For All Of U Who Are Good At Math..

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has anyone been able to prove it, cause ive been trying, and im going nuts... :eek:
 
hey i am new to this but here it goes.
proving that a parrallelogram is form. okay all the triangles are equillateral they have sides a,b,c,and d. all four sides are different. therefore, the triangles facing inward will never overlap themselves, thus the vertex points will exsist on those two trianlges.
also, the area of a trianlge = 1/2ab so this means that the trianlges vertex will always be 1/2 the lenght of the base line which is one of the sides of the quadralateral. now if you 90degree lines perpendicularlly to the quadrilarteral line the vertex with all of them will line up and also intersect in the middle of the quadrilateralWITH RESPECT THE THE ANGLES OF THE QUADRILATERAL LINES.

ther you have it in a very brief explanation.

by the way what clas is this for?
 
This actual proof, probably never.

The ability to think logically and reason through it. Quite often depending on what you end up doing (obviously working in ME, EE, or CS will require more logical thinking ability than flipping burgers at Wendy's).
 
this is for algebra/discrete mathematics course...does anyone else have a solution/proof for this question?
 
Originally posted by 22crazy@May 6 2003, 12:18 PM
this is for algebra/discrete mathematics course...does anyone else have a solution/proof for this question?
Click to expand...

damn? you had to do that for discrete math? my discrete math was just doing proofs for statements
 
:werd: mine was Turing machines and more Turing Machines. With a little condensing finite state automata for good measure.
 
yah but we did that, but my teacher decided to through me this question as a project, btw maybe i should add this is my final year of high school.
 
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