Adding this equations,

A B2 + BC2 = AC ×.AD + AC × CD.

187. The Area of a polygon is the measure of its surface. It is expressed in units, which represent the number of times the polygon contains the square unit that is taken as a standard. This unit is called the unit of surface, and is usually adopted arbitrarily.

188. The usual unit of surface is the square whose side is some unit of length; for example, a square inch, or a square foot. 189. Equivalent Polygons are those which have the same.

area.

190. The projection of a point upon a straight line of indefinite length is the foot of the perpendicular drawn from the point to the line. An entire line may be said to have a projection upon another line in the same way.

191. A medial line of a triangle is a straight line drawn from either vertex to the middle point of the opposite side.

192.

THEOREM LXI.

Two rectangles having equal altitudes are to each other as their basis.

Let A B C D and A BEF be two rectangles having equal altitudes.

Let A G be a common measure of A D and A F, and suppose it to be contained seven times in A D and four times in A F.

Then

AD 7

=

AF 4

(1)

Drawing perpendiculars to A D through the several points of division, the rectangle A B C D will evidently be divided into seven equal parts, of which the rectangle A B E F will contain four.

A B C D

A D

=

A F'

(things equal to the

Then by (1) and (2) ABE F

same thing are equal to each other, axiom.)

This may also be proved when A D and A F are incommensurable.

Corollary. Since either side of a rectangle may be taken as a base, it follows that two rectangles with equal bases are to each other as their altitudes.

THEOREM LXII.

193. Any two rectangles are to each other as the products of their bases by their altitudes.

Let A and B be any two rectangles leaving the altitudes a and a', and the bases b and b', respectively.

Let C be a third rectangle, having its altitude equal to the altitude of A and its base equal to the base of B.

Then the rectangles A and C, having equal altitudes, are to each other as their bases, (by Theorem LXI); that is,

A
C

h b'

(1)

And the rectangles C and B, having equal bases, are to each other as their altitudes (Theorem LXI, Corollary); that is,

Multiplying equations (1) and (2), we have

(2)

194.

THEOREM LXIII.

The area of a rectangle is equal to the product of its base and altitude.

Let a and b be the numerical measures of the altitude and base of the rectangle A, and let B be the square whose side is the unit of length.

To prove that the area of A, referred to B as a unit, is equal to a times b.

Since any two rectangles are to each other as the products of their bases by their altitudes (Theorem LXII), we have

195. Corollary. The area of a square is equal to the square

of one of its sides.

196.

THEOREM LXIV.

The area of a parallelogram is equal to the product of its base and altitude.

Let A B C D be a parallelogram leaving its altitude D F equal to a and its base A D equal to b.

To prove that A B C D a x b:

Draw A E perpendicular to A D and meeting B C produced at E.

Then A EFD is a rectangle having its altitude equal to a and its base equal to b.

In the right triangles A B E and D C F, A B D C and A EDF (Theorem XXIX).

Therefore, the triangles are equal (Theorem XX). Now if from the entire figure A E CD, we take the triangle A B E, there remains the parallelogram A B C D; and if we take the triangle DCF, there remains the rectangle A E F D.

Therefore, area A B C D = area A E F D.

But the area of a rectangle A EFD = a × 6 (Theorem LXIII).

Hence, area A B C D a x b.

THEOREM LXV.

197. The area of a triangle is equal to one-half the product of its base and altitude.

Let A B C be a triangle, having its altitude equal to a and its base equal to b.

To prove that area A BC=ax b:
Draw the lines A D and C D parallel

to B C and A B.

Then A B C D is a parallelogram having its altitude equal to

a and its base equal to b.

Now a diagonal of a parallelogram divides it into equal triangles (Theorem XXIX).

Therefore area A B C area A B C D.

=

But the area of the parallelogram A B C D is equal to a times b, (Theorem LXIV).

Hence, area A B C = } a × b.

198. Corollary I. Two triangles having equal bases and equal altitudes are equivalent.

199. Corollary II. Two triangles having equal altitudes are to each other as their bases; two triangles having equal bases are to each other as their altitudes; and any two triangles are to each other as the products of their bases by their altitudes.

200. Corollary III. A triangle is equivalent to one-half a parallelogram having the same base and altitude.

THEOREM LXVI.

501. The area of a trapezoid is equal to one-half the sum of its parallel sides multiplied by its altitude.

Let A B C D be a trapezoid, having its altitude equal to a, and its parallel sides equal to b and b' respectively.

To prove that area A B C D = a × √ (b + b') :

Draw the diagonal B D dividing the trapezoid into two triangles A B D and B CD, having the common altitude a and their bases equal to b and b' respectively.

Then, area A B C D area A B D + area B C D.

But since the area of a triangle is equal to one-half the product of its base by its altitude, (Theorem LXV),

area A B D = a xb,
area B CD = ax b'.

and

Therefore, area A B C D =

1 a × b + 1 a × b', = a × 3 (b + b').

202. Corollary. Since the line joining the middle points of the non-parallel sides of a trapezoid is equal to one-half the sum of the bases (Theorem XXXV), it follows that,

The area of a trapezoid is equal to the product of its altitude by the line joining the middle points of the non-parallel sides.