MENSURATION. 



*49 



And Trap. CB b' D : 

 '''"' Therefore the area of the space 'A a afbf B is nearly 



~ V "" = \ (AB+4a+a'6')B4. 



In like manner, it may be shewn that the space a'4'4" 

 a"' is nearly 



and that the space '"b" PQ 



= j(o-'6-4. 4 of- 6^+ PQ) B 6, 



and so on ; hence the area of the whole figure, which 

 U made up of these, is 



I + PQ 

 . 4 (a h + 



f 13. 



^ I 



a* was to be demonstrated. 



Ex. To find the area of the space ABC, supposing 

 it to be a quadrant of a circle, the radius of which is 

 = 1. 



Let the sector BCE be one- third of the quadrant, 

 draw ED perpendicular to AC, then CD=cos. SO>=J 

 Divide CD into four equal parti, and draw the 

 perpendiculars rt, pa, mm. 



Because CA=I, therefore CD= 4, Cr=J, C/>=', 

 C*=; ; benceDE=v'(I- : H)=tV3, and in like man- 

 ner, rtslStt, pg=;v'l5, M *=li/6s. Therefore, 



4- L=rl4-i 

 4E= 



USB of to- 



Th* u in 

 Multiply by 



The product = .4783 



Subtract the triangle CDE = .9163 



ThesactorCEB = .'is 

 The triple of which is the quad. ABC=.78J4 



SF.< no* n. 



mxst-RATioM or SOLIDS. 



PROBLEM I. 



To find the surface of a right prism or cylinder. 

 Sou. Multiply the perimeter of the end by the length 

 or height of the solid and the product will be the sur- 

 n.'. I i- face of all its sides. To this, when the whole surface of 

 the prism is required, the areas of the and* must be added. 



If the plane turfaces which form the sides of an up- 

 right pram were extended into one plane, it is mani- 

 feat, that the surface thus formed would be a rectangle, 

 having one of its sides equal to the height of the prism, 

 and its other aide equal to the perimeter of the end : 

 fiance, the truth of the rule U manifest in the case of 

 any right prism A cylinder may be regarded as the 

 limit of all prisms which can be inscribed in, or cir- 

 cumscribed about its base, therefore, hs surface will be 

 the limit of their aurfacea; now, the expression for the 

 limit is evidently the product of the circular base by the 

 height 



1 . How many square yards are in the surface of 

 the wall* of a room of any prismatic form, whose hejgfat 

 is 10 feet, and circumference 58 feet ? 



10 X 58 = MO sq. feet = 6*J sq. yards the answer. 

 VOL. xrv. rAarr i. 



Ex. 2. What is the convex surface of a cylindrical Mnsun- 

 pillar 12 feet long, and one foot in diameter ? 



The circumference of the base, is 3-1416 feet, (Prob. 

 7. Sect 1.) And the surface of the pillar 3.H16 X 

 12 = 37.699* sq. feet 



PaOBLKM II. 



To find the surface of a ripht pyramid or cone. 



Ht LE. Multiply the circumference of the base by the Pig. 16, IT. 

 slant height, and half the product will be the surface 

 of the side* ; to which the area of the end may be added 

 when the whole surface is required. 



The truth of the rule will be evident, if it be con- 

 sidered that the faces of the pyramid are equal triangles, 

 having the slant ride of the pyramid for their altitude, 

 and that a cone may be considered as a pyramid having 

 an infinite number of sides. 



Ex. 1. The slant height of a triangular pyramid 

 A BCD is 20 feet, and each side of the base 3 feet: Pig. U. 

 What is its upper surface ? 



Ex.*. _ 

 slant height 

 I ii-t-t. 



= 90 feet the surface. 



the convex surface of a cone, the 

 i feet, and the diameter of the base 



3 X 3-1416 = 9-4348 the circumference, 

 X 9-4*48 x 5 = 83-56* sq. feet the surface. 



PROBLEM III. 



of the two ends, Pig. 19. 

 height, and take 



To find the surface of the frustum of a right pyra- 

 mid or cone, that m the lower part, when the top is cut 

 off by a plane parallel to the base. 



RVLB. Add together the perimetei 

 and multiply the sum by the slant 

 half the product for the surface. 



For the surface is equivalent to a trapnoid, whose 

 parallel sidea art the perimeters of the ends, and height 

 is the slant surface. 



Ex. Find the surface of AC, the frustum of a square 

 pyramid, the slant bright M ln-ing 10 feet; each aide 

 of the greater end AC 3 feet 4 inches, and each side of 

 the leaser and EG X feet 2 inches. 



Here 5} x 4 = 13} the perim. of greater end. 



SJX4= 8f the perim. of leaser end 

 Their sum = K feet 



And MX 10 



- 



=110 feet, the answer. 



PKOBLCM IV. 



To find the solid content of any prism or cylinder. 



RULE. Multiply the arm of the base or end by the p ig . 14 j s 

 perpendicular height, ami the product will be the solid 

 content. 



Thi rule follow^ from Geou. PAT.. Sects. Prop. 

 11. and Sect 3. I'rop. 8. 



I . The sides of the bate of a triangular prism 

 are 3, 4 and 5 feet, and its height is 12 feet : What is its 

 solid content ? 



By rule 2 |of Problem 2. Sect. 1. the area of the 

 btsc ii 



X2x3)=6q. feet 

 Hence, the solid content =6x 12=72 cubic feet 

 Ex. 8. The Winchester bushel is a cylinder 18J 



inches in diameter, and 8 inches deep : What is its so- 



lid content? 



