# Thread: Parallelogram Question

1. ## Parallelogram Question

I feel really dumb asking this..but whatever

one length of the pgram is 7 the other is 9..the long diagonal is 14..whats the short diagonal

2. Hello, stones44!

We will need Trignometry for this one . . . the Law of Cosines.

One side of the parallelogram is 7, the other is 9.
The long diagonal is 14.
What is the short diagonal?
Code:
           A         9         B
* - - - - - - - - *
/              *  /
/           *     /
7/        *        / 7
/     *    14     /
/  *              /
* - - - - - - - - *
D         9         C
We have parallelogram $ABCD$ with: $AB = DC = 9,\;AD = BC = 7,\;BD = 14$
. . and we want the length of diagonal $AC.$

First, find $\angle A.$

. . In $\Delta ABD\!:\;\;\cos A \;=\;\frac{7^2+9^2-14^2}{2(7)(9)}\;=\;-\frac{11}{21} \quad\Rightarrow\quad A \:=\:121.5881355^o$

Then $\angle D \:=\:\angle ADC \:=\:180^o - 121.5881355^o \:=\:58.41186449^o$

In $\Delta ADC\!:\;\;AC^2\;=\;7^2+9^2-2(7)(9)\cos58.41186449^o \;=\;63.99999999$

Therefore: . $AC \:=\:\sqrt{63.99999999} \:\approx\:8$

3. Suppose that vectors $a,\,b$ are two adjacent sides of the parallelogram.
Then $a + \,b\quad \& \quad a - b$ would represent the two diagonals.
We know that $\left\| {a + b} \right\|^2 = \left\| a \right\|^2 + 2a \cdot b + \left\| b \right\|^2 \quad \& \quad \left\| {a - b} \right\|^2 = \left\| a \right\|^2 - 2a \cdot b + \left\| b \right\|^2$
So $\left\| {a + b} \right\|^2 + \left\| {a - b} \right\|^2 = 2\left\| a \right\|^2 + 2\left\| b \right\|^2.$

Use that to find the length of the second diagonal.