Please help with geometry!!!

- anonymous

Please help with geometry!!!

- Stacey Warren - Expert brainly.com

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- anonymous

can you help me first

- anonymous

|dw:1364254756475:dw| Find the area of ABCD.

- anonymous

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

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- anonymous

Ok like this??|dw:1364255189854:dw|

- anonymous

@Hoa

- anonymous

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????

- anonymous

b*h = Area for rectangle and parallelogram

- jim_thompson5910

are you sure that BC = 8?

- jim_thompson5910

it would make things a lot easier if BC = 8

- jim_thompson5910

i meant, BC = 6

- anonymous

Yea:) @jim_thompson5910 It does!!

- jim_thompson5910

ok one sec

- anonymous

Oh i know what ur saying... @jim_thompson5910

- jim_thompson5910

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

that length is 32
|dw:1364255489567:dw|

- anonymous

Yea...

- jim_thompson5910

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

- jim_thompson5910

since 22+10 = 32

- anonymous

Huh?

- anonymous

Nvm i know what urr doing:)

- jim_thompson5910

We now have this triangle
|dw:1364255559257:dw|

- anonymous

And so u have to do the pythagoreum theorm thing...

- anonymous

to find 'h'

- jim_thompson5910

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|

- anonymous

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

- jim_thompson5910

if x was this length
|dw:1364255675546:dw|

- jim_thompson5910

then 10-x must be this length here
|dw:1364255697460:dw|

- anonymous

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

- jim_thompson5910

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

- jim_thompson5910

alright sorry

- anonymous

|dw:1364255854635:dw| I dont get what u did here... @jim_thompson5910

- jim_thompson5910

I broke up that length of 10 into two pieces

- jim_thompson5910

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

- jim_thompson5910

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

- anonymous

So y can t u just leave it as 10??

- anonymous

@jim_thompson5910

- jim_thompson5910

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

- jim_thompson5910

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

- anonymous

k hold on..

- jim_thompson5910

alright

- anonymous

|dw:1364256139846:dw|

- jim_thompson5910

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

- anonymous

ok so now it makes sense...

- jim_thompson5910

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

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

- jim_thompson5910

with me so far?

- anonymous

Kinda, i still dont get y u did 10 - x ?? lol sry

- jim_thompson5910

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

- jim_thompson5910

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

- anonymous

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

- jim_thompson5910

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

- jim_thompson5910

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

- jim_thompson5910

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

- jim_thompson5910

so that's why we must go this route

- anonymous

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

- jim_thompson5910

no need for sorries, questions are good

- anonymous

lol

- jim_thompson5910

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

- jim_thompson5910

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

- jim_thompson5910

let's do substitution

- anonymous

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.

- anonymous

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

- jim_thompson5910

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)
(10-x)^2 + h^2 = 64
(10-x)^2 + (36 - x^2) = 64
Now FOIL out, expand and simplify
(10-x)^2 + (36 - x^2) = 64
100 - 20x + x^2 + 36 - x^2 = 64
100 - 20x + 36 = 64
136 - 20x = 64
-20x = 64-136
-20x = -72
x = -72/(-20)
x = 18/5
x = 3.6

- jim_thompson5910

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

- anonymous

k let me get this down...

- jim_thompson5910

alright

- anonymous

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

- anonymous

@jim_thompson5910

- jim_thompson5910

through the use of the pythagorean theorem

- jim_thompson5910

a^2 + b^2 = c^2

- jim_thompson5910

it's explained a bit more above

- anonymous

where did u get 36 from?

- jim_thompson5910

from squaring the 6

- anonymous

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

- anonymous

@jim_thompson5910 ??

- jim_thompson5910

I'm doing substitution

- anonymous

kk

- jim_thompson5910

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

notice how the h terms went away after that

- anonymous

kk i need to write this

- jim_thompson5910

alright

- anonymous

Sry it took so long... but is this 3.6??

- jim_thompson5910

good

- jim_thompson5910

x = 3.6

- jim_thompson5910

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

- anonymous

|dw:1364258514592:dw|

- anonymous

Is tht 3.6 ??

- jim_thompson5910

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

- jim_thompson5910

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

- anonymous

kk

- jim_thompson5910

So the height is 4.8

- anonymous

So is tht for this one??

- anonymous

|dw:1364258641203:dw|

- jim_thompson5910

no this one here
|dw:1364258674211:dw|

- jim_thompson5910

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

- anonymous

So, i thought u said that

- anonymous

|dw:1364258753145:dw| trhis length is 10 - x ??

- anonymous

*this

- jim_thompson5910

yes

- jim_thompson5910

but we found x, then used that to find h

- jim_thompson5910

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

- anonymous

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

- anonymous

@jim_thompson5910

- jim_thompson5910

exactly

- anonymous

k hold on let me get tht down..

- jim_thompson5910

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

- anonymous

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

- anonymous

@jim_thompson5910

- jim_thompson5910

ok that's a lot of work to find h

- jim_thompson5910

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

- jim_thompson5910

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

- anonymous

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

- jim_thompson5910

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

- jim_thompson5910

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

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

- anonymous

kk hod it bro

- jim_thompson5910

alright

- anonymous

So thts it?? @jim_thompson5910

- jim_thompson5910

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

- anonymous

No thank u, i perfer decimals:)

- jim_thompson5910

ok great

- jim_thompson5910

yeah decimals are more intuitive with area problems

- anonymous

THANK THANK YOU SOOOOSOSOOOO MUCH:D:D:D:D: :):) UR THE BEST :D:D:D

- jim_thompson5910

lol thx and glad to be of help

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