AREAS OF FIGURES.'] 



MATHEMATICS. MENSURATION. 



646 



A B x B C. Hence rectangle o c : rectangle B C : : a&X 

 be: ABxBC. 



Now suppose ab = 1, and cb = 1. Then the area ac is 

 the unit of area, i. e., is a square inch, or a square foot, 

 or a square yard, according as ab is an inch, foot, or 

 a vard. In this case 



liectangle BC = ABxBC. 



Hence, if a and b be the sides of a rectangle, its area is 

 ab, . e , it contains as many units of area as the product 

 of the number of units of length in one side, by the 

 number of units of length in the other. 



COK. If a be the side of a square, its area is a 1 . 



(2). To find the Area of a Triangle. 

 Let A B C be the triangle ; from A draw A N per- 

 pendicular to B C. Then, 

 since the area of ABC is 

 half that of a rectangle, 

 whose base is BC and 

 height AN, the area of tri- 



BCX AN 



angle = ^ 



COR. (1). If we have B 

 given AB. BC and angle B. 



, BC X AB sin. B 

 Area triangle = 



Since AX = AB sin. B. 

 (2). If we have given B C and the angles of triangle, 



e sin. C 

 Then, since _= 3 



a sin. A 



... a* sin. B sin. C 

 Area triangle a in. A 



(3). If we have all the sides given (by Trigon., 

 Art. 41) 



Area triangle ,/T~(i-a) (-&) (-c) 

 where 2 a + 6 -f-c. 



(3). To find the Area of a Parallelogram. 

 If a b be the sides of the parallelogram, and A the 

 angle contained by those sides, then by the last article 

 the area of half the parallelogram 



ab sin. A 



.'. Area of parallelogram = ab sin. A. 



Or, if p is the perpendicular distance between two 

 parallel sides, each of which = a, then 



Area of parallelogram = ap. 



(4). To find the Area of a Traptaoid. 

 Let A B C D be the trapezoid of which the side A B is 



parallel to the 

 N side C D. Join 



Fig. 8. 



B D, draw D M 



perpendicular to 

 A B, and B N 

 perpendicular to 

 D C, or D C pro- 

 duced, and let a= 



A B. b = D C and p = D M or B N. 

 Then Area triangle A B D = i op 



Area triangle B C D = X bp 

 .'.Area ABCD =*{;>(<* + &) 



Or the area of a trapezoid = ^ sum of parallel sides 

 plia the perpendicular distance between them. 



(5). To find the Area of a Trapezium. 

 Let ABCD (Fig. 9), be the trapezium ; join AC, from 



B and D let fall perpen- 

 diculars B m, D n, upon 

 AC. 



The area of triangle 

 ABC-* Bm. A C and 

 area of triangle A U C = 

 I Dn. A C. 



.'. Area of trapezium 



- J AC(l5m + 



Fig. 9. 



If instead of having the perpendiculars Bm and Dn 

 given, we have the sides A B. B C. C 1 >. D A. and a dia- 

 gonal, we must find the area of each triangle separately 

 by the formula. 



Area triangle = *J s. (s a) (s b) (3 c) 



Or, again, if we have the sides and one angle, as A, 



Fig. 10. 



given, we can proceed as fol- 

 lows. Let A B. = a. B C = b. 

 C D = c. L>A = d. Join BD 

 (Fig. 10), then we can calcu- 

 late B D from triangle A B D 

 by second case of oblique- 

 angled triangles, and having 

 calculated B D we can deter- 

 mine the areas of the triangles 

 as before. 



(6). To determine the Area of an irregular Polygon. 



Suppose ABCDEFto be an irregular polygon, its 

 area can be determined Fig. 11. 



b/ effecting the following 

 measurements ; join A D, 

 the longest distance across 

 the figure from B C E F, 

 draw perpendiculars to 

 AD., viz. B?n, Cn, Ep, 

 Fg, respectively ; measure 

 Am, mn, nD, Cn, Bm, 

 Ag, gp, pD, Ep, Fg. 



Then area of polygon = 

 ABm + BmnC + CnD + 

 DpE+EpgF + FgA. 



+ | Am X Bm -f- J mn 

 X (Bm + Cn) -f i Cn X nD + J. Dp +j>E + Jpg (E/)+ 

 Fg) + |Ag X gF. 



(7). To determine the Area of a regular Polygon. 



Various formulas have been already given for this 

 in Cliapter XI., on Trigonometry (Art. 45). It is there 

 shown that if n be the number of sides of the polygon, 

 and u the length of one of the sides, 



Area 

 Hence in case of pentagon, 



na s 180 



-, co tan. ' 



4 n 



Area 



ootan. 36 



[n case of hexagon, 



And so on in other cases. 



(8). To determine the Area of a Circle. 



If n is the number of sides of a regular polygon in- 

 scribed in a circle whose radius is r, then, as we have 

 already seen (Trig., Art. 45), 



Area polygon = ^- sin ^. 

 n 



sin. IT 



=r ' 7r 2]r 

 n 



Now if we increase the number of sides of the polygon, 

 it becomes more and more nearly equal to the circle, 

 and in the limiting case when the number of sides is 

 infinitely great, it becomes the circle ;* but when n= u> 



^5 = and (Trig., Art. 47) in this case sin. 

 n n 



o J- 



.'. Area of circle =r 2 IT. 



r'lr 1 rir 

 Con. Hence area of quadrant = -j-- = n ~n 



* Compare w th the ttatemeut in the text what if laid on p. 577. 



