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

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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.
It's impossible for a trapezoid to have three right angles. If three angles are right angles, the fourth must also be a right angle, making it a rectangle, which is a type of trapezoid.

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

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