LizzyLove<3
Please help with geometry!!!



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superraymond
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can you help me first

LizzyLove<3
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dw:1364254756475:dw Find the area of ABCD.

Hoa
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let CO is the height of the triangle, and let F is the point between A and O. redraw it. I guide you then

LizzyLove<3
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Ok like this??dw:1364255189854:dw

LizzyLove<3
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@Hoa

Hoa
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perfect, now, you have 2 regions. right? and you have formula to get the answer, try, the first region is parralelogram ADCF, the area formula is????

LizzyLove<3
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b*h = Area for rectangle and parallelogram

jim_thompson5910
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are you sure that BC = 8?

jim_thompson5910
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it would make things a lot easier if BC = 8

jim_thompson5910
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i meant, BC = 6

LizzyLove<3
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Yea:) @jim_thompson5910 It does!!

jim_thompson5910
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ok one sec

LizzyLove<3
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Oh i know what ur saying... @jim_thompson5910

jim_thompson5910
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anyways, we need CO, which will give us the height needed to find the area of the paralleogram
but before we find CO, we need FB

jim_thompson5910
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that length is 32
dw:1364255489567:dw

LizzyLove<3
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Yea...

jim_thompson5910
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this length is 22
dw:1364255516091:dw
so this remaining length here
dw:1364255533268:dw
must be 10

jim_thompson5910
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since 22+10 = 32

LizzyLove<3
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Huh?

LizzyLove<3
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Nvm i know what urr doing:)

jim_thompson5910
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We now have this triangle
dw:1364255559257:dw

LizzyLove<3
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And so u have to do the pythagoreum theorm thing...

LizzyLove<3
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to find 'h'

jim_thompson5910
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it's not drawn to scale because this is actually a right triangle with hypotenuse of 10
so it really looks like this
dw:1364255624494:dw

LizzyLove<3
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But u cant do tht, if it doesnt say in my HW...

jim_thompson5910
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if x was this length
dw:1364255675546:dw

jim_thompson5910
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then 10x must be this length here
dw:1364255697460:dw

LizzyLove<3
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Hold on there bro, let me see what ur doing, plz ... @jim_thompson5910

jim_thompson5910
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so what this means is that we can say
x^2 + h^2 = 6^2
and
h^2 + (10x)^2 = 8^2

jim_thompson5910
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alright sorry

LizzyLove<3
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dw:1364255854635:dw I dont get what u did here... @jim_thompson5910

jim_thompson5910
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I broke up that length of 10 into two pieces

jim_thompson5910
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one piece is x units long
the other is 10x units long

jim_thompson5910
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both add back up to 10
(x) + (10x) = x + 10x = 10

LizzyLove<3
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So y can t u just leave it as 10??

LizzyLove<3
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@jim_thompson5910

jim_thompson5910
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because breaking up this length of 10 will help us find the height h

jim_thompson5910
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it's not the only way to find h, but it's probably one of the easier ways

LizzyLove<3
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k hold on..

jim_thompson5910
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alright

LizzyLove<3
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dw:1364256139846:dw

jim_thompson5910
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yes you break up the triangle like that and that will help you find h

LizzyLove<3
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ok so now it makes sense...

jim_thompson5910
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this first triangle here
dw:1364256244802:dw
has legs of x and h, with hypotenuse 6, so
a^2 + b^2 = c^2
x^2 + h^2 = 6^2
x^2 + h^2 = 36

jim_thompson5910
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this triangle here is a right triangle with
leg1 = 10x
leg2 = h
hypotenuse = 8
which means
a^2 + b^2 = c^2
(10x)^2 + h^2 = 8^2
(10x)^2 + h^2 = 64

jim_thompson5910
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with me so far?

LizzyLove<3
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Kinda, i still dont get y u did 10  x ?? lol sry

jim_thompson5910
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well if you break up 10 into two pieces, then one piece is x units long
the other must be 10x units long

jim_thompson5910
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example
if x was say x = 2 inches, then the first piece would be 2 inches
the remaining bit would be 10x = 102 = 8 inches

LizzyLove<3
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Yea i get it but y can u just split it in half??

jim_thompson5910
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they must add back up to 10, and they do
x + (10x)
2 + (8)
10

jim_thompson5910
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because we do NOT know whether the length h cuts the side 10 in half or not

jim_thompson5910
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it could, but it could not (in this case, it doesn't)
we just don't know

jim_thompson5910
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so that's why we must go this route

LizzyLove<3
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ok... sry for tht, its just i need to know how to do this, thxss :) so wats next?

jim_thompson5910
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no need for sorries, questions are good

LizzyLove<3
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lol

jim_thompson5910
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ok we have these two equations
x^2 + h^2 = 36
(10x)^2 + h^2 = 64

jim_thompson5910
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we can either use substitution or elimination to solve for x and h

jim_thompson5910
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let's do substitution

LizzyLove<3
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ok, i think there is a way simpler way... im pretty sure tht u can split tht in half, my teacher said tht we wouldn't need to use sub. or eli.

