Geometry Help Please!

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XxMontanaxX

XxMontanaxX

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Will somebody explain this to me? Like very detailed...because I'm slow.

gank9.jpg


ABCD is a rectangle.

1. If AC = 2x + 13 and DB = 4x - 1, find x.

2. If AC = x + 3 and DB = 3x - 19, find AC.

3. If AE = 3x + 3 and EC = 5x = 15, find AC.

4. If DE = 6x - 7 and AE = 4x + 9, find DB.

5. If m<DAC = 2x + 4 and m<BAC = 3x + 1, find x.

6. If m<BDC = 7x + 1 and m<ADB = 9x -7, find m<BDC.

7. If m<ABD = x^2 - 7 and m<CDB = 4x + 5, find x.

8. If m<BAC = x^2 + 3 and m<CAD = x + 15, find m<BAC.



And m< means "measure of angle"

and x^2 is "x squared"

I didn't know how to get the symbols for those.

Thanks to anyone who helps!

And I have a lot more problems so everyone keep comming back and checking pleaseee!
 
Montana i feel for you i really do, were just finishing quadratic equations and moving onto algebra with a HUGE test on tuesday. Good luck with the questions but sadly i suck at geometry so i cant really help you.
 
girl please you are NOT slow. i HATE GEOMETRY PASSIONATELY!!

1. AC and DB are the same length. So....

2x+13 = 4x-1

i would subtract 2x from both sides, which gives you 2x-1=13. from there i'm assuming you can do algebra....x=7.

2. Again, AC and DB are the same length SO...

x+3 = 3x - 19

i would subtract x from both sides...2x-19=3. using simple algebra again....x = 11.

Now plug 11 in as x. so, 11 + 3 = 14. So AC = 14.


3. AE = EC just like AC = DB.

YOU wrote 3x+3=5x=15. I don't know if you meant 5x+15 or 5x-15 so i'm going to just assume +15.

3x+3=5x+15

i would subtract 3x from both sides, leaving you with 2x+15=3. simple algebra again gives us x = -6.

SO 3(-6)+3= -18+3=-15...
AND 5(-6)+15= -15....

that would equal -30. but that doesn't make sense because the line is clearly there, it's not a negative number! so if you meant....

3x+3=5x-15

i would subtract 3x again from both sides. 2x-15=3. simple algebra, gives us x = 9.

SO 3(9)+3= 30
AND 5(9)+15 = 60

Add both of those segments together to give you fully AC (AE + EC = AC) and you get 90!

AC = 90!


4. DE and AE are both the same length as well. Because of the 90 degree angles, everything is going to be equal in there....

SO...knowing THAT...

6x-7 = 4x+9

subtract 4x from both sides... 2x-7=9. simple algebra...x = 8.

Now DE = 6(8) - 7 = 41.

Now, AE is the same as EB, so just AE's given equation of 4(8)+9 = 41.

41 + 41 = 82. (DE + EB = DB)

DB = 82 !!!

Yawn...you get the picture here yes?!?!?! I hope I helped you lady! PM me if you need anything else or reply here. I won't be around tomorrow much, but good luck!! :)

Tracy


 
I am so sorry if any of these are wrong! And btw, I was very slow in Math and nearly failed grade 12 math :shock:

1. If AC = 2x + 13 and DB = 4x - 1, find x.

2X + 13 = 4x - 1
-2x +1 -2x+1
---------------------
0 + 14 = 2x + 0

You must subtract to have only one value on each side: An "x" value which equals a number value. The only way to do that is to make some of the values 0. Because 2x is smaller than 4x, it only makes sense to eliminate "2x", so we subtract 2x from 2x: 2x - 2x = 0. Whatever you subtract from one side, you must do to the other. So we subtract 2x from 4x: 4x - 2x = 2x.

Now that you've eliminated a value one the left side, we have to do it on the right, also since the x value is located on the right.
If you add a +1 to the - 1, that results in a net value of 0. And remember, whatever you do to one side, you must do to the next. So we had a +1 to + 13, making it +14

14 = 2x
7 = x

2. If AC = x + 3 and DB = 3x - 19, find AC.

Same applies to #2! Subtract so that there is only one value on one side. I usually start with the x values - I eliminate the smaller of the two.

x + 3 = 3x - 19
-x + 19 -x + 19
-----------------------
0 + 22 = 2X + 0

22 = 2X
11 = x

3. If AE = 3x + 3 and EC = 5x = 15, find AC.

To do this, we need to find "x"

3X + 3 = 5x = 15 If this is the question, then you need to move 15 to the other side.

5x = 15.
- 15 - 15
-------------
5x - 15 = 0

So now:

3X + 3 = 5X -15
-3x +15 -3x +15
------------------------
0 + 18 = 2x + 0

18 = 2x
9 = x

So now that we have that value of X, we can plug it in!

AE: 3(9) + 3 = 30
EC: 5(9) - 15 = 30

AE + EC = AC or 60. I got a different answer than Tracy..

EDIT: Wait, Tracy. You're original equation was identical to mine "5X - 15" And then you later switched to EC = 5(9) "+" 15 = 90. How/why did you switch from a " -15 " to a "+15" I is confused!

4. If DE = 6x - 7 and AE = 4x + 9, find DB.

6x - 7 = 4x + 9
-4x +7 - 4x + 7
------------------------
2x + 0 = 0 + 16

2x = 16
x = 8

DE: 6(8) - 7 = 41
AE: 4(8) + 9 = 41

The length of AE is identical to BE. So, DE + AE (41 + 41) = 82!

