Parallelogram...

**PTL** · August 5th 2009, 10:46 PM

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...

**red_dog** · August 5th 2009, 10:59 PM

Let $O(0,0), \ A(1,2t), \ B(2t,1), C(1+2t, 1+2t)$

$Area[OACB]=2\cdot Area[OAB]=2\cdot\frac{1}{2}|\Delta|$ where

$\Delta=\begin{vmatrix}0 & 0 & 1\\1 & 2t & 1\\2t & 1 & 1\end{vmatrix}=1-4t^2$

Therefore $f(t)=|1-4t^2|$

**alunw** · August 6th 2009, 03:08 AM

Another way to do it is using the cross product of v and w. That also works with 3D coordinates.

**Soroban** · August 6th 2009, 06:49 AM

Hello, PTL!

Yet another approach . . .

Did you make a sketch?

Let $f(t)$ be the area of the parallelogram with vertices:

$\vec v\,=\,(1,2t),\;\vec w\,=\,(2t,1),\;(0,0)$ , and $\overrightarrow{v+w} \:=\:(2t+1,2t+1).$

Find a formula for $f(t).$

Code:

        |
   2t+1 + - - - - - - - - - - o v+w
        |                 *  *:
        |             *     * :
        |     v   *        *  :
     2t + - - o           *   :
        |    *           *    :
       1+   *           o w   :
        |  *        *   :     :
        | *     *       :     :
        |*  *           :     :
    - - o - - + - - - - + - - * - -
      (0,0)   1        2t    2t+1

We have a rhombus.

If $d_1,\:d_2$ are diagonals of a rhombus,

. . the area of the rhombus is: . $A \:=\:\tfrac{1}{2}d_1d_2$

The long diagonal $d_1$ goes from $(0,0)$ to $(2t+1,2t+1)$

. . $d_1 \:=\:\sqrt{(2t+1)^2 + (2t+1)^2} \;=\;\sqrt{2}\,|2t+1|$

The short diagonal $d_2$ goes from $(1,2t)$ to $(2t-1)$

. . $d_2 \;=\;\sqrt{(2t-1)^2 + (1-2t)^2} \;=\;\sqrt{2}\,|2t-1|$

Hence: . $A \;=\;\tfrac{1}{2}\bigg[\sqrt{2}\,|2t+1|\bigg]\,\bigg[\sqrt{2}\,|2t-1|\bigg] \;=\;|4t^2-1|$

Therefore: . $A \;=\;\begin{Bmatrix}4t^2-1 & &\text{for } |t| > \frac{1}{2} \\ \\[-4mm] 0 && \text{for }t \:=\:\frac{1}{2} \\ \\[-4mm] 1-4t^2 && \text{for }|t| < \frac{1}{2} \end{Bmatrix}$

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