anonymous
  • anonymous
Please help with geometry!!!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
can you help me first
anonymous
  • anonymous
|dw:1364254756475:dw| Find the area of ABCD.
anonymous
  • 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Ok like this??|dw:1364255189854:dw|
anonymous
  • anonymous
@Hoa
anonymous
  • 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
  • anonymous
b*h = Area for rectangle and parallelogram
jim_thompson5910
  • jim_thompson5910
are you sure that BC = 8?
jim_thompson5910
  • jim_thompson5910
it would make things a lot easier if BC = 8
jim_thompson5910
  • jim_thompson5910
i meant, BC = 6
anonymous
  • anonymous
Yea:) @jim_thompson5910 It does!!
jim_thompson5910
  • jim_thompson5910
ok one sec
anonymous
  • anonymous
Oh i know what ur saying... @jim_thompson5910
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
  • jim_thompson5910
that length is 32 |dw:1364255489567:dw|
anonymous
  • anonymous
Yea...
jim_thompson5910
  • jim_thompson5910
this length is 22 |dw:1364255516091:dw| so this remaining length here |dw:1364255533268:dw| must be 10
jim_thompson5910
  • jim_thompson5910
since 22+10 = 32
anonymous
  • anonymous
Huh?
anonymous
  • anonymous
Nvm i know what urr doing:)
jim_thompson5910
  • jim_thompson5910
We now have this triangle |dw:1364255559257:dw|
anonymous
  • anonymous
And so u have to do the pythagoreum theorm thing...
anonymous
  • anonymous
to find 'h'
jim_thompson5910
  • 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
  • anonymous
But u cant do tht, if it doesnt say in my HW...
jim_thompson5910
  • jim_thompson5910
if x was this length |dw:1364255675546:dw|
jim_thompson5910
  • jim_thompson5910
then 10-x must be this length here |dw:1364255697460:dw|
anonymous
  • anonymous
Hold on there bro, let me see what ur doing, plz ... @jim_thompson5910
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
  • jim_thompson5910
alright sorry
anonymous
  • anonymous
|dw:1364255854635:dw| I dont get what u did here... @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
I broke up that length of 10 into two pieces
jim_thompson5910
  • jim_thompson5910
one piece is x units long the other is 10-x units long
jim_thompson5910
  • jim_thompson5910
both add back up to 10 (x) + (10-x) = x + 10-x = 10
anonymous
  • anonymous
So y can t u just leave it as 10??
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
because breaking up this length of 10 will help us find the height h
jim_thompson5910
  • jim_thompson5910
it's not the only way to find h, but it's probably one of the easier ways
anonymous
  • anonymous
k hold on..
jim_thompson5910
  • jim_thompson5910
alright
anonymous
  • anonymous
|dw:1364256139846:dw|
jim_thompson5910
  • jim_thompson5910
yes you break up the triangle like that and that will help you find h
anonymous
  • anonymous
ok so now it makes sense...
jim_thompson5910
  • 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
  • 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
  • jim_thompson5910
with me so far?
anonymous
  • anonymous
Kinda, i still dont get y u did 10 - x ?? lol sry
jim_thompson5910
  • 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
  • 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
  • anonymous
Yea i get it but y can u just split it in half??
jim_thompson5910
  • jim_thompson5910
they must add back up to 10, and they do x + (10-x) 2 + (8) 10
jim_thompson5910
  • jim_thompson5910
because we do NOT know whether the length h cuts the side 10 in half or not
jim_thompson5910
  • jim_thompson5910
it could, but it could not (in this case, it doesn't) we just don't know
jim_thompson5910
  • jim_thompson5910
so that's why we must go this route
anonymous
  • anonymous
ok... sry for tht, its just i need to know how to do this, thxss :) so wats next?
jim_thompson5910
  • jim_thompson5910
no need for sorries, questions are good
anonymous
  • anonymous
lol
jim_thompson5910
  • jim_thompson5910
ok we have these two equations x^2 + h^2 = 36 (10-x)^2 + h^2 = 64
jim_thompson5910
  • jim_thompson5910
we can either use substitution or elimination to solve for x and h
jim_thompson5910
  • jim_thompson5910
let's do substitution
anonymous
  • 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
  • anonymous
Can we just assume, b/c it's a right angle??
