Logo for MightySolve
Diagonals in a parallelogram

The lengths of the diagonals in a parallelogram can be computed using the law of cosines for the constructed triangle including the sides aa and bb and the angle α\alpha or β\beta.

The diagonal ff, from the upper left corner to the lower right corner of the parallelogram, uses therefore the following formulaf=a2+b22abcos(α). f = \sqrt{a^2 + b^2 - 2ab \cos\left(\alpha\right)}. The other diagonal ee, from the lower left corner to the upper right corner of the parallelogram, includes in the constructed triangle the angle β\beta. The angles in a parallelogram have the property thatβ=180°α(=πα). \beta = 180 \degree - \alpha \left(= \pi - \alpha\right). So we get the following formula for computing the length of the diagonal eee=a2+b22abcos(β)=a2+b22abcos(πα)=a2+b2+2abcos(α). \begin{aligned} e &= \sqrt{a^2+b^2 -2ab \cos \left(\beta\right)}\\ &= \sqrt{a^2+b^2 -2ab \cos \left(\pi -\alpha\right)} \\ &= \sqrt{a^2+b^2 +2ab \cos \left(\alpha\right)}. \end{aligned}

Both diagonals intersect at half of their length, i.e. they bisect each other.