GEOMETRY. 



the seventh decimal place. It is usual to represent 

 this number by the symbol ; 



.'. circumference = * X diameter. 



If we convert the expression for * into a series 

 of continued fractions,* we shall find for it 



3 323 355 IQ3993 &c 

 i' 7' 106' 113' 33102' 



Let 



C denote the circumference, 

 D the diameter, 

 r the radius. 



We then have the following formulae : 

 C = <rD; .-. D=-. 



C = 2 ,,;,., = C == D 



2<r 2 



OF SURFACES. 

 The Rectangle. 



PROP. VIII. To find the area of a rectangle, 

 having given two adjacent sides. 



Thus, let it be required to find the area of the 

 rectangle ABCD, of 

 which the side AB is 

 7 feet, and BC 4 feet. 

 If we divide AB into 

 seven equal lengths, 

 and BC into four, and 

 through the points of 

 division draw lines 



parallel to the sides ; then it is clear that for 

 every division of AB we have a row, and each row 

 contains as many compartments as there are 

 divisions in BC. But we have seven divisions in 

 AB ; hence we have seven rows and four compart- 

 ments in each row, therefore there are in all 



7 X 4, or 28 compartments, 

 but each compartment is i square foot 

 .*. area = 7X4 square feet, or 

 28 square feet. 



Hence, to find the area, we mul- 

 tiply one side by the adjacent 

 one. 



To find the area of a square, 

 we multiply the side by itself. 



Any Parallelogram. 



PROP. IX. To find the area of any parallelo- 

 gram. 



Let ABCD be the parallelogram. Drop DP 

 and AQ perpendicular 

 to CB. Then the paral- 

 lelogram ADCB is equal 

 in area to the rectangle 

 ADPQ, since they stand c - 

 on the same base AD, 

 and between the same parallels AD and CB. But 

 the area of ADCB is found by multiplying AD by 

 DP ; hence the area of the parallelogram is found 

 by multiplying one side by the perpendicular on 

 it from the opposite side. 



Se Munn's Tktory of Arithmttie. 



The Triangle. 



PROP. X. To find the area of a triangle : 



First, When a side is given, and the perpen- 

 dicular on it from the oppo- 

 site angle. 



Thus in the triangle ABC, 

 let the base BC and the per- 

 pendicular AF from A on BC 

 be given. 



Then BC X AF denotes 

 the area of the rectangle EBCD, but the given 

 triangle is half of this ; 



/. area of triangle - 52L^ 



Hence the area is found by multiplying the base 

 by the perpendicular on it from the opposite angle, 

 and taking the half of this product 



Second, When the three A 



sides are given, 



Let the three sides AB, BC, 

 CA be given ; then it has 

 already been shewn (Prop. II., 

 p. 633) how we may find the 

 perpendicular AD upon BC. 

 Then, by last proposition, 



area 



BC X AD 



Let BC = a, CA = b, AB = c ; 



:. area = - X AD. 



2 



But AD = - a Js(s a)(s J)(j - c). (See Prop. 

 II., p. 633) ; 



1*1 



Hence to find the area of the triangle, when the 

 three sides are given, we have the following rule : 

 Add the three sides together ; take half the sum. 

 From half the sum subtract each side separately. 

 Multiply half the sum and the three remainders 

 together, and take the square root of the pro- 

 duct. 



In the case of the isosceles or equilateral tri- 

 angle, this is very much simplified ; thus for the 

 area of the equilateral triangle, we take the fourth 

 of the square of the 

 side and multiply it by 

 the V3- 



PROP. XI. To find 

 the area of a trapezoid. 



Let ACBD be the given 

 figure. On AB drop the 

 perpendiculars DE and 

 CF ; then the 



area-~XDE 



therefore the area is found by multiplying the 

 diagonal b.y half the sum of the perpendiculars 

 on it 



