Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

6. If a cone and a sphere be inscribed in a cylinder, their volumes are as 1, 2, and 3.

7. The volume of a sphere is to the volume of the inscribed

[blocks in formation]

8. The surface of an inscribed cube is equal to eight times the square of the radius of the sphere.

9. Draw on a spherical blackboard the polar triangle of a triangle with a right angle; with two right angles; with three right angles; with two obtuse angles.

10. A sphere may be inscribed in a regular polyedron.

11. The volume of a regular polyedron is equal to its surface multiplied by one third the radius of the inscribed sphere.

12. Compare the convex surfaces of a sphere and the circumscribed cylinder and cone, the cone having an angle of 60° at the vertex.

13. Compare the volumes of a sphere and its circumscribed cube, cylinder and cone, the cone having an angle of 60° at the vertex.

14. A square prism is inscribed in the cylinder of (IX. 26). Compare the volumes of sphere and prism.

15. Knowing the radii of two parallel sections of a sphere, and their distance from each other, to find the radius of the sphere.

16. If from a point in space we draw secants to a given sphere, the product of the distances from this point to the two points of intersection of each secant, with the sphere, is con

stant.

MENSURATION OF SOLIDS.

SECTION I.

1. To find the volume of a prism.

Multiply the area of the base by the altitude. (VIII. 9.)

2. To find the surface of a prism.

Multiply the perimeter of the base by the altitude, and add

the areas of the two ends. (VIII. 10.)

3. To find the volume of a pyramid.

Multiply the area of the base by one third the altitude. (VIII. 17.)

4. To find the surface of a pyramid.

Multiply the perimeter of the base by one half the slant height,

and add the area of the base.

(VIII. 11.)

5. To find the volume of a frustum of a pyramid.

Add together the areas of the two ends and a mean proportional between them; multiply the sum by one third the altitude. (VIII. 19.)

6. To find the surface of a frustum of a pyramid.

Add together the perimeters of the two ends; multiply the sum by one half the slant height; to this add the areas of the ends. (VIII. 11, Sch.)

7. To find the volume of a cylinder.

Multiply the area of the base by the altitude. (VIII. 20.)

8. To find the surface of a cylinder.

Multiply the perimeter of the base by the altitude, and add

the areas of the two ends. (VIII. 20.)

9. To find the volume of a cone.

Multiply the area of the base by one third the altitude. (VIII. 21.)

10. To find the surface of a cone.

Multiply the perimeter of the base by one half the slant height, and add the area of the base. (VIII. 21.)

11. To find the volume of a frustum of a cone.

Add together the squares of the radii of the ends and their product, and multiply the sum by, and by one third the altitude of the frustum. (VIII. 22, Sch.)

12. To find the surface of a frustum of a cone.

Add together the radii of the two ends, and multiply the sum by and by the slant height; to this add the areas of the ends. (VIII. 22, Sch.)

13. To find the surface of a regular polyedron.

I. Multiply the area of one of the faces by the number of faces.

II. Multiply the square of one of the edges by the surface of a similar polyedron whose edge is unity. (VIII. 23.)

14. To find the volume of a regular polyedron.

I. Multiply the surface by one sixth the perpendicular distance between the opposite faces, if the faces be opposite; if not, by one third the radius of the inscribed sphere.*

II. Multiply the cube of one of the edges by the volume of a similar polyedron whose edge is unity. (VIII. 18, Cor. 3.)

* This rule is proved by supposing the polyedron to be made up of pyramids whose common vertex is at the centre, and whose bases are the faces of the polyedron.

[blocks in formation]

15. To find the surface of a sphere.

I. Multiply the circumference of a great circle by its diameter. (IX. 19.)

II. Multiply the square of the radius by 4π. (IX. 19, Cor. 1.)

16. To find the volume of a sphere.

I. Multiply its surface by one third the radius. (IX. 24.)
II. Multiply the cube of the radius by . (IX. 24, Cor. 1.)
III. Multiply the cube of the diameter by ᅲ. (IX. 24, Cor. 1.)

17. To find the surface of a spherical zone.

Multiply the altitude of the zone by the circumference of a great circle of the sphere. (IX. 18.)

18. To find the volume of a spherical segment of one base. Apply the formula V=A2(R-A);

R being the radius of the sphere, and A the altitude of the segment. (IX. 25.)

19. To find the volume of a spherical segment of two bases. Subtract the volumes of two spherical segments of one base. (IX. 25, Sch.)

20. To find the volume of a spherical sector.

Multiply the surface of the zone which forms its base, by one third the radius of the sphere. (IX. 24, Sch.)

21. To find the surface of a cylindrical ring.*

Add the thickness of the ring to the inner diameter; multiply this by the thickness of the ring and by π2.

22. To find the volume of a cylindrical ring.

Add the thickness of the ring to the inner diameter; multiply the sum by the square of one half the thickness and by π2.†

23. To find the volume of an irregular solid. Immerse the solid in water, and measure the rise of the water. If now we know the size of the vessel, we may calculate the volume of displaced water, and consequently the volume of the solid.‡

EXAMPLES.

1. What is the lateral surface of a hexagonal prism, each of the sides of its base being 2 feet and its altitude 8 feet? Ans. 96 sq. ft.

2. What are the volume and lateral surface of a cylinder circumscribing the prism of last example? Αης. 32π; 32π.

3. The pyramid of Cheops is 764 feet square at the base and 480 feet in perpendicular height. What is its volume? Ans. 93391360 cu. ft.

4. If the pyramid of last example were cut off by a plane parallel to its base at a distance of 320 feet from the ground, what would be the volume of the upper portion? Ans. 2 of the whole.

* A cylindrical ring is formed by bending a cylinder so that its axis is a circle. If R be the inner radius of the ring, and r the radius of the cylinder, the length of the axis is 2(R+r), and this, multiplied by the perimeter of the cylinder, 2πr, will be the surface, or,

S=2π(R+r)×2nr=π2(2R+2r)2r, which agrees with the rule.

V=2π(R+r) × πr2 = π2(2R+2r)r2.

‡ This is only an approximate value, on account of the porosity of most solids.

« ΠροηγούμενηΣυνέχεια »