A New way, The Area of Trapezium by Piyush Goel

mykar2un shared this question 4 years ago

A New way, The Area of Trapezium

Piyush Goel Discovered a new way of "Area of Trapeium" while working on "Pythagoras Theorem".

untitled-drawingLot of mathematicians have proved Pythagoras theorem in their own ways. If you google it you will indeed found hundred of ways.

Meanwhile I was also sure that maybe one day I could find something new out of this incredible Pythagoras theorem and Recently I got something which I would like to share with you.

To Prove: Deriving the equation of area of trapezium using Arcs

Proof: There is a triangle ABC with sides a b and c as shown in the figure.


Area of ∆ BCEG = Area of ∆ BDC +Area of ⌂ DCEF + Area of ∆ EFG
c^2=ac/2+ Area of ⌂ DCEF + (c-b) c/2
(2c^2– ac –c^2+ bc )/2=Area of ⌂ DCEF
(c^2– ac+ bc )/2=Area of ⌂ DCEF
c(c– a+ b)/2=Area of ⌂ DCEF
Area of ⌂ DCEF=BC(DE+CF)/2

Comments (2)



this is a forum related to GeoGebra. Since we are getting more posts from you with other math proofs, but don't contain any question or else related to GeoGebra, I'm not going to approve them, because they're off topic.

And a copyright to a theorem, sorry if I say this, it's quite hilarious.


ok next time ask question this time sorry for and help me.

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