Mensuration for beginners [With] Answers

three sides of the base are respectively 11.2 ft., 1·5 ft., and 11.3 ft.; how many c. ft. are there in its volume?

(9) A triangular pyramid has each side of the base 5 ft.; the height is 30 ft.; what is the solid content?

[Find the area of the base by Art. 94.]

(10) The altitude of an upright pyramid whose base is an equilateral triangle is 16 ft., and each side of the base is 27 in.; required the solid content in c. ft.

(11) Required the content of a pentagonal pyramid of which each side of the base is 2 ft. 6 in., and height 12 ft.

[Find the area of the base by Art. 94.]

(12) Required the solidity of an hexagonal pyramid, each of the equal sides of the base being 30 ft., and the perpendicular height 20 ft.

(13) A vessel in the form of an hexagonal pyramid is 9 ft. deep, and each side of its base is 5 ft.; how many c. ft. of water will it hold?

(14) Find the weight, in tons, etc., of an hexagonal pyramid of stone 18 ft. high, each side of the base being 18 in., and a c. ft. of stone weighing 250 lbs.

(15) Find the weight of an hexagonal pyramid of marble, each side of the base of which is 5 ft., and the vertical height 18 ft., the weight of the marble being 180 lbs. per c. ft.

(16) Each side of the base of an upright heptagonal pyramid is 10 in., and the altitude is 8 ft.; required the solid content. (17) The spire of a parish church in Lincolnshire is a hollow octagonal pyramid, each side of the base being 4 ft., and its perpendicular height 45 ft.; also each side of the cavity or hollow part at the base is 3 ft., and its perpendicular height 36 ft. How many solid feet of stone are there in the spire?

[From the volume of a pyramid whose height is 45 ft., and each side of the base 4 ft., subtract the volume of a pyramid whose height is 36 ft. and each side of the base 3 ft.; the remainder will be the solid content of the hollow pyramid.]

(18) A monument of Caen stone in a certain cemetery consists of a square pyramid standing upon a square prism. Each side of the base of the pyramid and of the prism is 3 ft. 3 in.; the height of the whole monument is 18 ft., and of the pyramid 6 ft. Find the value of the whole, at 3s. per c. ft.

[The height of the square prism will be 18 ft. -6 ft.=12 ft.]

164. The whole surface of a right pyramid consists of the area of the base and the areas of the triangular faces.

If the base be a regular polygon and the pyramid

upright, the triangular faces will be all equal; the perpendicular distance of the vertex from the base of any of these triangles will be the slant height of the pyramid. The area of each triangle will therefore be obtained by multiplying its base by half this slant height (Art. 66), so that the sum of the areas will be found by multiplying the perimeter of the base of the pyramid by half the slant height; this product added to the area of the base will give the whole surface.

If the perpendicular height of the pyramid be given, the slant height may be found by Art. 71. (See also worked example 2 in Ex. LXX.)

165. To find the whole surface of a right pyramid.

RULE. Multiply the perimeter of the base by half the slant height, and to the product add the area of the base.

[To find the volume of a pyramid it is necessary to know the perpendicular height; but to find the surface we require to know the slant height.]

Ex. 1. Find the area of the whole surface of a pyramid on a square base; each side of the base is 27 yds., and the slant height is 48 yds.

Perimeter of base = 4 × 27=108 yds.

Area of triangular faces=× 108 × 48=2592 sq. yds.
Area of base-27 x 27-729 sq. yds.

.. Whole surface=2592+729-3321 sq. yds. Ex. 2. A square pyramid is 20 yds. high, and 30 yds. along each side of the base; find its whole surface.

Here EF (see fig. Art. 162)=20 yds., and FG=15 yds., being half the length of a side of the base; hence (Art. 71) the square of E G=400+225=625, and E G=the square root of 625-25 ft.

Perimeter of base = 4 × 30=120 yds.

