Let f(t) be the area of the parallelogram with vertices v=(1,2t), w = (2t,1), (0,0), and v+w = (2t+1,2t+1). Find a formula for f(t).
I'm assuming one can't simply 'stand up' the parallelogram so that it's a rectangle...
Hello, PTL!
Yet another approach . . .
Did you make a sketch?
Let be the area of the parallelogram with vertices:
, and
Find a formula forWe have a rhombus.Code:| 2t+1 + - - - - - - - - - - o v+w | * *: | * * : | v * * : 2t + - - o * : | * * : 1+ * o w : | * * : : | * * : : |* * : : - - o - - + - - - - + - - * - - (0,0) 1 2t 2t+1
If are diagonals of a rhombus,
. . the area of the rhombus is: .
The long diagonal goes from to
. .
The short diagonal goes from to
. .
Hence: .
Therefore: .