And the rest... do you have to convert your answers into degrees or radians?

 
:huh

I uhh...... nope.... no help here. Sorry.

Tho my son said you have to get "x" to one side? I think that's what he said earlier......

OMG I am so bad at math....
 
Yeah, I meant -15 in that one equation.

Thank you so much for helping me and explaining! :hug:

But could someone explain 7 and 8 for me too please?
 
OMG! Montana I wish I could help :pI'm having to study this stuff again because I need to refresh my math for some courses I am taking now. I HATE math. I try to tell myself everyday that I don't so I can keep a *positive* attitude lol, but it doesn't work. Good luck!
 
I know! That's me! I'm like "Today I'm gonna try really hard and I'm going to understand this!" Then five minutes later i'm like "blahh".



Can someone help me with this too?

ggcb3.jpg


PRST is a rectangle. Find each measure if m<1 = 50

m<2 -

m<3 -

m<4 -

m<5 -

m<6 -

m<7 -

m<8 -

m<9 -
 
well m<1 is 50...and it's an equilateral triangle, so m<7 and m<9 are 50 degrees as well!

so is m<4 because it's equal in size to the other triangle.

annnnnd that's about as far as i can gurantee you. don't hate me <3.

Tracy
 
WAIT although maybe not....because don't all angles of a triangle have to equal 180 degrees? then they'd all have to be 60 degrees...

forget i even tried. for gods sake....you're bringin' back my nightmares girl!! :)

Tracy
 
oOo geezzz... I'm getting nightmares too =P

I'll give it a shot, I can't guarantee that they're right :?

M>1 and M> 7 appear equal, so I will assume that M>7 is 50 as well. I never trust an equilateral triangle unless I measure all three sides exactly! And after measuring, the length of M>9 is 1 mm longer than M>1 and M>7, so it is not equilateral.

In a triangle, everything must add up to 180 in total. So, 180 - 50 - 50 = 80. And so, I think that M>9 = 80.

M>4 looks equal to M>1 and M>7 as well, so M>4 = 50.

P,T,R,S are all 90 degree angles.. so I think that to find M>2, you'd have to subtract M>1 from 90 degrees: 90 - 50 = 40. M>2 = 40.

M>3, M>5, and M>6 are identical to M>2... so they all equal 40.

For M>8, again all triangles must equal 180. 180 - 40 - 40 = 100. M>8 = 100.

M>1 = 50
M>2 = 40
M>3 = 40
M>4 = 50
M>5 = 40
M>6 = 40
M>7 = 50
M>8 = 100
M>9 = 80

Also, when there is a line.. such as M>8 and M>9 create a line. The line always had to add up to 180. 80 + 100 do equal 180.

Again, I'm sorry if they're wrong. It's been a year since I've been in math. Argh.. how I hated math.
 
Okay, lets see if I can remember my geometry, I haven't done this since high school. But if I can do advanced calculus, I should be able to figure this out.

So, you know that every corner of a rectanble is 90 degrees. So any angles in a rectangle corner will add up to 90 degrees.

>1=50 means that >2=90-50=40 degrees

There are also rules about oposite and equal angles. Which gives you this relationship:

>2=>3=40 degrees

>4=90->3=90-40=50

>5=>2=40

>6=>5=40

>7=90->6=90-40=50

Now, you know that all angles in a triangle sum up to 180, so use that to find 8 and 9.

180->6->5=180-40-40=100=>8

180->7->1=180-50-50=80=>9

Geometry follows a very specific set of rules. The trick is starting in the correct place and they should all fall out. I would write out all of the angle rules and have them easily accesable to you when you do your homework. It's all logic, and once you get it you shouldn't really forget it.

--Dawn


 
Thank goodness for Pixie and Dawn. I'm glad I posted my disclaimer and said ya can't trust my math! BUT, I think the angles stuff I did a few days ago for ya, Montana, is correct!! :)

Tracy
 
Yayy I got the same answers as Dawn, so I'm not that brain dead!
 
You guys are amazing!

:hugsquish:

I think I'm understanding this a little better. A few minutes reading what you guys posted makes me understand it then a whole year of Mr Casto's class...that's sad.
 
Geometry is pretty easy once you learn the rules. If you have a text book, I would go through the chapter and write out each rule and make sure you understand it. Then once you have all the rules written out, you can just refer to them when you do your homework.

I find that's one of the most successful way of studying, well for me it is. I write out all the formulas and rules then do some practice problems. It helps to clarify the rules in my mind.

--Dawn
 
I'll bring home my geometry book tommorow and try that. That's a good idea!

And here are some more if you guys don't care to help me.

yywv4.jpg


Use rhombus DKLM with AM = 4x, AK = 5x - 3, and DL = 10

1. Find X

2. Find AL.

3. Find m<KAL

4. Find DM.



So to find x, do you just do 4x = 5x -3?

And to find AL, is it the same as AM?
 
I had to go and look up exactly what a rhombus was, but I got it figured out.

I'm going to try to explain it to you so you can do this one yourself.

Take a look at these two links:
http://en.wikipedia.org/wiki/Rhombus
http://www.mathsisfun.com/quadrilaterals.html

So from the properties of a rhombus, we know that:

1) AM=AK, and from that you can solve for X.

2) Same property of above, DA=AL and DL=DA+AL

3) Look at the second link, the answer is in the description of a rhombus, scroll down and read that section.

4)Use pythagorus theorum to find DM, which is your hyponenus.

I hope that helps.

--Dawn


 
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