Who wants to do my Geometry homework??

realityisterror · Jan 6, 2005, 02:34 PM

i can't get this problem..

using this diagram, find the area of pentagon BNREC...
http://bellsouthpwp.net/a/r/arminv/storage/homework.gif

i can get the area of triangle BEC easily... i just can't get the BNRE part because of that little bend in there...

also, the drawing isn't quite to scale, but none of the sides touch a point unless they are obviously on it...

thanks,
reality

wordmunger · Jan 6, 2005, 02:42 PM

I think I got it. You can easily get the area of the rectangle abnz, where z is an imaginary new point on line segment be. z is exactly 3 units above f. the area of triangle bnz is 1/2 area abnz. Now imagine a point x on segment vr, which forms square zfrx. It's easy to find the area of triangle xrn, so your job is done.

AppleMatt · Jan 6, 2005, 03:01 PM

Legend.

AppleMatt

realityisterror · Jan 6, 2005, 03:19 PM

Quote:

Originally Posted by wordmunger

I think I got it. You can easily get the area of the rectangle abnz, where z is an imaginary new point on line segment be. z is exactly 3 units above f. the area of triangle bnz is 1/2 area abnz. Now imagine a point x on segment vr, which forms square zfrx. It's easy to find the area of triangle xrn, so your job is done.

only on MR people... only on MR...

you are a god wordmunger... i didn't really care if i missed one question... if was just driving me crazy..

thanks again,
reality

MacNeXT · Jan 6, 2005, 04:03 PM

373,5

noel4r · Jan 6, 2005, 05:56 PM

i got 369, i grouped them by 4 triangles and 1 square got the area then added them up.

Daveway · Jan 6, 2005, 06:06 PM

:-B

I could've done this problem.

themadchemist · Jan 6, 2005, 06:46 PM

The way I see you doing it is getting the area of everything outside of the desired area (besides the bec part, which you already have). That stuff is simple enough to figure out because of the lengths given on the side. When you subtract it all, you are left with the area of the weird shape for which you're looking.

Of course, wm beat us all to the punch.

hcuar · Jan 6, 2005, 08:53 PM

Pah... that's not Geometry homework... Where's your T column proofs? heh...

Now that's Geometry homework.

Daveman Deluxe · Jan 6, 2005, 09:11 PM

I think I have an even easier way (that is, one that doesn't require the creation of new points).

Take the area of triangle bce, which is 162 units square. Then you place he same imaginary point z (I guess it does take one imaginary point) and then find the area of triangle bnz, which is 121.5 units squared. Then find the area of trapezoid nrfz, which is 22.5 square units, then find the area of triangle rfe, which is 4.5 units square. The total area is 310.5.

That said, I agree with hcuar.

Edit: I just realized that themadchemist's solution is even more elegant than mine is.

Onishenko · 09:55 PM

Quote:

Originally Posted by noel4r

i got 369, i grouped them by 4 triangles and 1 square got the area then added them up.

Agreed. I did it differently, using some of the method that wordmunger used. But came to the same conclusion.

369. On the money.

Edit: Agreed with post above. Madchemist's method is alot simpler. Wish I saw that. Probably spent hours in 8th grade meditating over these types of questions when I could have used this simpler method.

scem0 · Jan 6, 2005, 11:29 PM

Quote:

Originally Posted by hcuar

Pah... that's not Geometry homework... Where's your T column proofs? heh...

Now that's Geometry homework.

Thinking about those hurts my soul

.

Geometry is something I want to leave far, far behind. Algebra II was a bunch of fun, though

.

scem0

oldpismo · Jan 7, 2005, 01:50 AM

I used the same method as wordmunger, and came up with 373.5, as did MacNext

We seem to be getting alot of different answers to a relatively simple question.

Peter

jasylonian · Jan 7, 2005, 02:56 AM

Quote:

Originally Posted by oldpismo

I used the same method as wordmunger, and came up with 373.5, as did MacNext

We seem to be getting alot of different answers to a relatively simple question.

373.5 looks right to me. 162+162+36+9+4.5.

Daveman Deluxe · Jan 7, 2005, 08:45 PM

Quote:

Originally Posted by oldpismo

I used the same method as wordmunger, and came up with 373.5, as did MacNext

We seem to be getting alot of different answers to a relatively simple question.

