maths project on basic proportionality theorem
BASIC PROPORTIONALITY THEOREM: BASIC PROPORTIONALITY THEOREM - THALES OF MILETUS
THALES OF MILETUS : THALES OF MILETUS Thales of Miletus -Socratic Greek philosopher from Miletus in Asia Minor, and one of the Seven Sages of Greece. Many, most notably Aristotle, regard him as the first philosopher in the Greek tradition. According to Bertrand Russell, "Western philosophy begins with Thales. ‘’Thales attempted to explain natural phenomena without reference to mythology and was tremendously influential in this respect. In mathematics, Thales used geometry to solve problems such as calculating the height of pyramids and the distance of ships from the shore. He is credited with the first use of deductive reasoning applied to geometry, by deriving four corollaries to Thales' Theorem. As a result, he has been hailed as the first true mathematician and is the first known individual to whom a mathematical discovery has been attributed. Also, Thales was the first person known to have studied electricity.
PowerPoint Presentation: If a line intersects two sides of a triangle and is parallel to the third side, then it divides the first two sides proportionally . Basic Proportionality Theorem
RESTATEMENT OF THE THEOREM: RESTATEMENT OF THE THEOREM If a line ( EF ) intersects two sides ( AB & AC ) of a triangle ( ABC ) and is parallel to the third side( BC ), then it divides the first two sides proportionally.
PowerPoint Presentation: THUS , BE : EA = BF : FC
OTHER PROPORTIONS ARE : : OTHER PROPORTIONS ARE : 2. BE : BA = BF : BC 3. BA : EA = BC : FC 4. BE : BF = EA : FC 5. FC : EA = BC : BA 6. EF : AC = BF : BC 7. EF : AC = BE : BA
VERIFYING BE : EA = BF : FC :-: VERIFYING BE : EA = BF : FC :- BE : EA = BF : FC . 15 : 5 = 12 : 4 By Simplifying , 3 : 1 = 3: 1
PROOF OF BASIC PROPORTIONALITY THEOREM: PROOF OF BASIC PROPORTIONALITY THEOREM
PowerPoint Presentation: GIVEN : In triangle ABC , DE ll BC , DE bisects AB at D and AC at E . TO PROVE : AD/BD = AE/CE . CONSTRUCTION : Join CD and BE , Draw EF _l_ AD and AE _l_ DG . PROOF : Ar ( ADE) = ½ x B x h= ½ x AD x EF . Ar (BDE) = ½ x BD x EF . Ar (ADE)/ Ar (BDE) = ½ x AD x EF / ½ x BD x EF = AD / BD . -------1 . Ar (ADE) = ½ x AE x DG . Ar (CDE) = ½ x CE x DG .
Ar (ADE) / Ar (CDE) = ½ x AE x DG / ½ x CE x DG = AE / CE . As, Triangle BDE and Triangle DEC are on the same base DE and between the same parallels DE ll BC . Therefore , Ar (BDE) = Ar (DEC ) . Implies , Ar(ADE) / Ar(BDE) = Ar(ADE) / Ar(DEC) . Implies , AD / BD = AE / EC . Hence , Proved the B.P.T. . : Ar (ADE) / Ar (CDE) = ½ x AE x DG / ½ x CE x DG = AE / CE . As, Triangle BDE and Triangle DEC are on the same base DE and between the same parallels DE ll BC . Therefore , Ar (BDE) = Ar (DEC ) . Implies , Ar (ADE) / Ar (BDE) = Ar (ADE) / Ar (DEC) . Implies , AD / BD = AE / EC . Hence , Proved the B.P.T. .
PowerPoint Presentation: GIVEN BY – MRS . RUPAM BHATNAGAR MAM .
PowerPoint Presentation: T H A N K Y O U MADE BY – ADITYA KALWAY . CLASS – X – ‘B’ . ROLL NUMBER – 25 . SCHOOL – K.V.PRAGATI VIHAR
To get full information or details of maths project on basic proportionality theorem please have a look on the pages
http://pioneermathematicsbasic-proportio...rmula.html
http://mathematicsworld.blogspot2009/09/...ality.html
if you again feel trouble on maths project on basic proportionality theorem please reply in that page and ask specific fields in maths project on basic proportionality theorem