help
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WITH THE POWER OF LOVE YOU WILL FIND THE SOLUTION TO EVERY DOUBT IN YOUR MIND
well what do you need help with?
close the thread
yeah , close the question to get it off the main page
ok, name a number and ill help , same kind of stuff as yesterday
YOU CAN FIND YOUR WAY WITH THE POWER OF LOVE.
The congruence theorems are pg2 #17 - 20, that you can use to prove congruent
For a) the two triangles share that side AC so that is equal to itself, and all three sides are the same, side-side side-SSS
b) same reason, SSS, DC = AB from rectangle property, AD=BC as marked, and the diagonal equals itself... side-side-side
you just need to look for one of those orders, ASA, SSS, SAS, AAS
c) you can use a number of them, SSS, SAS, ASA, AAS, any really, since it is a rectangle you can show te angles and sides are equal
rectangles have equal opposite sides that are parallel, the angles are 90 at the corners, you can show these things from being a rectangle
all 3 corresponding sides are the same, and all 3 angles the same for both
for d) EB and BC are equal corresponding sides, (rotate one trignale 180 to line up the right sides) , angle C and E are marked the same corresponding angles, and you can also say, the angles at B are equal vertical angles, thus you have ASA
yes that could be shown
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the quick thing to see here is that this has two equal sides, isocelese, and the angles across from equal sides are also equal.... |dw:1453427373608:dw|
then, just remember the angles for a triangle total 180 degrees x + x + 68 = 180
2x = 180-68 = 112 x = 56
so for 32) same idea, except they gave you equal angles, the sides across them are equal the same as the last triangle was
3x - 5 = x + 11 2x = 16 x=8
Where to now?
34 - 37 are just like ones from yesterday, you can re-read those on there
just remember 2 sides have to be greater than the third side for any sides of a triangle
okay
You good with the proof things like 30)
my pc died
okay
the proofs not very good at it
ill do that one real quick... those things include a bunch of the concepts from many other probs, so if you can do this , should be able to do most others
okay
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For the most part, you start with somethign Given and go from there... so the first line... AT = GS; AT || GS -------> Givens
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