840 



MATH KM \T1CS.-MKXSU RATK'V 



[AREAS OP FIODRBS. 



Now - - the length of arc on which quadrant stands. 



.'. Area of quadrant- j X arc on which quadrant stands. 



(9). To find Area of a Sector of a Circle. 



Let A O B bo any sector of a circle, 



O A r. A O B (in circular measure), 



.'.AB-r0. Hg.lJ. 



Now if A O B were a quad- B 



rant, A B would equal -*- and 

 the area of the quadrant 

 I?". A) 



But in equal circles sectors are to each other as the 

 arcs on which they stand. 



.'. Area sector : area quadrant : : r : ~^-. 



.'. Areasector : - : : : I. 

 4 Z 



/. Area sector - 



B. r X (arc on which sector stands). 



(10). To find the Area of a Segment of a Circle. 

 Let O A C B be a sector of a circle. Join A B. We 

 are required to find the area H 



of the segment ABC. Draw 

 O N C perpendicular to A B. 

 LetAOB = 0. OB-r. 







.'. ON=rcos. s-AN= 



/.area A BC^r 2 sin. X 



cos. g =2" r * sin - e ' 



Now area AC B0=i-r*0. 

 SI 



.'. area of segment = ( sin.0). 



COB. From A draw A n perpendicular to O B. Then 

 A n=r sin. 9, also arc A C B =r 0. 

 Hence the area of segment 



^-(rO rsin. 9) 

 JL X (ACB An) 



(11). To find the space between two Concentric Circles, 



viz., ABOEDF. 

 Fig. 14. LetOA=rOB = r v 



Then area of interior 

 circle = )r r 2 , and area of 

 exterior circle=wr l 8 

 ; Area of the space be- 

 tween the circles 



On O B as a diameter, 

 describe a circle O P B 

 cutting the interior circle 

 inP;joinOP.PB, then 

 O P B is a right angle, 

 BP'-OB* < ' 



also since O B P is a right 

 angle, B P touches the circle, or the area of the space in 

 question is equal to the area of a circle whoso radius is 

 the length of the line drawn from any point in the exterior 

 circle to touch the interior circle. 



The cited cases of areas of plane figure* are the clii.-f 

 of th<MO which belong to Elementary Mathematics. 

 The determination of areas bounded by curvrd Inn's 

 belongs, of right, to the Integral Calculus ; the following 

 proposition*, however, are best given hero, though the 

 reasoning is not of a strictly elementary character. 



(12). To find the Area of an Ellipse.* 



Fig. li. 



Let nB A bo a semi- 

 ellipse. 



OA its semi-major 

 aiis=a, 



O 13 its semi-minor 

 axis 6. 



On oA as a diameter 

 describe a semicircle 

 aCA. In AB take 

 any point P and draw an ordinate P N perpendicular to 

 O A, and produce it to meet the circle in Q, and draw 

 another ordinate q;in parallel and near to Q P N, and 

 complete the parallelograms nP, nQ 

 Now, by a property of the ellipse, 



NQ : NP :: o : 6 

 .'. nQ : nP : : o : 6 



and hence, if wo suppose a series of parallelograms to be 

 described in circle and ellipse, the same proportion will 

 hold good between each of these, and therefore 



Sum of parallelograms in semicircle : sum of parallelo- 

 grams in semi-ellipse : : a : b. ; and this being true, how- 

 ever great the number may be, is true in the limit. 



Now the semi-ellipse is the limit of the parallelograms 

 inscribed in it, and the semicircle is the limit of paral- 

 lelograms inscribed in it. 



.'. semicircle : semi-ellipse : : a : 6 

 .'. circle : ellipse : : it a* : ir a 6 



But area of circle = TT a* 

 .'. area of ellipse=jr 06. 



(13). Let qPQbea portion of a Parabola. It is re- 

 quired to find its Area. (Fig. 17)- 



Bisect Q<j- in V, draw the diameter P V. Through P 

 draw r P R parallel to Q V q, and draw Q R and ?r 

 parallel to PV. Then the area gPQis two- thirds of 

 grRQ. 



Take p and p,, two points in PQ, and through p, 

 draw p, m,, p, n, parallel to 1 J V and Q V, and through 

 p draw pm pn also parallel to P V, Q V, and produce 

 them to meet p^ n, in q^ and j>, m 1 in 5 respectively. 

 Then 



Parallelogram pm, : parallelogram pn t : : p n X HH, : 

 pm X mm,. 



::pnx (P, Pn) : Ptv X Cp, n, pn) 



Let AP Q be an area included by two traiht lines A B. A P, and the 

 eurvc P Q ; divide A Q into equal parta An, at, lie cd, dd, and on these 

 lines draw the rectangles Ap, aq, tr, ct tcithin Uie curved areas, and 

 complete the parallelograms Ap', aq', kr' , ci, dt. Then it is plain that 

 the difference between the interior parallelograms ( \p, aq, kr, a) and the 

 exterior parallelograms (Ap', no', kr 1 , cs', i<') will equal Ap'. Now, by 

 making the number of parallelograms very large, Aa, and .'. Ap' will 



Fig. 16. 



become Tery small, and my be made lew than any magnitude that may 

 be assigned. Now the curvilinear area is clearly greater than the mt. r...r 

 and lew than the exterior parallelograms, and therefore diffrv tram tlic 



. by a quantity less than Ap' ; i.e.. a quantity th 



can he made lew than any that can bo assigned : and therefore the curvi- 

 linear area is the limit to which the sum of the interior parallelograms 

 continually approaches when their number U increased. 



