PROOF 151

Prove.

Derivation of the trapezoid area formula

On the Euclidean plane.

A way to read this proof

First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.

What you may use

A1area of a triangle = ½bh
area of a triangle (Proof 150)
A2height = from vertex to base
base and height of a triangle — the length of the perpendicular from the vertex to the line through the base (Proof 150)
A3the distance between parallel lines is constant
from any point of one line, the distance to a parallel line is the same (a consequence of the parallel postulate)
A4a diagonal splits a quadrilateral in two
diagonal decomposition of a convex quadrilateral — the two triangles do not overlap and fill the quadrilateral
A5non-overlapping pieces: areas add
additivity of area
A6every polygon has an area
existence of area
A7opposite sides of a parallelogram are equal
Euclid I.34
A8associativity, commutativity, distributivity
arithmetic of the reals
Ea = b ⟹ same operation on both sides
properties of equality

D1AB∥DC, a, b, h
the trapezoid and its height
L1both heights equal h
Lemma 1 — the heights of the two triangles
T[ABCD] = ½(a+b)h
theorem — area of a trapezoid
Cparallelogram = bh
corollary — area of a parallelogram

Nothing outside this list may be cited.

Definition D1

A convex quadrilateral ABCD with AB parallel to DC is called a trapezoid. Let AB = a, DC = b, and call the distance between the parallel lines AB and DC the height, written h.

Proof.

(1)The distance from any point of line AB to line DC is the same, so h is uniquely determined.A3

(2)Write [X] for the area of a figure X. Polygons have areas.A6, convention

∎

Lemma L1

The height of △ABC on base AB and the height of △ACD on base DC are both h.

Proof.

(1)C is on line DC, and line DC is parallel to line AB, so the distance from C to line AB is h.D1, A3

(2)That distance is precisely the height of △ABC on base AB, so that height is h.(1), A2

(3)A is on line AB, so for the same reason the distance from A to line DC is h, and the height of △ACD on base DC is also h.D1, A3, A2

∎

Theorem T

For a trapezoid ABCD with top side a, bottom side b, and height h, [ABCD] = ½(a + b)h

Proof.

(1)The diagonal AC divides the trapezoid ABCD into △ABC and △ACD, and the two triangles do not overlap and fill the trapezoid.D1, A4

(2)△ABC has base AB = a and height h, so [ABC] = ½ah(1), D1, L1, A1

(3)△ACD has base DC = b and height h, so [ACD] = ½bh(1), D1, L1, A1

(4)[ABCD] = [ABC] + [ACD] = ½ah + ½bh = ½(a + b)h(1), (2), (3), A5, A8, E

∎

Corollary C

For a convex quadrilateral ABCD with both pairs of opposite sides parallel, if DC = b and the height is h, then [ABCD] = bh

Proof.

(1)AB and DC are parallel, so ABCD is a trapezoid.assumption, D1

(2)Opposite sides of a parallelogram are equal, so a = AB = DC = b(1), assumption, A7

(3)[ABCD] = ½(b + b)h = bh(1), (2), T, A8, E

∎

Remark

Can A1 be removed from this proof?

Separate the task of cutting the trapezoid into two pieces from the task of writing the pieces' areas as numbers.