Kevin claims he can draw a trapezoid with three right angles. Is this possible? Explain.

1. Let the quadrilateral have vertices \( A \), \( B \), \( C \), and \( D \) in order. 2. Suppose Kevin claims that three of the angles are right angles. Without loss of generality, let \[ \angle A = \angle B = \angle C = 90^\circ. \] 3. Since the sum of the interior angles of any quadrilateral is \[ 360^\circ, \] the fourth angle must be \[ \angle D = 360^\circ - (90^\circ + 90^\circ + 90^\circ) = 90^\circ. \] 4. Thus, all four angles are right angles; the quadrilateral is a rectangle. 5. A rectangle (which has four right angles) does have a pair of parallel sides (in fact, two pairs), so according to the broad definition of a trapezoid as a quadrilateral with at least one pair of parallel sides, a rectangle is technically a trapezoid. 6. However, if the intended definition of a trapezoid is a quadrilateral with exactly one pair of parallel sides (excluding rectangles), then a quadrilateral with three right angles is impossible because three right angles force the fourth angle to be a right angle as well. Therefore, if Kevin means a non-rectangular trapezoid with exactly three right angles, it is impossible.