# A parallelogram has sides 12cm and 18cm and a contained angle of 78 degrees. Find the shortest diagonal?

Feb 3, 2016

The length of the shorter diagonal is $\approx 19.45 \text{cm}$.

#### Explanation:

Let the sides of your parallelogram be

$a = 12 \text{cm}$ and $b = 18 \text{cm}$.

$\gamma = {78}^{\circ}$, the angle between $a$ and $b$ (or $b$ and $a$)

As two adjacent angles in a parallelogram are suplementary, we know that the adjacent angle to $\gamma$ is

$\beta = {180}^{\circ} - \gamma = {180}^{\circ} - {78}^{\circ} = {102}^{\circ}$

Now, we can use the law of cosines to compute both diagonals in the parallelogram.

Let $d$ be the diagonal opposite to $\gamma$ and $e$ be the diagonal opposite to $\beta$.

The law of cosines states:

${d}^{2} = {a}^{2} + {b}^{2} - 2 a b \cdot \cos \left(\gamma\right)$

${e}^{2} = {a}^{2} + {b}^{2} - 2 a b \cdot \cos \left(\beta\right)$

Thus, we can compute the lengths of the diagonals as follows:

${d}^{2} = {12}^{2} + {18}^{2} - 2 \cdot 12 \cdot 18 \cdot \cos \left({78}^{\circ}\right)$

$= 144 + 324 - 432 \cdot \cos \left({78}^{\circ}\right)$

$\approx 468 - 432 \cdot 0.208$

$\approx 378.18$

$\implies d \approx 19.45 \text{cm}$

And for the other diagonal,

${e}^{2} = {12}^{2} + {18}^{2} - 2 \cdot 12 \cdot 18 \cdot \cos \left({102}^{\circ}\right)$

$= 468 - 432 \cdot \left(- 0.208\right)$

$\approx 557.82$

$\implies e \approx 23.62 \text{cm}$

Thus, the length of the shorter diagonal is $\approx 19.45 \text{cm}$.