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

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Answer 1 · 2016-02-03T12:51:04

The length of the shorter diagonal is $~~19.45"cm"$ .

Let the sides of your parallelogram be

$a = 12 "cm"$ and $b = 18 "cm"$ .

and your angle be

$gamma = 78^@$ , 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

$beta = 180^@ - gamma = 180^@ - 78^@ = 102^@$

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:

$d^2 = a^2 + b^2 - 2ab * cos(gamma)$

$e^2 = a^2 + b^2 - 2ab * cos(beta)$

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

$d^2 = 12^2 + 18^2 - 2 * 12 * 18 * cos(78^@)$

$= 144 + 324 - 432 * cos(78^@)$

$~~ 468 - 432 * 0.208$

$~~ 378.18$

$=> d ~~ 19.45 "cm"$

And for the other diagonal,

$e^2 = 12^2 + 18^2 - 2 * 12 * 18 * cos(102^@)$

$= 468 - 432 * (-0.208)$

$~~ 557.82$

$=> e ~~ 23.62 "cm"$

Thus, the length of the shorter diagonal is $~~19.45"cm"$ .