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can you help me first
|dw:1364254756475:dw| Find the area of ABCD.
let CO is the height of the triangle, and let F is the point between A and O. redraw it. I guide you then
Ok like this??|dw:1364255189854:dw|
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????
b*h = Area for rectangle and parallelogram
are you sure that BC = 8?
it would make things a lot easier if BC = 8
i meant, BC = 6
Yea:) @jim_thompson5910 It does!!
ok one sec
Oh i know what ur saying... @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
that length is 32 |dw:1364255489567:dw|
this length is 22 |dw:1364255516091:dw| so this remaining length here |dw:1364255533268:dw| must be 10
since 22+10 = 32
Nvm i know what urr doing:)
We now have this triangle |dw:1364255559257:dw|
And so u have to do the pythagoreum theorm thing...
to find 'h'
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|
But u cant do tht, if it doesnt say in my HW...
if x was this length |dw:1364255675546:dw|
then 10-x must be this length here |dw:1364255697460:dw|
Hold on there bro, let me see what ur doing, plz ... @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
|dw:1364255854635:dw| I dont get what u did here... @jim_thompson5910
I broke up that length of 10 into two pieces
one piece is x units long the other is 10-x units long
both add back up to 10 (x) + (10-x) = x + 10-x = 10
So y can t u just leave it as 10??
because breaking up this length of 10 will help us find the height h
it's not the only way to find h, but it's probably one of the easier ways
k hold on..
yes you break up the triangle like that and that will help you find h
ok so now it makes sense...
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
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
with me so far?
Kinda, i still dont get y u did 10 - x ?? lol sry
well if you break up 10 into two pieces, then one piece is x units long the other must be 10-x units long
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
Yea i get it but y can u just split it in half??
they must add back up to 10, and they do x + (10-x) 2 + (8) 10
because we do NOT know whether the length h cuts the side 10 in half or not
it could, but it could not (in this case, it doesn't) we just don't know
so that's why we must go this route
ok... sry for tht, its just i need to know how to do this, thxss :) so wats next?
no need for sorries, questions are good
ok we have these two equations x^2 + h^2 = 36 (10-x)^2 + h^2 = 64
we can either use substitution or elimination to solve for x and h
let's do substitution
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.
Can we just assume, b/c it's a right angle??
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
This is why I asked if BC was 8 or not If BC = 6, then you could split the 10 into two equal pieces
k let me get this down...
So how did u get these two equations?? (10-x)^2 + h^2 = 64 x^2 + h^2 = 36
through the use of the pythagorean theorem
a^2 + b^2 = c^2
it's explained a bit more above
where did u get 36 from?
from squaring the 6
So is it sub or elim tht ur doing ? @jim_thompson5910
I'm doing substitution
I isolated h^2 in the first equation and then replaced the other h^2 in the second equation with 36 - x^2
notice how the h terms went away after that
kk i need to write this
Sry it took so long... but is this 3.6??
x = 3.6
btw I showed how it's x = 3.6 above lol but I'm glad you got that on your own
Is tht 3.6 ??
yep, that length you are pointing to is 3.6 units long
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
So the height is 4.8
So is tht for this one??
no this one here |dw:1364258674211:dw|
that's what we wanted we really don't need x to find the area we only needed x to find h
So, i thought u said that
|dw:1364258753145:dw| trhis length is 10 - x ??
but we found x, then used that to find h
we don't need to know what 10-x is we just want h
So the last equation u put was to find 'h' right??
k hold on let me get tht down..
I started with x^2 + h^2 = 36, plugged in x = 3.6, then solved for h
OK!! lol i get what we did, so whats next?? Do we just do 32 * 4.8??
ok that's a lot of work to find h
like I said, it would be a lot easier if BC = 6
but it's still doable to find h even if BC wasn't 6
so was i right?? or is there a different way to do this?
anyways, this is a trapezoid with base1 = 32 base2 = 22 height = 4.8
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
A = h*(b1+b2)/2 A = 4.8*(32+22)/2 A = 4.8*(54)/2 A = 4.8*27 A = 129.6
kk hod it bro
So thts it?? @jim_thompson5910
yep pretty much, unless you want to convert that into a fraction
No thank u, i perfer decimals:)
yeah decimals are more intuitive with area problems
THANK THANK YOU SOOOOSOSOOOO MUCH:D:D:D:D: :):) UR THE BEST :D:D:D
lol thx and glad to be of help