Please help?
The bases of a trapezoid have lengths 50 and 75. Its diagonals have lengths 35 and 120. Find the area of the trapezoid.
is there a picture?
Sure just a sec
|dw:1385240010377:dw| Sorry it's not all that accurate xD
hmmm.... how about a screenshot?
There's no picture, sorry :P
ohh.. ok
is it an isosoles trapezoid
hmmm... if it were an isosceles trapezoid... how come the diagonals differ? one is 35 and the other 120..... doesn't sound like one
No, because the diagonals are different like jdoe said
The textbook told us to use similar triangles, and I only found the top and bottom triangles to be similar. I'm not sure how that would help us find the area though..
hmm ok .. so I'm thinking that |dw:1385241489796:dw|
you cant just assume that it has right angles
It doesn't show any right angles..
so you just need to find "h" or height, in this case at the bottom of it , using pythagorean theorem, keeping in mind the \(\bf \textit{Area of a Trapezoid}=\cfrac{h}{2}(base1+base2)\)
well, you do have to assume because shape-wise for the given values, is what fits, and to be a trapezoid, it just have to have 2 parallel sides, the other 2 sides don't have to be slanted or equal
im sorry i cant help because i havent done this type of geometry in a while
I guess that makes sense, thank you so much! :D
Noooooooooooooooo
You cant assume. I remember doing problems like this It can be done without assuming that the angles are 90
the only proble is that i forgot
with an isosceles trapezoid is simple since there's a property to use for the diagonals, with non-isosceles there isn't
ya i know it isnt an isoseles
Lol, I think that's the only way we can think of right now xD Thanks though, hopefully it's correct
have you guys talked about adding things extra parts in class
extra parts?
Well the textbook told us to use similar triangles to find the length of the concurrency to each corner
im talking about adding midlines
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