MB 



MATHEMATICS. MENSURATION. 



PARKAS OF FIOT-IIF.S. 



'. .tc. A prism is called a right prism when the planes 

 of tin- i>:inillflograros are IXT]-II- 

 dicuUr to the plane* of the base (i.e., 

 abtde in the accompanying figure). 



DBF. 2. A pyramid is a solid 

 bounded by any plane rectilinear 

 figure, and by triangles having a 

 common vertex, and for bases the 

 aides of the rectilinear figure respeo- 



Thiw (Tig. 21) PABCD is a 



pyramid, on a quadrilateral base, 

 D. 



DBF. 3. A cylinder is a solid 

 whose surface is traced out by a 

 straight line which always moves 

 parallel to its first position, and whose extremity is 

 guided br a given curve. 



Thus (frig. 22) if aeb is a circle, 

 and Ao a straight line perpen- 

 dicular to the plane of the circle, 

 then ABC abc is a cylinder which 

 is traced out by a line C. 

 that is always parallel and 

 equal to Aa. This cylinder 

 is strictly defined as a " right 

 cylinder with a circular base ;" it is, however, in elemen- 

 tary treatises generally called " a cylinder." 



DBF. 4. A cone is a solid, the surface of which is 

 traced out by a straight lino, one end of which passes 

 through a fixed point, and the other end through a given 

 plane curve, called its base. 



Fig. 21 Fig. 23. 



If B C D (Fi#. 23) is a circle, and A a fixed point, the 

 1 surface ABCD is called a cone on a circular base. If 

 ' O is the centre of the circle, join A ; then draw straight 

 j lines from A to different points in circumference of circle, 

 ! making equal angles with A O ; or if (which is the same 

 thing) A O is perpendicular to plane of circle, A B D is 

 strictly defined as "a right cone with a circular base ;" 

 or, as it is more generally called in elementary treatises, 

 a " riijht cone." 



N. B. It is manifest that if a regular polygon be in- 

 scribed in the circle BCD, and its angular points are 

 joined with A, that the resulting solid will be a pyramid 

 inscribed in the cone ; and also, that if we increase the 

 number of sides in the polygon, the inscribed pyramid 

 will approach more nearly to the cone ; and, since the 

 circle is the limit of the inscribed polygon, the cone will 

 be the limit of the inscribed pyramid. 



iilarly, the cylinder will be the limit of the inscribed 

 prism. 



(15). To find the Area of a right Prism. 

 In figure (20) AEea is a rectangle. Since the planes 

 A6, \t are perpendicular to the base, their cone of in- 

 tersection An is also perpendicular to the base, and 

 therefore angle Aoe is a right angle. Hence tho area of 

 AEea is Aa X A E, similarly of all the other parallelo- 

 grams. Hence the area of the parallelograms is Aa 

 (AK + ED + DC + ... ) =AEX (the perimeter of 

 the base). Hence the whole area = height X perimeter 

 of base + area* of the two ends. 



(16). To find the Area of the surface of a right Cylinder. 

 Since the surface of the cylinder is tho limit of the 

 turf ace of the inscribed prism, and the area of the prism 

 height x perimeter of base + 2 X base, whatever be 



the number of sides to the base, this will bo true in the 

 limit when base is a circle. 



.'. Area of a cylinder height X perimeter of base + 

 2 X base. 



Hence, if h = height, and a = radius of base. 



Area of cylinder 2 ir ah -)- 2 ira? > 2 v a (a + h). 



(17). To find ihe Area of the curved turf ace of a right 

 Cone. 



Let P A C D B be the cone, A C D B its circular base, 

 O the centre of the circle, then 1' () is at right angles to 



plane of the circle. In the circle 

 inscribe any regular polygon, 

 A CD, <kc., and join PA, 1'C, 

 PD, etc. Hisect CD in n, join 

 Pn. Then Pn is perpendicular 

 to CD, and .*. the area of tri- 

 angle PCD =$ CD, Pn. How 

 the line joining P with the bi- 

 section of any other side of the 

 polygon is equal to P n. 



Hence area 

 scribed in cone 



Fig. 24. 

 f 



of pyramid 



= P N X perimeter of polygon. 



Now, this is true whatever be the number of sides tho 

 polygon may have, and hence is true in the limit ; when 

 perimeter of polygon = circumference of circle, and P n 

 is drawn to a point in circumference, or is = slant side 

 of cone. 



.'. Area of curved surface of cone = slant side X 

 circumference of base. 



COR. If we suppose that the surface of a cone is 

 capable of being unwrapt, it is plain that its surface will 

 be a sector of a circle whose radius Fig. 25. 



is the slant side of cone, and base c 



of the same length as the cir- 

 cumference. It is plain (Men- 

 suration, Art. 9), that the area 

 of this sector is, as it should be, 

 the same as that of the surface g .CI 1^.6 

 of cone. 



DBF. The frustum of a cone 

 or pyramid is the portion cut off A 



by a plane parallel to the base ; " " 



thus, A B b a is a frustum of the cone CAB. 



(18). To find the Area of the Frustum of a Cone. 



Suppose a sector, OPQ, (Fig. 26) of a circle 

 described with radius O P 

 = CA (Fig. 25), and if its 

 base P Q = circumference of 

 AB, we have seen that the 

 area of P O Q is the same 

 as that of the cone CAB. 

 Take Oq Ca, and describe 

 the arc Oqp. Then as be- 

 fore, area of opq = area of 

 Ca6, and .'. the area of frus- 

 tum -f ?QPp> 



Let angle POQ = 0. 



.'. Area Opq . 



.'. Area OPQ - 



.'. Area qOPp = 



Fig 20. 



Bisect Q in t and draw arc tt? 

 Then '-i - tf and r-r 1 = $Q. 



.'. Area of frustum = rectangle between Qq and U' or 

 between slant side of cone and circumference of mean 

 section of frustum. 



(19). To find the Area of a portion of the surface of a 



Sflun. 

 Let A B be a quadrant of a cirle whose radius is A B 