Area of triangular faces=120 × 25=1500 sq. yds.
Area of base =30x30 900 sq. yds.

Whole surface=1500+900=2400 sq. yds.

Ex. 3. Find the area of the whole surface of a pyramid on a square base; each side of the base is 16 ft., and each of the other edges is 17 ft.

Here, the three sides of each triangular face being 16 ft., 17 ft., and 17 ft. respectively, the area will be found, by Art. 69, to be 120 sq. ft.

Areas of triangular faces 4× 120=480 sq. ft.

(1) Find the whole surface of a square pyramid, each side of the base of which is 12 ft., and its slant height 15 ft.

(2) A pyramid stands on a square base which is 10 ft. long; its slant height is 12 ft. 6 in.; find the whole surface.

(3) In a square pyramid the side of the base is 5 ft. 8 in., and the slant height 8 ft.; calculate the whole surface.

(4) The perpendicular height of a square pyramid is 12 ft., and each side of the base is 10 ft. long; find the whole surface.

[See worked example 2 above.]

(5) The perpendicular height of a square pyramid is 24 ft., each side of the base being 14 ft.; what is the whole surface? (6) A square tower, 20 ft. on each side, has a pyramidal roof whose perpendicular height is 24 ft., covered with lead, at 1s. 6d. per sq. ft.; find the cost.

(7) A square pyramidal stone whose perpendicular height is 16 ft., and each side of the base 24 ft., is to be sold, at 2s. per c. ft.; the polishing of its triangular faces will cost 8d. per sq. ft.; required the whole expense when finished.

(8) A pyramid stands on a square base which is 8 ft. long, and each of the four faces which meet at the vertex is an equilateral triangle; find the area of the whole surface.

[The area of each triangular face may be found by Art. 91.]

(9) Find the area of the whole surface of a pyramid on a square base; each side of the base is 4 ft. 6 in., and each of the other edges is 10 ft. 3 in.

[In this case the three sides of each triangular face are 4 ft. 6 in., 10 ft. 3 in., and 10 ft. 3 in. respectively; the area will be found by Art. 69. See worked example 3 above.]

(10) The slant height of a triangular pyramid is 10 ft., and each side of the base is 4 ft.; what is its whole surface?

(11) The four faces of a triangular pyramid are equilateral triangles, the edge of each being 6 ft. ; find the area of the who surface.

(12) What is the area of the side faces of an hexagonal pyramid, each side of its base being 20 ft., and slant height 24 ft.?

(13) The slant height of an hexagonal pyramid is 16 ft., and each side of its base 5 ft.; find the cost of tooling the side faces, at 1s. 6d. per sq. ft.

(14) The base of a pyramid is a regular octagon, each side being 18 ft., and each of the other edges of the pyramid 41 ft.; find the upright surface.

THE RIGHT CONE.

166. A Cone is a solid of which the base is a circle, and whose curved surface tapers to a point, called, as in the pyramid, the vertex of the solid.

Thus A B C represents a cone, of which A is the vertex.

The straight line A D drawn from the vertex A to the centre D of the base is called the axis of the cone.

When the axis is perpendicular to the base of a cone, it is said to be a right cone. The cones mentioned in this book are all right cones.

A D is the perpendicular height, E and A B the slant height.

Suppose a cylindrical vessel ABCD to be filled with water, and to contain three pints. If now a right cone of the same base and height be carefully dropped into it, as in the case of the pyramid one-third of the contents will be found to overflow, thus showing that the volume of a right cone is equal to onethird of the volume of a cylinder whose base and height are the samc. Hence A the rule.

167. To find the volume of a right cone,

RULE. Multiply the area of the base by the perpendicular height, and one-third of the product will be the volume.

Here, the three sides of each triangular face being 16 ft., 17 ft., and 17 ft. respectively, the area will be found, by Art. 69, to be 120 sq. ft.

Areas of triangular faces-4 × 120=480 sq. ft.