Assume I have a trapezoid and I know all its sides:
$AB = 10$
$CD =6$
$AC = 3$
$BD = 5$
I need to know the angle between $AB$ and $AC$.
AB and CD are parallel
[Math] How to know the angles of a trapezoid by its sides
geometry
Related Solutions
Hint: for any trapezoid, show there is a rectangle of equal area and smaller perimeter. Therefore, by stretching, there is a rectangle of equal perimeter but greater area.
Given the lengths $\ell_1, \ell_2$ of two diagonals and the angle $\alpha$ between them, let $2a$ be the required sum of lengths of two adjacent sides.
- Construct three helper points $A, B, C$ such that $AB = \ell_1, AC = \ell_2$ and $\measuredangle BAC = \alpha$.
($AB$ and $AC$ are the two blue line segments in diagram below) - Construct an ellipse having $A, B$ as foci with semi-major axis $a$.
(The red ellipse in diagram below) - Start from the two line segments $AB$ and $AC$, construct parallelogram $ABEC$ and let the line $AE$ intersect the ellipse at $F$.
- Start from the line segments $AC$ and $CF$, construct parallelogram $AGFC$.
Take $AB$ and $GF$ as the two desired diagonal. The quadrilateral $AGBF$ will be a trapezoid one seek. To see this.
- By construction, $|AB| = \ell_1$. Since $AGFC$ is a parallelogram, $|GF| = |AC| = \ell_2$ and angle between $AB$ and $GF$ is the angle $\measuredangle BAC = \alpha$.
- Since $F$ lies on the ellipse having $A$, $B$ as foci. $|AF| + |BF| = 2a$.
- Since $ABEC$ and $AGFC$ are parallelograms, $GBEF$ is a parallelogram too. As a result, $GB \parallel FE \parallel AF$.
Best Answer
I assume that $AB\parallel CD$. Then the line parallel to $BD$ passing through $C$ intersects $AB$ in a point $P$ such that we know the lengths of all sides in triangle $APC$ as $AP=10-6=4, PC=BD=5, CA=3$.