contains the unit of sw’ 
unit ABCD as often as the numeral product pq con- 
* tains unity, consequently the product pq reall wepeene 
the area of the rectangle, or indicate how often it 
perficies.. Thus, if EF contain 
the linear unit AB four times,:and EH contain it three 
times, the area, EFGH: will be 3x 4—= 12, that is, 
twelve times a,square.whose side is AB = 1. 
In consequence of the surface of a rectangle EFGH 
being expressed by the product of its sides, the rect- 
le, or its area, may be denoted by the symbol EF x 
FG; or thus, EF.FG,.in conformity. to the. manner 
of expressing a in arithmetic. 
However, instead of expressing the area of a square 
— on a line AB thus, AB x AB, it is thus expressed, 
AB*. 
Note. A rectangle is said to be contained by two of 
its sides about any one of ‘its angles. 
Prop. V. Tueor. 
The area of a parallelogram is equal to the product 
”_. of its base by its altitude. 
For the parallelogram ABCD is equivalent to the 
rectangle ABEF, which has the same base AB and the 
same altitude (1.) ; and this last is measured by AB x 
BE, or ABX AF; that is, by the product of the base 
of the parallelogram and its altitude (4.) 
Cor. Paralle of the same base are to one an- 
other as their altitudes ; and parallelograms of the same 
altitude are to one another as their bases. For in the 
former case, Eo B for their common base, and A and 
A’ for their altitudes ; then we have Bx A: Bx A’:: 
A: A’. And in the latter, put A for their common al- 
titude, <3 Band B’ for their bases ; then Bx A: B’x 
A::B:B’. 
Prop. VI. Tueror. 
The area of a trian, - to the product of its 
base by half its altitude. ' 
For the triangle ABC is half the parallelogram ABCE, 
which has the same base BC, and the same altitude AD 
2.) ; but the area of the am is BC x AD 
+49 therefore the area of the triangle is }BC x AD, or 
x 4AD. 
Cor. Two triangles of the same base are to one an- 
other as their altitudes, and two triangles of the same 
altitude are to one another as their bases, 
Prop. VII. Taeor. 
. The area of a trapezoid ABCD is equal to the 
duct of half the sum of its parallel sides AB, DC by its 
altitude EF. 
I, the middle of the side BC, draw KL pa- 
rallel to t ite side DA, and produce DC until 
it meet KL. pnesesetes IBL,; ICK, the side [IB = 
IC, the angle B= C (2 Cor, 21. 1.), the angle BIL= 
CIK ; therefore the triangles are equal (7. 1.), and the 
side CK=BL. Now, the ogram ALKD is the 
_ sum of the polygon ALICD and the triangle CIK, and 
‘the trapezoid ABCD is the sum of the same polygon 
and a triangle equal to Bil. ; therefore the trapezoid 
ABCD is equal to the logram ALKD, and has 
for its measure AL x EF. And because AL = DK and 
BL=CK, therefore AB+CD=AL 4+ DK =2AL; 
3 
GEOMETRY. 
215 
and hence AL is half the:sum of the parallel sides AB, 
CD: therefore the area of the trapezoid is.equal to of F 
3 (AB + CD) x EF. 
Prop. VIII.’ Turor. 
If a straight line AC be divided into any two parts Fis: 97 
AB, BC, the square made on the whole line AC is equal 
to the squares on its two-parts AB,,BC, together with 
twice the rectangle contained by these parts, Or the 
hones may. be briefly expressed thus; AC? or 
(AB 4+ BC)? AB? + BC* 42 AB x BC. 
Construct the square ACDE ; take AF = AB ; draw 
FG parallel to AC, and BH_ parallel to AE. 
The square ACDE is composed settee «A the 
first ABIF is the square on. AB, because AF = AB ; 
the second IGDH is the squareon BC ; for since AC= 
AE, and AB=AF, the difference AC —:AB is equal 
to the difference AE — AF, that is, BC—EF; but 
because of:the parallels, BC= 1G, and EF=HI, there- 
fore HIGD is the square on BC. . These two parts be» 
ing taken from the whole square,  thereremains the 
two rectangles BCGI, EFIH, which are each equal to 
AB xX BC; so that the truth of the proposition is»evi- 
dent. 
Prop. IX. .Turor. 
If aline AC be the difference of two lines AB, BC, 
the square of AC shall be equal to the excess of the 
squares of AB and BC above twice the rectangle con- 
tained by AB and BC; that is, 
AC? or (AB — BC)?= AB? 4+ BC*— 2 AB x BC. 
Construct the weer ABIF, take AE = AC, draw 
CG parallel to BI,-HK parallel to AB, and complete 
the square FEKL. 
The two rectangles CGIB, GLKD are each equal to 
AB x BC. If these be taken from the figure ABILKEA, 
which is equal to AB? BC?, there will evidently re- 
main the square ACDE; that is, the square of AB— 
BC, 
Prop. X. TuHeror. 
The rectangle contained by the sum and the diffe- 
rence of two lines, is equal to the difference of the 
squares of these lines; that is, 
(AB + BC) x (AB— BC) =AB?— BC?. 
Construct upon AB and AC the squares ABIF, 
ACDE, produce AB so that BK = BC, and complete 
the rectangle AKLE, 
The base AK of the rectangle is the sum of the two 
lines AB, BC; the altitude AE is the difference of the 
same lines ; therefore the rectangle AKLE = (AB + 
BC) x enna Gt But the same rectangle is made 
up of two parts ABHE + BHLK ; and the part BHLK 
_is equal to the rectangle EDGF ; for BH = DE, and 
BK = EF; therefore AKLE=ABHE+ EDGF. But 
these two parts form the excess of the square ABIF 
above the square DHIG, which is the square of BC ; 
therefore (AB 4+ BC) x (AB— BC) = AB* — BC’. 
Prop. XI, TueEor. 
In any right angled triangle, the square which is de- 
scribed on the. side opposite to. the right angle is equal. 
Fig. 98, 
Fig. 99. 
