I have yet another question on triangles? I will not tolerate'Do your homework'as an answer. Read on ...
914532
Let the trianglas be ABC & BCD\nAD forms 2 triangles, ABD & ACD\nAB=AC since ABC is isisceles\nBC+BD since BCD is isosceles.\nangle ABC=angle ACB\nangle DBC=angls DCB so angls ABD=angle ACD so ABD is congruent to ACD by SAS\n\ncall the point where AD crosses BC E\n\nso both vertex angles are bisected.\nlook at ABC.\nAE bisects BAC & AE is a commin side for BAE & CAE.\nsince AB=AC, the triangles are congruent by SAS\ntherefore BE=CE\n& angle AEB= angle AEC which are supplements so they must be right angles.\nQED
243764
im a geo student, but i havent done proofs for a long time sorry. ok\n2. false, because squares are parallelograms with 4 congruent sides and 4 right angles, while kites have 2 pairs of congruent sides and only 1 pair of congruent angles that are NOT right angles\n3. false, because a rhombus is a parallelogram with 4 congruent sides, while a kite is any thing BUT a parallelogram. enuff said\n4. false, because kites are quadrilaterals with only two pairs of congruent sides, while the quadrilateral mentioned has 3 congruent sides\n5. false, for example- do trapezoids have 4 congruent sides, and 4 right angles, and is a parallelogram (characteristic of squares)? i think NOT\n6. read above- trapezoids are not parallelograms.
875394
Very understandable....I just posted a question about my pre-calculus homework and i didnt ask for the answer, i just need help finding how to get it....i'd like to know how to do it myself, i dont just want someone to tell me the answer....does me no good @ all!
1110486
It comes from the fact that the "hypotenuse" of the two "triangles" is not a straight line segment.  It is actually two segments of similar, but not equal, slope.
36413
The previous answers are either wrong/incomplete/mischievious.\nFirstly there is nothing called parallel angles! You probably meant to ask what happens if the angles of two triangles are same. In that case there is no congruence/similarity between the two triangles.\nBut if two angles are equal betwen them, they are SIMILAR (same shape) and not congruent.\nIf in addition they have a side equal, they become congruent!
1000630
Pythagoras Theorem asserts that for a right triangle with short sides of length a and b and long side of length c \na2 + b2 = c2 \n\nor c2-a2 = b2 \n\nhere is a link with all the necessary thing you have to know about this theorem . I hope it helps , just read it carefully and everything is going to be fine :)
286401
Oh, come on. I answered your first one, and now you flood the board with your homework? Are you really that lazy to think it's easier to post your homework here? These problems are all the same, with different numbers. Do it yourself, I already showed you how.
471036
Why do some questions not have answers? I guess that is what you mean. I went to answer one just a minute ago and it said I need help with my geometry but that was all it said. Now how can you answer that. If the question is really vulgar I don't answer them I figure someone will.
224697
Whether any one likes it or not, the sum of the squares of the sides of a right angled triangle will always like to be equal to the square of the hypotenuse.\n\nPlease, from today say to yourself that I like Pythagoras Theorm. Soon you will begin to like it. It will lead you to many wonders of science
331507
sorry hunny you need at least 2 side lengths in order for us to help you figure it out... plus once you have that the formula for figuring it out is using the pathagorean theorem! a2+b2=c2 (s = squared) say for instance... the 2 sides of your triangle are 7 and 3inches. 3>2+7>2=c>2\nwhich would turn into 9+49=c>2~~~~~~~ then 58=c>2 then to get c by itself(which is needed) you would take the square root of 58! your hypotenuse would equal 7.6 or rounded, 8!\nnow if you use my steps you should be able to do your problem! I'm a 10th grader in geometry and it's the most lovely subject!
303035
If the triangle is a right triangle, you can find the other side by Py's Th and then by the formula,(A=1/2bh).\nIf it is not, you should know the angle they make each other or an adjacent angle. Only if, you can find the area by sine laws or cosine laws.
607226
1.What are the applications for proving 2 Triangles  similar?\n2.What is the formula that is used to calculate the area of a i}Trapezium\n  ii}Hexagon\n3.How can you prove 2 circles are similar.\n\n\nHope this helps..I hope this is what u were asking for?
1030296
for it to be a triangle the angles have to measure to 180, so if you have a 90 degree andlge off one, you could subtract 90 from that, take the remaining 90 and work with that, since you know your need a 1:1 of the legs , then you know you need an equal angle.. angle has to equal 90/2... and thats where i get stumped...\n      gl
902207
I'm assuming you are referring to the disguised triangles in the attached link... (link 1).  I'm sure there are some rigorous mathematical proofs you could supply, or automated ways to count the triangles, but I basically decided to just use the "brute force" method of counting.  The hard part is recognizing all the different triangles while not double counting...\n\nWhat I finally did was to create several copies of the diagram and tried to color the triangles.  You can color the individual triangles (24), right off the bat.  Now if you try to pair up triangles, you can pair them one way (12) or another using the flipped image (12).  Grouping them further in groups of three or four triangles, I counted another set of (7) and the flip for another (7).  Then I noticed four other ways to group four triangles (3), (3), (3), (3).  Lastly, there are two large triangles composed of the inner sixteen triangles (1) and (1).\n\nAltogether I got seventy (76) triangles.  You can double check my work using the attached image (link 2).  I think I got them all and didn't double count...
754643
im taking Trig. try finding a homework helper web page. or go to the book that ur learning from web page because the majority of the times they have practice help on there.
104581
It takes practice!  And its like a puzzle.  Just use the main trig identities as reference and a good math tutor.  Unfortunately, this place isn't the best place to ask that kind of question because we are limited by text explainations and trig identities need a lot more than that!
651802
The two "triangles" are not exact triangles, but rather are quadrilaterals.  The error in representation actually adds up to be one unit.  I have done the caculations in AutoCAD, and it turns out to be one square unit missing causing the extra space.