Prisms.

Definition. Figures the sides of which are parallelograms, and the ends equal and parallel.

NOTE.-When the ends are triangles, they are called triangular prisms; when they are square, they are called square or right prisms; when they are pentagons, pentagonal prisms, &c., &c.

To ascertain the Surface of a Prism (Figs. 38 and 39).

RULE. Find the areas of the ends and sides as by the rules for the mensuration of squares, triangles, &c., and add them together; the sum will be the surface of the figure.

Or, 2 a+a=s, where a represents the area of the ends, and a' the area of the sides.

EXAMPLE.-The side a b, Fig. 38, of a square prism is 12 inches, and the length, b c, 30; what is the surface?

12x12=144-area of one end.
144x2=288=area of both ends.
12x30=360-area of one side.

360x4=1440-area of four sides.

Then, 288+1440=1728 inches, the surface required. Ex. 2. What is the surface of a triangular prism, the sides a b, bc, and c a, Fig. 39, being each 12 inches, and the length, cd, 30 inches?

=

12÷2=6, and V62-122-10.3923 width of prism. Hence, 10.3923 × 12÷2=62.3538-area of each end.

Then,

12 × 30=360=area of one side. 62.3538×2=124.7076=area of ends, 360 × 3 1080= =area of sides.

and 124.7076+1080-1204.7076 inches surface required. Ex. 3. What is the surface of a rhomboidal prism, the depth of it being 5 feet 9 inches, the width 7 feet, and the length 10 feet? Ans. 402.5 feet.

Centre of Gravity.

When the ends are parallelograms, it is

in their geometrical centre.

When the ends are triangles, trapeziums, etc., it is in the middle of their length at the same distance from the base as that of the triangle or trapezoid which is a section of them.

Wedge.

Definition. A wedge is a prolate triangular prism, and its surface is found by the rule for that of a right prism.

EXAMPLE. The back of a wedge, a b c d, Fig. 40, is 20 by 2 inches, and its end, e ƒ, 20 by 2 inches; what is its surface? 202+2÷1401=sum of the squares of half the base, a f, and the height, e f, of the triangle, ef a.

-2

√/401=20.025 square root of above sum-length of e a. Then, 20.025 × 20×2=801 area of sides.

And 20×2=40=area of back, and 20×2÷2×2=40=area of ends.

Hence, 801+40+40=881-surface required.

Centre of Gravity. See rule for prisms.

Prismoids.

Definition. Figures alike to a prism, but having only one pair of their sides parallel.

To ascertain the Surface of a Prismoid (Fig. 41).

RULE. Find the area of the ends and sides as by the rules for squares, triangles, &c., and add them together.

EXAMPLE. The ends of a prismoid, e f g h and a b c d, Fig. 41, are 10 and 8 inches square, and its slant height 25; what is its surface?

10+8
2

Then,

×25=225, and 225 ×4=1000=area of sides.

100+64+1000=1164-surface required.

Ex. 2. The ends of a prismoid are 15 and 12 inches square, and its slant height 40; what is its surface?

Ans. 2529 inches.

Ex. 3. The ends of a prismoid are 12x16 and 14x18 inches, and its vertical height is 33; what is its surface?

Ans. 2424 inches.

Centre of Gravity. Is at the same distance from its base as that of the trapezoid or trapezium which is a section of it.

Ungulas.

Definition. Cylindrical ungulas are frustrums of cylinders. Conical ungulas are frustrums of cones.

To ascertain the Curved Surface of an Ungula, Figs. 42, 43, 44, and 45.

1. When the Section is parallel to the Axis of the Cylinder, Fig. 42.

RULE.-Multiply the length, a b c, of the arc line of one end by the height, b d, and the product will be the curved surface required.

Or, cxh=s, where c represents length of arc line.

Fig. 42. a

b

d

EXAMPLE.-The diameter of a cylinder from which an ungula is cut is 10 inches, its length 50, and the versed sine or depth of the ungula is 5 inches; what is the curved surface of it?

10÷2 5 radius of cylinder.

Hence the radius and versed sine are equal; the arc line, therefore, of the ungula is one half the circumference of the cylinder, which is 31.416÷2=15.708,

and

15.708×50=785.400 inches, Ans.

Ex. 2. The base line of the section of a cylindrical ungula is 48 inches, the height or versed sine of the arc is 20, and * For mensuration of conical ungulas, see Conic Sections, p. 253.

the length of the ungula is 20.5 feet; what is its curved surface? Ans. 109.8388 feet.

2. When the Section passes obliquely through the opposite Sides of the Cylinder, Fig. 43.

RULE.-Multiply the circumference of the base of the cylinder by half the sum of the greatest and least heights, d b and e a, of the ungula, and the product will give the curved surface required.

EXAMPLE.-The diameter of a cylindrical ungula is 10 inches, and the greater and less heights are 25 and 15 inches; what is its surface?

Hence,

10 diameter 31.416 circumference.

25+15=40, and 40÷2=20. 31.416×20=62.8320 inches, Ans.

Ex. 2. The circumference of an ungula is 60.75 inches, and the mean height of it 13 feet; what is its surface?

Ans. 65.8125 feet.

3. When the Section passes through the Base of the Cylinder and one of its Sides, and the Versed Sine does not exceed the Sine, Fig. 44. RULE.-Multiply the sine, a d, of half the arc, d 9, of the base, d gf, by the diameter, e g, of the cylinder, and from this product subtract the product* of the arc and cosine, a o. Mul* When the cosine is 0, this product is 0.