jim_thompson5910
  • 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
  • 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
  • anonymous
k let me get this down...
jim_thompson5910
  • jim_thompson5910
alright
anonymous
  • anonymous
So how did u get these two equations?? (10-x)^2 + h^2 = 64 x^2 + h^2 = 36
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
through the use of the pythagorean theorem
jim_thompson5910
  • jim_thompson5910
a^2 + b^2 = c^2
jim_thompson5910
  • jim_thompson5910
it's explained a bit more above
anonymous
  • anonymous
where did u get 36 from?
jim_thompson5910
  • jim_thompson5910
from squaring the 6
anonymous
  • anonymous
So is it sub or elim tht ur doing ? @jim_thompson5910
anonymous
  • anonymous
@jim_thompson5910 ??
jim_thompson5910
  • jim_thompson5910
I'm doing substitution
anonymous
  • anonymous
kk
jim_thompson5910
  • 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
  • jim_thompson5910
notice how the h terms went away after that
anonymous
  • anonymous
kk i need to write this
jim_thompson5910
  • jim_thompson5910
alright
anonymous
  • anonymous
Sry it took so long... but is this 3.6??
jim_thompson5910
  • jim_thompson5910
good
jim_thompson5910
  • jim_thompson5910
x = 3.6
jim_thompson5910
  • jim_thompson5910
btw I showed how it's x = 3.6 above lol but I'm glad you got that on your own
anonymous
  • anonymous
|dw:1364258514592:dw|
anonymous
  • anonymous
Is tht 3.6 ??
jim_thompson5910
  • jim_thompson5910
yep, that length you are pointing to is 3.6 units long
jim_thompson5910
  • 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
  • anonymous
kk
jim_thompson5910
  • jim_thompson5910
So the height is 4.8
anonymous
  • anonymous
So is tht for this one??
anonymous
  • anonymous
|dw:1364258641203:dw|
jim_thompson5910
  • jim_thompson5910
no this one here |dw:1364258674211:dw|
jim_thompson5910
  • 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
  • anonymous
So, i thought u said that
anonymous
  • anonymous
|dw:1364258753145:dw| trhis length is 10 - x ??
anonymous
  • anonymous
*this
jim_thompson5910
  • jim_thompson5910
yes
jim_thompson5910
  • jim_thompson5910
but we found x, then used that to find h
jim_thompson5910
  • jim_thompson5910
we don't need to know what 10-x is we just want h
anonymous
  • anonymous
So the last equation u put was to find 'h' right??
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
exactly
anonymous
  • anonymous
k hold on let me get tht down..
jim_thompson5910
  • jim_thompson5910
I started with x^2 + h^2 = 36, plugged in x = 3.6, then solved for h
anonymous
  • anonymous
OK!! lol i get what we did, so whats next?? Do we just do 32 * 4.8??
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
ok that's a lot of work to find h
jim_thompson5910
  • jim_thompson5910
like I said, it would be a lot easier if BC = 6
jim_thompson5910
  • jim_thompson5910
but it's still doable to find h even if BC wasn't 6
anonymous
  • anonymous
so was i right?? or is there a different way to do this?
jim_thompson5910
  • jim_thompson5910
anyways, this is a trapezoid with base1 = 32 base2 = 22 height = 4.8
jim_thompson5910
  • 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
  • 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
  • anonymous
kk hod it bro
jim_thompson5910
  • jim_thompson5910
alright
anonymous
  • anonymous
So thts it?? @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
yep pretty much, unless you want to convert that into a fraction
anonymous
  • anonymous
No thank u, i perfer decimals:)
jim_thompson5910
  • jim_thompson5910
ok great
jim_thompson5910
  • jim_thompson5910
yeah decimals are more intuitive with area problems
anonymous
  • anonymous
THANK THANK YOU SOOOOSOSOOOO MUCH:D:D:D:D: :):) UR THE BEST :D:D:D
jim_thompson5910
  • jim_thompson5910
lol thx and glad to be of help

Looking for something else?

Not the answer you are looking for? Search for more explanations.