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can you help me first

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

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

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

that length is 32
|dw:1364255489567:dw|

Yea...

since 22+10 = 32

Huh?

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'

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

alright sorry

|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..

alright

|dw:1364256139846:dw|

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

ok so now it makes sense...

with me so far?

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

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

lol

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

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

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...

alright

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

kk

notice how the h terms went away after that

kk i need to write this

alright

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

good

x = 3.6

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

|dw:1364258514592:dw|

Is tht 3.6 ??

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

kk

So the height is 4.8

So is tht for this one??

|dw:1364258641203:dw|

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 ??

*this

yes

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??

exactly

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

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

alright

So thts it?? @jim_thompson5910

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

No thank u, i perfer decimals:)

ok great

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