ABCD = AB × AD:

hence, the area of a rectangle is equal to the product of its base and altitude; that is, the number of superficial units in the rectangle, is equal to the product of the number of linear units in its base by the number of linear units in its altitude.

Scholium 2. The product of two lines is sometimes called the rectangle of the lines, because the product is equal to the area of a rectangle constructed with the lines as sides.

PROPOSITION V. THEOREM.

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

Let ABCD be a parallelogram, AB its base, and BE its altitude: then will the area of ABCD be equal to AB × BE.

For, construct the rectangle ABEF, having the same base and altitude: then will the rectangle be equal to the parallelogram (P. I.); but the area of the rectangle is equal to AB × BE:

F D

E

A

B

also equal to

hence, the area of the parallelogram is
is also
AB × BE; which was to be proved.

Cor. Parallelograms are to each other as the products of their bases and altitudes. If their altitudes are equal, they are to each other as their bases. If their bases are equal, they are to each other as their altitudes.

PROPOSITION VI. THEOREM.

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

Let ABC be a triangle, altitude: then will the area BC × AD.

BC its base, and AD its
of the triangle
of the triangle be equal to

E

A

For, from C, draw CE parallel to BA, BA, and from A, draw AE parallel to CB. The area of the parallelogram BCEA is BC × AD (P. V.); but the triangle ABC is half of the parallelogram BCEA: hence, its area is equal to BC × AD; which was to be proved.

B

D

Cor. 1. Triangles are to each other, as the products of their bases and altitudes (B. II., P. VII.). If their altitudes are equal, they are to each other as their bases. If their bases are equal, they are to each other as their altitudes.

Cor. 2. The area of a triangle is equal to half the product of its perimeter and the radius of the inscribed circle. For, let DEF be a circle inscribed in the triangle ABC. Draw OD, OE, and OF, to the points of contact, and OA, OB, and OC, to the verti

B

E

D

ces.

The area of OBC will be

A

F

equal to OE × BC; the

area of OAC will be equal to OFX AC; and the area

of OAB will be equal to OD × AB; and since OD, OE, and OF, are equal, the area of the triangle ABC (A. 9), will be equal to OD (AB + BC + CA).

The area of a trapezoid is equal to the product of its altitude and half the sum of its parallel sides.

Let ABCD be a trapezoid, DE its altitude, and AB and DC its parallel sides: then will its area be equal to ·DE × (AB + DC).

For, draw the diagonal AC, forming the triangles ABC and ACD. The altitude of each of these triangles is equal to DE The area of ABC is equal to AB x DE (P. × VI.); the area of ACD is equal to ¿DC × DE: hence, the sum of the triangles, is and DC × DE, or to to be proved.

D

AE

×

B

area of the trapezoid, equal to the sum of DE × {(AB + DC)

which is the AB X DE

which was

The square described on the sum of two lines is equal to the sum of the squares described on the lines, increased by twice the rectangle of the lines.

allel to AE; lay off AF equal to AB,

E

H

I

F

G

and from F draw FG parallel to AC: then will IG and IH be each equal to BC; and IB and IF, to AB. The square ACDE is composed of four parts. The part ABIF is a square described on AB; the part IGDH is equal to a square described on BC; the part BCGI is equal to the rectangle of AB and BC; and the part FIHE is also equal to the rectangle of AB and BC and

A

B

because the whole is equal to the sum of all its parts (A. 9), we have,

AC2 AB2 + BO2 + 2AB × BC ;

which was to be proved.

will also be equal: hence, the equal to four times the square

Cor. If the lines AB and BC are equal, the four parts of the square on AC square described on a line is described on half the line.

The square described on the difference of two lines is equal to the sum of the squares described on the lines, dimin ished by twice the rectangle of the lines.

Let AB and BC be two lines, and AC their difference: then will

ᎪᏟ = AB2 + BO2

2AB × BC.

On AB construct the square ABIF; from

C draw

CG parallel to
to BI; lay off CD equal to

AC, and

from D draw DK parallel and equal to BA; complete

the square EFLK: then will EK be equal to BC, and EFLK will be equal to the square of BC.

The whole figure ABILKE is

equal to the sum of the squares described on AB and BC. The part CBIG is equal to the rectangle of AB and BC; the part DGLK is also equal to the rect

L F

G I

K

E

D

A

C

B

angle of AB and BC. If from

the whole figure ABILKE, the two parts CBIG and DGLK be taken, there will remain the part

which is equal to the square of AC: hence,

2

A02 AB2 + BC2 2AB X BC;

which was to be proved.

ACDE,

The rectangle contained by the sum and difference of two lines, is equal to the difference of their squares.

Let AB and BC be two lines, of which AB is the greater then will

(AB + BC) (AB – BC) On AB, construct the square ABIF; prolong AB, and make BK equal to BC; then will AK be equal to AB+ BC; from K, draw KL parallel to BI, and make it equal to AC; draw LE parallel to KA, and CG parallel to BI: then DG is equal to

AB2 – BC2.

I

F

G

H

E

D

BC, and the figure DHIG is equal to the square on

BC, and EDGF is equal to BKLH.