Properties of Trapezoids (US) / Trapeziums (UK)
A trapezoid, or trapezium, is a quadrilateral which has a pair of parallel sides. The two parallel sides are called the trapezoid's bases, and the two non-parallel sides are referred to as the legs.
Contents
Summary
Angles
ParallelAngleBisector
For any trapezoid, the following two correspond:
(1) Like any other quadrilateral, the degree measures of the four angles add up to 360 degrees. Thus the trapezoid \(ABCD\) shown in the figure above satisfies the following:
\[ \angle A + \angle B + \angle C + \angle D = 360^\circ. \]
(2) Two angles on the same side of a leg are always supplementary, that is, they add up to 180 degrees. Thus for the trapezoid \(ABCD\) shown above, we have
\[ \begin{align} \angle A + \angle B &= 180^\circ \\ \angle C + \angle D &= 180^\circ. \end{align} \]
Area
The area of a trapezoid is found using the formula \( \frac{ 1 } {2} h \times (a + b ) \), where \( a \) and \( b \) are length of the parallel sides and \( h \) is the perpendicular height of the trapezoid.
Isosceles Trapezoids
An isosceles trapezoid is a trapezoid that has congruent legs.
Thus, for the isosceles trapezoid \(ABCD\) in the figure above, the following correspond:
(1) The bases are parallel.
(2) The legs are equal in length. Thus, \(\lvert\overline{BA}\rvert=\lvert\overline{CD}\rvert.\)
(3) The angles the two legs make with a base are equal. Thus, \(\angle BAD=\angle CDA\) and \(\angle ABC=\angle DCB.\)
Example Problems
If \( \angle A + \angle B + \angle C = 290^\circ \) in the trapezoid above, what is \( \angle D?\)
Since a trapezoid is a quadrilateral, the sum of its four internal angles must equal \( 360^\circ .\) Hence we have
\[ \begin{align} \angle D &= 360^\circ - (\angle A + \angle B + \angle C) \\ &= 360^\circ - 290^\circ \\ &= 70^\circ. \ _\square \end{align} \]
ParallelAngleBisector
The figure above depicts an isosceles trapezoid. \[\]
If the length of \( \overline{AB} \) is 5, what is the length of \( \overline{CD} ?\)
An isosceles trapezoid is a trapezoid that has congruent legs. Thus, the lengths of \(\overline{CD}\) and \( \overline{AB}\) are equal, i.e.
\[ \lvert \overline{CD} \rvert = \lvert \overline{AB} \rvert = 5. \ _\square \]
ParallelAngleBisector
The above figure depicts a trapezoid where \( \overline{AB} \parallel \overline{CD} .\) If \( \angle A=120^\circ,\) then what is \( \angle D?\)
Since two angles on the same side of a trapezoid's leg add up to \( 180^\circ,\)
\[ \begin{align} \angle A + \angle D = 180^\circ, \\ \angle B + \angle C = 180^\circ. \end{align} \]
Thus, \[ \angle D = 180^\circ - \angle A = 180^\circ - 120^\circ = 60^\circ. \ _\square \]
The above figure depicts an isosceles trapezoid. If \( \angle B=65^\circ ,\) then what is \( \angle D ?\)
In an isosceles trapezoid, the angles on either side of the bases are the same. Thus, it follows that
\[ \angle B = \angle C = 65^\circ. \]
Since two angles on the same side of a trapezoid's leg are supplementary, we know that
\[ \begin{align} \angle C + \angle D &= 180^\circ \\ \Rightarrow \angle D &= 180^\circ - \angle C \\ &= 180^\circ - 65^\circ \\ &= 115^\circ. \end{align} \]
Thus, \( \angle D=115^\circ. \ _\square \)
ParallelAngleBisector
The above figure shows an isosceles trapezoid where \( \overline{AD} \parallel \overline{BC}.\) If \( \lvert \overline{AB} \rvert = \lvert \overline{AD} \rvert\) and \( \lvert \overline{BC} \rvert = 2 \lvert \overline{AD} \rvert ,\) then what is \( \angle ABD ?\)
ParallelAngleBisector
Let points \( A^{'}\) and \(D^{'}\) be the perpendicular foots on \(\overline{BC}\) from points \(A\) and \(D,\) respectively. Then we have \(\lvert\overline{AD}\rvert=\lvert\overline{A'D'}\rvert.\)
Since the problem states that \(ABCD\) is an isosceles trapezoid, we know that \(\lvert\overline{BA'}\rvert=\lvert\overline{CD'}\rvert.\) Given that the length of \(\overline{BC}\) is twice the length of \(\overline{AD},\) we have \[ \begin{align} \lvert \overline{BA^{'}} \rvert &=\frac{1}{2}\cdot \lvert \overline{A^{'}D^{'}} \rvert \end{align} \] and since the length ratio of the hypotenuse and the base of \( \triangle ABA^{'} \) is \( 2:1 ,\) \( \angle ABA^{'} \) is \( 60^\circ . \qquad (1) \)
We know that \( \triangle ABD \) is an isosceles triangle, so \( \angle ABD=\angle ADB,\) and since \( \overline{AD} \parallel\overline{BC},\) it follows that \( \angle DBC=\angle ADB.\)
Thus, \[ \angle ABD = \angle ADB = \angle DBC . \qquad (2) \]
From (1) and (2), we know that \( \angle ABD=\angle DBA^{'}\) and \( \angle ABA^{'} = 60^\circ.\) Hence we have
\[ \begin{align} \angle ABD + \angle DBC = 60^\circ \\ 2\cdot \angle ABD = 60^\circ \\ \angle ABD = 30^\circ. \end{align} \]
Thus, \( \angle ABD \) is \( 30^\circ.\ _ \square\)