LizzyLove<3
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Can we just assume, b/c it's a right angle??

jim_thompson5910
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Start with equation 1 and isolate h^2 (not h, just h^2)
x^2 + h^2 = 36
h^2 = 36  x^2

Now replace h^2 in equation 2 with 36  x^2 (since h^2 = 36  x^2 above)
(10x)^2 + h^2 = 64
(10x)^2 + (36  x^2) = 64
Now FOIL out, expand and simplify
(10x)^2 + (36  x^2) = 64
100  20x + x^2 + 36  x^2 = 64
100  20x + 36 = 64
136  20x = 64
20x = 64136
20x = 72
x = 72/(20)
x = 18/5
x = 3.6

jim_thompson5910
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This is why I asked if BC was 8 or not
If BC = 6, then you could split the 10 into two equal pieces

LizzyLove<3
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k let me get this down...

jim_thompson5910
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alright

LizzyLove<3
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So how did u get these two equations??
(10x)^2 + h^2 = 64
x^2 + h^2 = 36

LizzyLove<3
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@jim_thompson5910

jim_thompson5910
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through the use of the pythagorean theorem

jim_thompson5910
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a^2 + b^2 = c^2

jim_thompson5910
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it's explained a bit more above

LizzyLove<3
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where did u get 36 from?

jim_thompson5910
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from squaring the 6

LizzyLove<3
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So is it sub or elim tht ur doing ? @jim_thompson5910

LizzyLove<3
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@jim_thompson5910 ??

jim_thompson5910
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I'm doing substitution

LizzyLove<3
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kk

jim_thompson5910
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I isolated h^2 in the first equation
and then replaced the other h^2 in the second equation with 36  x^2

jim_thompson5910
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notice how the h terms went away after that

LizzyLove<3
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kk i need to write this

jim_thompson5910
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alright

LizzyLove<3
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Sry it took so long... but is this 3.6??


jim_thompson5910
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x = 3.6

jim_thompson5910
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btw I showed how it's x = 3.6 above lol
but I'm glad you got that on your own

LizzyLove<3
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dw:1364258514592:dw

LizzyLove<3
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Is tht 3.6 ??

jim_thompson5910
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yep, that length you are pointing to is 3.6 units long

jim_thompson5910
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so
x^2 + h^2 = 6^2
x^2 + h^2 = 36
(3.6)^2 + h^2 = 36
12.96 + h^2 = 36
h^2 = 36  12.96
h^2 = 23.04
h = sqrt(23.04)
h = 4.8

LizzyLove<3
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kk

jim_thompson5910
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So the height is 4.8

LizzyLove<3
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So is tht for this one??

LizzyLove<3
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dw:1364258641203:dw

jim_thompson5910
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no this one here
dw:1364258674211:dw

jim_thompson5910
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that's what we wanted
we really don't need x to find the area
we only needed x to find h

LizzyLove<3
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So, i thought u said that

LizzyLove<3
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dw:1364258753145:dw trhis length is 10  x ??

LizzyLove<3
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*this


jim_thompson5910
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but we found x, then used that to find h

jim_thompson5910
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we don't need to know what 10x is
we just want h

LizzyLove<3
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So the last equation u put was to find 'h' right??

LizzyLove<3
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@jim_thompson5910

jim_thompson5910
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exactly

LizzyLove<3
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k hold on let me get tht down..

jim_thompson5910
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I started with x^2 + h^2 = 36, plugged in x = 3.6, then solved for h

LizzyLove<3
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OK!! lol i get what we did, so whats next?? Do we just do 32 * 4.8??

LizzyLove<3
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@jim_thompson5910

jim_thompson5910
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ok that's a lot of work to find h

jim_thompson5910
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like I said, it would be a lot easier if BC = 6

jim_thompson5910
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but it's still doable to find h even if BC wasn't 6

LizzyLove<3
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so was i right?? or is there a different way to do this?

jim_thompson5910
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anyways, this is a trapezoid with
base1 = 32
base2 = 22
height = 4.8

jim_thompson5910
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so we can find the area using this formula
A = h*(b1+b2)/2
where in this case
h = 4.8
b1 = 32
b2 = 22

jim_thompson5910
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A = h*(b1+b2)/2
A = 4.8*(32+22)/2
A = 4.8*(54)/2
A = 4.8*27
A = 129.6

LizzyLove<3
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kk hod it bro

jim_thompson5910
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alright

LizzyLove<3
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So thts it?? @jim_thompson5910

jim_thompson5910
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yep pretty much, unless you want to convert that into a fraction

LizzyLove<3
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No thank u, i perfer decimals:)

jim_thompson5910
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ok great

jim_thompson5910
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yeah decimals are more intuitive with area problems

LizzyLove<3
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THANK THANK YOU SOOOOSOSOOOO MUCH:D:D:D:D: :):) UR THE BEST :D:D:D

jim_thompson5910
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lol thx and glad to be of help