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.

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

$$\begin{aligned} \angle A + \angle B &= 180^\circ \\ \angle C + \angle D &= 180^\circ. \end{aligned}$$

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.

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 $\angle ABC=\angle DCB.$

If $\angle A + \angle B + \angle C = 290^\circ$ in the trapezoid above, what is $\angle D?$

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Since a trapezoid is a quadrilateral, the sum of its four internal angles must equal $360^\circ .$ Hence we have

$$\begin{aligned} \angle D &= 360^\circ - (\angle A + \angle B + \angle C) \\ &= 360^\circ - 290^\circ \\ &= 70^\circ. \ _\square \end{aligned}$$

ParallelAngleBisector

The figure above depicts an isosceles trapezoid. $$$$

If the length of $\overline{AB}$ is 5, what is the length of $\overline{CD} ?$

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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 $\angle D?$

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Since two angles on the same side of a trapezoid's leg add up to $180^\circ,$

$$\begin{aligned} \angle A + \angle D = 180^\circ, \\ \angle B + \angle C = 180^\circ. \end{aligned}$$

Thus, $$\angle D = 180^\circ - \angle A = 180^\circ - 120^\circ = 60^\circ. \ _\square$$

ParallelAngleBisector

The above figure depicts an isosceles trapezoid. If $\angle B=65^\circ ,$ then what is $\angle D ?$

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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

$$\begin{aligned} \angle C + \angle D &= 180^\circ \\ \Rightarrow \angle D &= 180^\circ - \angle C \\ &= 180^\circ - 65^\circ \\ &= 115^\circ. \end{aligned}$$

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 $\angle ABD ?$

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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 $$\begin{aligned} \lvert \overline{BA^{'}} \rvert &=\frac{1}{2}\cdot \lvert \overline{A^{'}D^{'}} \rvert \end{aligned}$$ 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

$$\begin{aligned} \angle ABD + \angle DBC = 60^\circ \\ 2\cdot \angle ABD = 60^\circ \\ \angle ABD = 30^\circ. \end{aligned}$$

Thus, $\angle ABD$ is $30^\circ.\ _ \square$