Math Help

[RevBrown]

I have an iscoles trapezoid problem

ad is the bottom
bc is the top
ab and cd are the legs


The top is 9'2"
The bottom is 13'-2"
The sides are 2'-9"
There is a line going thru the middle points e and f and two lines coming down from the center line to h and g
I need to find the distance between a g and h d

Also it says I need to find the distance between ab and cd to the nearest inch


If you can understand this help me out and show me how you reached your answer.

[nippinout]

You haven't given enough information. Where are g and h. e and f are mid points of what?

By the way, Pythagoras is rolling in his grave.

[Will Wood]

42 - How? It's simple. It's the answer to life, the universe, and everything.

[RevBrown]

e is midway between ab
f is midway between cd

if you draw a line straight down from e to ad and from f to ad you have points g and h
it makes right triangles

[Kevmaster]

ok....

here we go...

trapeziod ABCD

midpoinds of ABand CD are E, F respectively

the perp bisectors of AD that run through E, F are named G, H respectively

i make the points I, J that fall on the line AD so that they are perp bisectors of AD and run through B, C respectively.

so...looking at the left side, you have right triangles AEG and ABI. They are similar because of the Angle-Angle-Angle theorem:
both triangles share the angle at A. both have a right angle on the bottom. Thus the angle at the top of both must be the same.

Thus all their sides must be similar...in this case 2:1 ratio (because AE is 1/2 AB because E bisects AB)

The same is true for triangles DFH and DCJ

so....

since I and J bisect BC and AD at 90*, the distance between IJ must be the same as BC. This is 9'2"

you know the entire AD is 13'2"

Thus, AI and JD must be 2" each.

So, if AI is 2', AG must be 1'. Same for DJ and DH (DH = 1')

So the distance b/w G and H is 13'2" - 1' - 1'....or 11' 2"

I **THINK** thats the answer you'relooking for. I might have misinterpereted your problem...but i dont think so


nothing like a little geometry coming back to mesome 6 years later...boo yah!

[markoger]

well i tried real quick to make a scketch but i can't get it in here. change everything into inches first. i have figured that the distance a-g and h-d are both 12" 158"- 110"= 48". 48/2(2 representing each side extening past the top) is 24. half of 24 is 12". because it is half the distance of the 24" end. the distance between ab and cd should be 110" (the length of the top).if it is the distance between e and f it would be 134". the 158 minus the 24" from the triangles. let me know if this makes sense(i am kind of tired)

marco

[RevBrown]

well i tried real quick to make a scketch but i can't get it in here. change everything into inches first. i have figured that the distance a-g and h-d are both 12" 158"- 110"= 48". 48/2(2 representing each side extening past the top) is 24. half of 24 is 12". because it is half the distance of the 24" end. the distance between ab and cd should be 110" (the length of the top).if it is the distance between e and f it would be 134". the 158 minus the 24" from the triangles. let me know if this makes sense(i am kind of tired)

marco

Thats what I thought but I was thinking that was to simple

[markoger]

lol... nice job Kev, i hope one of us is right... i guess i was working on this longer than i thought. i need some sleep. good luck revbrown.

Marco

[markoger]

It might be too simple, but it seems right. being that i have not done this for 12 years. lmao

[Xyxyll]

I started doing the problem just for the heck of it. Then I saw Kevmaster's post and looked at my notes... the same, lol. Fun stuff to brighten up your thoughts at 12:30am!

[RevBrown]

Thanks all

[Kevmaster]

It might be too simple, but it seems right. being that i have not done this for 12 years. lmao

mark....i think we have the same answer...you just got there quicker... i think i went all the way through proving your reasoning