Solve a Trapezoid Given its Bases and Legs

Using the trapezoid above, draw \(BB'\) parallel to \(CD\).

Using triangle \(ABB'\), apply the cosine rule:

\[ c^2 = a^2 + (d - b)^2 - 2a(d - b)\cos(\angle BAD) \] \[ \cos(\angle BAD) = \frac{a^2 + (d - b)^2 - c^2}{2a(d - b)} \] \[ \angle BAD = \arccos\left(\frac{a^2 + (d - b)^2 - c^2}{2a(d - b)}\right) \]

In the same figure, \(\angle BB'A\) and \(\angle CDA\) are equal. Using triangle \(ABB'\) again:

\[ a^2 = c^2 + (d - b)^2 - 2c(d - b)\cos(\angle BB'A) \] \[ \cos(\angle BB'A) = \frac{c^2 + (d - b)^2 - a^2}{2c(d - b)} \] \[ \angle CDA = \angle BB'A = \arccos\left(\frac{c^2 + (d - b)^2 - a^2}{2c(d - b)}\right) \]

Since \(AD \parallel BC\), consecutive angles are supplementary:

\[ \angle ABC = 180^\circ - \angle BAD, \quad \angle DCB = 180^\circ - \angle CDA \]

\[ h = a \sin(\angle BAD) = a \sin(\angle ABC) \] \[ \text{Area} = \frac{1}{2}(b + d)h \]

Apply the cosine rule in triangles \(DAB\) and \(BCD\):

\[ BD^2 = a^2 + d^2 - 2ad \cos(\angle BAD) \] \[ BD = \sqrt{a^2 + d^2 - 2ad \cos(\angle BAD)} \] \[ AC^2 = c^2 + d^2 - 2cd \cos(\angle CDA) \] \[ AC = \sqrt{c^2 + d^2 - 2cd \cos(\angle CDA)} \]

Given: Bases \(d > b\), legs \(a\) and \(c\), with \(AD \parallel BC\).

Angles:
\(\displaystyle \angle BAD = \arccos\left(\frac{a^2 + (d - b)^2 - c^2}{2a(d - b)}\right)\)
\(\displaystyle \angle CDA = \arccos\left(\frac{c^2 + (d - b)^2 - a^2}{2c(d - b)}\right)\)
\(\angle ABC = 180^\circ - \angle BAD\)
\(\angle DCB = 180^\circ - \angle CDA\)

Height:
\(h = a \sin(\angle BAD)\)

Area:
\(\displaystyle A = \frac{1}{2}(b + d)h\)

Diagonals:
\(\displaystyle BD = \sqrt{a^2 + d^2 - 2ad \cos(\angle BAD)}\)
\(\displaystyle AC = \sqrt{c^2 + d^2 - 2cd \cos(\angle CDA)}\)