thedarktiger (2014-01-11 17:33:13)


Good. You can read.

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#9 2014-01-11 20:31:32

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,528

Re: Difficult geometry prob

hi Bob,

433 looks good to me.

To hide:

square bracket hide="optional message in box"  square bracket  "what you want to hide here" then end with square bracket /hide square bracket.

For me, the key to the solution was when I drew the lines AE and CF.  The software I use (Sketchpad) lets me put on lines that are truly parallel and maintains all the 90 angles.  So I was able to experiment by varying the parallels.  It looked like AE = BF so I thought I'd try to prove it.  Then everything fell into place nicely.

With geometry I usually find the diagram is the key thing.  Get that right and the route to an answer often becomes visible.  I needed to put on two lines to show the 12 and 17 distances and it seemed it would keep things simpler if I drew these from A and from C. 

Hope that helps.  smile

Bob


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

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