Well, I realized after seeing this that I did my arithmetic wrong. It is 373.5 units square.

Good thing I check my own tests, I suppose.

Ajmbc · Jan 8, 2005, 01:38 AM

Quote:

Originally Posted by hcuar

Pah... that's not Geometry homework... Where's your T column proofs? heh...

Now that's Geometry homework.

Heh, Geometry is not geometry without the legendary 2 column proofs :-P

-ajmbc

Jan 6, 2005, 02:34 PM	#1
realityisterror macrumors 65816 Join Date: Aug 2003 Location: Snellville, GA	Who wants to do my Geometry homework?? i can't get this problem.. using this diagram, find the area of pentagon BNREC... http://bellsouthpwp.net/a/r/arminv/storage/homework.gif i can get the area of triangle BEC easily... i just can't get the BNRE part because of that little bend in there... also, the drawing isn't quite to scale, but none of the sides touch a point unless they are obviously on it... thanks, reality __________________ </end>
	0

Jan 6, 2005, 02:42 PM	#2
wordmunger macrumors 603 Join Date: Sep 2003 Location: North Carolina	I think I got it. You can easily get the area of the rectangle abnz, where z is an imaginary new point on line segment be. z is exactly 3 units above f. the area of triangle bnz is 1/2 area abnz. Now imagine a point x on segment vr, which forms square zfrx. It's easy to find the area of triangle xrn, so your job is done. __________________ ResearchBlogging.org Science-Based Running
	0

Jan 6, 2005, 03:01 PM	#3
AppleMatt macrumors 68000 Join Date: Mar 2003 Location: UK	Legend. AppleMatt __________________ Avatar by edesignuk. Fold for team MacRumors.
	0

Jan 6, 2005, 04:03 PM	#5
MacNeXT macrumors 6502 Join Date: Jun 2004	373,5 __________________ Oh, and by the way, Micro$oft suxorz!!!
	0

Jan 6, 2005, 06:06 PM	#7
Daveway macrumors 68040 Join Date: Jul 2004 Location: New Orleans / Lafayette, La	:-B I could've done this problem. __________________ Spare me my life.
	0

Jan 6, 2005, 05:56 PM	#6
noel4r macrumors 6502a Join Date: Jul 2002 Location: Los Angeles	i got 369, i grouped them by 4 triangles and 1 square got the area then added them up.
	0

Jan 6, 2005, 06:46 PM	#8
themadchemist macrumors 68030 Join Date: Jan 2003 Location: Chi Town	The way I see you doing it is getting the area of everything outside of the desired area (besides the bec part, which you already have). That stuff is simple enough to figure out because of the lengths given on the side. When you subtract it all, you are left with the area of the weird shape for which you're looking. Of course, wm beat us all to the punch. __________________ Sometimes, things get worse before they get better. Sometimes, they get better before they get worse. Almost never do they stay the same.
	0

Jan 6, 2005, 08:53 PM	#9
hcuar macrumors 65816 Join Date: Jul 2004 Location: Dallas	Pah... that's not Geometry homework... Where's your T column proofs? heh... Now that's Geometry homework. __________________ iMac 27" i5 Mid 2011 \| 1TB HD \| 16GB RAM iPhone 4 32 Gig
	0

Jan 6, 2005, 09:11 PM	#10
Daveman Deluxe macrumors 68000 Join Date: Jun 2003 Location: Corvallis, Oregon	I think I have an even easier way (that is, one that doesn't require the creation of new points). Take the area of triangle bce, which is 162 units square. Then you place he same imaginary point z (I guess it does take one imaginary point) and then find the area of triangle bnz, which is 121.5 units squared. Then find the area of trapezoid nrfz, which is 22.5 square units, then find the area of triangle rfe, which is 4.5 units square. The total area is 310.5. That said, I agree with hcuar. Edit: I just realized that themadchemist's solution is even more elegant than mine is.
	0

Jan 7, 2005, 01:50 AM	#13
oldpismo macrumors member Join Date: Aug 2003 Location: UK	same method different answers I used the same method as wordmunger, and came up with 373.5, as did MacNext We seem to be getting alot of different answers to a relatively simple question. Peter __________________ 20" G5 iMac, late 2008 2.4GHz MacBook
	0


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