A

Sch. 1.-The trapezoids BF, CG, etc., composing the convex surface of a frustum of a regular pyramid, are equal to each other, for they are the differences between the equal triangles AEF, AFG, etc., and the equal triangles ABC, ACD, etc.

BR

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Sch. 2.-Since the area of the trapezoid BF is equal (Cor. 2, Theo. XV, Book II) to its mean breadth, LM multiplied by RS, which is the slant height of the frustum, and since the same is true of each of the other trapezoids, it follows that the convex surface of a frustum of a regular pyramid is equal to its slant height multiplied into the perimeter of a middle section between its two bases.

THEOREM XXIV.

Two triangular pyramids of equal bases and altitudes are equivalent.

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N

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B

Let the two pyramids have their bases in the same plane, and DC equal to BC. Conceive a plane to cut the two solids parallel to the plane of their bases, making the triangular sections FGH and LMN. Now, since the vertices R and S are, by hypothe

sis, equidistant above the plane of the bases, a third plane may pass through these two points parallel with the other two planes (Cor., Theo. II). Then, the lines HL and RS, being the intersections of two parallel planes by a third plane CRS, will be parallel (Def. 7,

Sec. XIV); hence, the triangles CRS, CHL, are similar (Cor., Theo. XI, Book II). In the same manner it may be shown that the triangles CRB, HRG, are similar; also CSD and LSM.

But CB is, by hypothesis, equal to CD; therefore, HG is equal to LM. In the same way, it may be shown that the other sides of the triangle FHG are respectively equal to the other sides of the triangle LMN; and these triangles are consequently equal throughout. Hence, at equal heights, the sections parallel to the bases are equal; and the two pyramids may be conceived to be applied to one another so as to coincide at all equal heights successively, from their bases to their vertices. They are, therefore, equivalent. That is, two triangular pyramids, etc.

THEOREM XXV.

A triangular pyramid is one third of a triangular prism of the same base and altitude.

Let ABCDEF be a triangular prism. Join AF, BF, and BD. Now, the pyramid B-DEF, cut off by the plane of the triangle BDF, is equivalent (Theo. XXIV) to the pyramid F-ABC, cut off by the plane of the triangle ABF; for they have equal bases, DEF and ABC (Def. 11, Sec. XVI), and the same altitude, namely, the altitude of the prism.

F

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one third, etc.

But the pyramid F-ABC is equivalent to the third pyramid B-ADF; for they have equal bases, ADF and FCA (Cor. 1, Theo. XX, Book I), and the same altitude, namely, the perpendicular distance of their common vertex B above the plane of their bases ADFC. Hence, the pyramid B-DEF is one third of the prism ABCDEF.

That is, a triangular pyramid is

Cor. 1.-Hence, the volume of a triangular pyramid is equal to one third of the product of its base by its altitude (Theo. XXI).

Cor. 2.-The volume of any pyramid whatever, is equal to one third the product of its base by its altitude; for, by dividing its base into triangles, and passing planes through the lines of division and the vertex, the pyramid can be divided into a number of triangular pyramids; and the sum of their volume will be equal to one third the product of the sum of their bases by their common altitude.

THEOREM XXVI.

The convex surface of a cone is equal to half the product of the circumference of the base by the slant height; and its volume is equal to one third the product of the area of the base by the altitude.

Let ABC be a cone, having a regular pyramid inscribed in it. If the number of sides of the polygon constituting the base of the pyramid be indefinitely increased, its perimeter will ultimately coincide with the circumference of the base of the cone. Then, the

slant height of the pyramid will be equal to the slant height of the cone, and the convex surface and solidity of the pyramid to the convex surface and solidity of the cone. But the convex surface of the pyramid will be equal to half the product of the perimeter of the base by the slant height (Theo. XXIII); and the volume of the pyramid will be equal to

one third the product of the area of the base by the altitude (Cor. 2, Theo. XXV).

Therefore, the convex surface of a cone, etc.

Sch. In the same manner, it may be shown that the convex surface of a frustum of a cone is equal to the product of the slant height into the circumference of a middle section between the two bases (Sch. 2, Theo. XXIII). And as this will hold true however small the upper base may be, it will hold true of the cone itself, which may be treated as a frustum whose upper base is nothing.

EXERCISES.

1. If the edge of a cube is x, what is the volume? The surface? The diagonal?

2. The altitude of a right pyramid is a, its base is an equilateral triangle described about a circle whose radius is r. Find the volume and surface.

3. The slant height of a cone is s, and the radius of base r. Find the volume and the entire surface.

4. Prove that the vertex of a regular pyramid (Def. 2, Sec. XVII) is in a line perpendicular to the base at its

center.

Ques. 1.-What is true of all the lateral edges of a regular pyramid?

E. G.-13.

Ques. 2.-What is the locus of all the points equally distant from every point in the circumference of a circle?

Ques. 3.-Why is every regular polygon capable of being inscribed in a circle?

5. The edges of a regular tetraedron are each 2e. Show that the surface is 4e2/3.

6. The square of a diagonal of a rectangular parallelopiped is equal to the sum of the squares of three edges meeting at a common vertex.

7. Two prisms, or pyramids, having equivalent bases, are to each other as their altitudes. Having equal altitudes, they are proportional to their bases.

8. The lateral surface of a pyramid or cone is greater than the base.

E

9. The portion of a tetraedron cut off by a plane parallel to any face is a tetraedron similar to the given one.

If the plane BCD is parallel to EFG, the tetraedron A-BCD is similar (Def. 6, Sec. XVI) to A-EFG (?).

10. Two pyramids are equal when the base and two adjacent sides of each are respectively equal. 11. The plane which bisects a diedral angle of a tetraedron divides the opposite edge in the ratio of the areas of the adjacent faces.

The plane SAE bisects the diedral AE, and so divides the edge OJ that EAO EAJ = OS : JS.

Let fall the perpendiculars JX and OY upon the plane SAE. They are, therefore, parallel, and (1) OY: JX = OS: JS (?).