Math PROBLEM

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theprince786

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I am sort of stumped on this Math Problem for my Geometry/Algebra 2 class (hey, im in ninth grade, lol). We are supposed to be able to do it in 3 minutes and have to combine geometry and algebra, but the way I was doing it was purely algebraic and takes a REALLLYYY long time to do. I was wondering if anyone could help me here. Sorry the figure is so rough. And the line segment signs are underlines. If a variable is to the power of something, it is followed by ^, then the exponent.

The Problem:
tri-1.jpg


In the figure above, AC bisects BD. Also, let the following be true:
- BD = 5x-2
- MB = x^2-7
- AM = 16-x
- CM = 2x+4
- AD = 3x+3.

Is BD the bisector of AC?

Thx in advance for the help...

Peace brothers and sisters...

:w:​
 
salaams

2 difficult 4 me

i know some1 who is like the best in maths in the country if u stiil dont know?
 
Assalamu Alaikum

lol im working on it bro

so far i figured out that AM is equal to CM *both equal to 4*
im i little rusty in my geometry but isnt that all that u need to know whether its a bisector?

fi aman Allah
w'salaam
 
theprince786 said:
In the figure above, AC bisects BD.
Is BD the bisector of AC?
Click to expand...

wait a min if AC bisect BD then ofcourse BD is the bisector of AD right??
because to bisect mean to like cut in half so if BD bisects AD then obviously AD will bisect BD
 
:sl: brother
i reely wish i cud help bu..im sori i failed in maths big tym..evn failed ma resit!:embarrass ..ur diagrams enough to confuse me:confused: !


al da best..hope derz someone hu can help more than myslf! :w:
 
*charisma* said:
Assalamu Alaikum

lol im working on it bro

so far i figured out that AM is equal to CM *both equal to 4*
im i little rusty in my geometry but isnt that all that u need to know whether its a bisector?

fi aman Allah
w'salaam
Click to expand...

:sl: Thanks but how did u get that? how do u get the value of X?:w:

Gandalf said:
wait a min if AC bisect BD then ofcourse BD is the bisector of AD right??
because to bisect mean to like cut in half so if BD bisects AD then obviously AD will bisect BD
Click to expand...

:sl: Not necessarily.... here is an example where that wont apply...
bisect-1.jpg


You see, the midpoint of AB is the blu point, which i forgot to label. The midpoint of CD is the redpoint. Bisect is when a line, segment, or ray goes through a line segment's midpoint. So can you see now, that AB bisects CD but is not bisected by CD? Hopefully that will clear stuff up for u....

And to others, thanx for helpin or tryin to....

:w:
 
AOA i forgot.... I see how u got X charisma...

the thing is, all thats given is:
AC bisects BD. Also, let the following be true:
- BD = 5x-2
- MB = x^2-7
- AM = 16-x
- CM = 2x+4
- AD = 3x+3.

It seems that you are assuming that AM=CM. It might not be true, because it is not a given fact. If you can prove that it is, thatll be great, cuz thats what I was trying to do originally, but it took waaayyy over 3 mins.

thanks agin...

WASA
 
Assalamu Alaikum

lol.. alhemdulilah that u got it cuz i was bout to go and look back at my geometry notes lol...i forgot a lot of that stuff..but it was a pleasure to help, so anytime ekhi :) math will get fun when u get into polynomial functions and parabolas haha

fi aman Allah
w'salaam
 
:sl: In the diagram above line BD bisects line AC because BD is perpindicular to AC therefore it creates a cross or plus sign, however you may call it.:w:
 
:sl:
Sister Halima, The problem is that it is not given that the line BD is perpendicular to AC. we learned in class that the diagrams may decieve you.... dont trust them... trust the measurements. For example, it was not given that angle BAC was a right angle, or that the two segments were perpendicular. Therefore, you do not know that for sure.... Thanks anyway.... your nswer was right tho

Ok so for you who are interested, here is how my dad helped me solve the problem. Sister Charisma was doing it right. The x=4 could be easily checked with the other equations. My mstake was that I thought you were not allowed to assume something then prove it. Turns out that you are.

Ok so here it is. Let's assume that AM = CM. This way you get the equation 16-x=2x+4. You solve for X and get 4. You plug this into the equation for DM and MB, both of wich are x^2-7, because the midpoint of DB is revealed by the bisecting segment AC. so you get 16-7=16-7=9=9. You have the measure of AD, which is the hypotenuse of triangle DMA. When you get the measure of the hypotenuse squared (AD^2=225), use the pythagorean theorem to check your answer like so: DM^2+AM^2= 81+144 = 225. IT WORKS.... SO, to get the other one, BC, you use the pythagorean theorem. You end up with AD=BC. You will end up wth AM = CM, after all these checks. So you see that M is the midpoint of both segments. The answer is yes.

Hopefully that helped. Im not so sure I was clear, but oh well. Thanks evry1...


:w:
 

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