A Pyramid is a solid bounded by a polygon called the base, and by triangles called lateral faces, which all terminate in one point called the vertex.

A pyramid is triangular, hexagonal, etc., according as the base is a triangle, a hexagon, etc.

A Right Pyramid is a pyramid all of whose lateral faces are isosceles triangles. If the base of a right pyramid is a regular polygon, the vertex is in the perpendicular erected at the centre of the base.

A Cone is a solid bounded by a circle called the base, and by a curved surface called the lateral surface, which terminates in a point called the vertex.

A Right Cone is a cone whose vertex is in the perpendicular erected at the centre of the base. This perpendicular is called the axis of the right cone.

The height of a pyramid or cone is the distance from the vertex to the base. The slant height of a right cone is the distance from the vertex to the circumference of the base.

The lateral surface of a right cone is found by multiplying the circumference of the base by half the slant height.

Volume of a pyramid or cone =

base height.

(All the following questions refer to the right pyramid and right cone.)

Find the volumes of the following pyramids : 432. Area of base 64 sq. in., height 1 ft. 9 in.

433. Base an equilateral triangle, side 4 ft., height 15 ft. 434. Base a regular hexagon, side 6 ft., height 30 ft. 435. Base a regular octagon, side 24 ft., height 74 yds.

Find the volume of a square pyramid, having given: 436. Side of base 3 ft. 4 in., height 9 ft.

437. Diagonal of base 78 yds., height 30 yds. 438. Altitude of one face 18 in., height 13 in. 439. Lateral edge 29 in., height 17 in.

Find the whole surface of a square pyramid, having given: 440. Side of base 8 ft., altitude of each face 20 ft.

441. Side of base 26 ft., altitude of each face 84 ft.

Find the volume and also the curved surface of the following cones :

442. Radius of base 7 in., height 3 ft. 6 in.

443. Diameter of base 7 ft., height 10 ft.

444. Radius of base 3 ft. 6 in., slant height 5 ft. 10 in. 445. Radius of base 12 ft., slant height 37 ft. 446. Circumference of base 44 ft., slant height 25 ft. 447. Find the cost of covering a conical spire measuring 40 ft. round the base, and whose slant height is 30 ft., with lead of an inch thick, if 1 cu. in. of lead weighs 6 oz., and lead is worth 12 cents per pound.

448. How much canvas is needed to make a conical tent 8 ft. high, and 7 ft. in diameter at the base?

449. If 132 sq. ft. of canvas are formed into a conical tent whose slant height is 12 ft., find how much ground this tent will cover.

450. What length of canvas of a yard wide is required to make a conical tent 12 ft. in diameter and 8 ft. high?

451. A right triangle whose sides are 3 in., 4 in., and 5 in. in length, is made to describe a cone by turning round on the side 4 in. as an axis. Find the curved surface, and the volume of the cone thus generated.

A Frustum of a pyramid or cone is the part which remains after the top has been cut off by a plane parallel to the base. The base and the section made by the cutting plane are called the bases of the frustum.

The height of a frustum is the distance between the bases. The slant height of the frustum of a right cone is the distance from a point in the circumference of one base to the circumference of the other base.

The lateral surface of the frustum of a right pyramid is composed of trapezoids. The lateral surface of the frustum of a right cone is found by multiplying the sum of the circumferences of the bases by half the slant height.

To find the volume of a frustum, add together the areas of its bases and the square root of their product; then multiply the sum by one-third of the height.

Find the volumes of the following frustums:

452. Square pyramid, sides of bases 21 and 15 yds., height 84 yds.

453. Square pyramid, sides of bases 45 and 25 yds., height 96 yds.

454. Square pyramid, sides of bases 36 and 18 yds., height 120 yds.

455. Right cone, radii of bases 86 and 68 yds., height 57 yds.

456. Right cone, radii of bases 8 ft. and 6 ft., height 15 ft.

457. Right cone, radii of bases 5 ft. and 4 ft., height 6 ft.

458. How many square feet of tin will be required to make a funnel with the radii of top and bottom 14 in. and 7 in. respectively, and the height 24 in. ?

459. A church spire has the shape of a frustum of a regular hexagonal pyramid; each side of the base is 5 ft., and of the top 2 ft; the altitude of each trapezoidal face is 20 ft. How many square feet of tin roofing are required to cover the lateral faces and the top?

460. Find the expense of polishing the curved surface of a marble column in the shape of a frustum of a right cone, slant height 12 ft., radii of bases 3 ft. 6 in. and 2 ft. 4 in., at 60 cents per square foot.

461. A round stick of timber is 20 ft. long, 3 ft. in diameter at one end, 2.6 ft. at the other. How many cubic feet does it contain?

462. A bucket is 16 in. deep, 18 in. wide at the top, and 12 in. wide at the bottom. How many gallons of water will it hold, reckoning 7 gals. to the cubic foot?

Find its

463. The mast of a ship is 51 ft. high, and the circumferences of its ends are 5 ft. 6 in. and 1 ft. 10 in. value at 60 cents per cubic foot.

464. Find the volume of the frustum of a square pyramid, the sides of its bases being 40 ft. and 16 ft., and the altitude of a lateral face being 20 ft.

465. The chimney of a factory has the shape of the frustum of a square pyramid; its height is 180 ft., and the sides of its upper and lower bases are 10 ft. and 16 ft. respectively; the section of the flue is throughout the entire length a square whose side is 7 ft. How many cubic feet of brick does the chimney contain?

A Sphere is a solid bounded by a curved surface all points of which are equally distant from a point within called the centre.

A straight line drawn from the centre to the surface is called a radius, and a straight line drawn through the centre and terminated each way by the surface is called a diameter.

The circumscribing cylinder of a sphere is a cylinder whose height and the diameter of whose base are both equal to the diameter of the sphere (see Fig.).

It can be proved that the surface of a sphere is just equal to twothirds the total surface of the circumscribing cylinder, and that the volume of a sphere is just equal to two-thirds the volume of the circumscribing cylinder. Hence it follows that (assuming to be equal to 2,2),

Find the surface of a sphere, if the radius is:

Find the volume of a sphere, if the diameter is: