If the whole figure thus formed revolve about DC, as a fixed axis, the figure DCBE will generate a cylinder, (Def. 7), the ACDE will generate a cone, (Def. 8), and the quadrant will generate a hemisphere; now these figures may be conceived to be made up of an infinite number of lamina, represented in thickness by the line GI or the line KM, the solids generated by the several parts of the line GI will be as the squares of their generating lines; but the generating lines are in the cone GN, in the circle GH, and in the cylinder GI. Now the square of GI is = the sum of the squares of GH and GN, for GN is equal to GC, and the squares of CG and GH are equal to the square of the radius = the square of GI. .. the lamina thus added to the cone and sphere are together equal to the lamina added to the cylinder, and the same is evidently true at any other point. Hence the cone and hemisphere together are equal to the cylinder; but the cone was shown (Prop. 104) to be one-third of the cylinder, therefore the hemisphere is two-thirds of the cylinder; .. the whole sphere will be equal to two-thirds of a cylinder, having its altitude double of the line DC, that is, equal to the diameter of the sphere; and it is evident, that the diameter of the base being the double of CD, is also equal to the diameter of the sphere.

Cor. 1. The portion of the sphere, together with the portion of the cone lying between the lines CB and GI, are together equal to the portion of the cylinder lying between the same lines.

Cor. 2. Any portion of the sphere, together with the corresponding portion of the cone, is equal to the corresponding portion of the cylinder.

END OF SOLID GEOMETRY.

GEOMETRICAL EXERCISES.

1. If a straight line bisect another at right angles, every point of the first line will be equally distant from the two extremities of the second line.

2. If straight lines be drawn, bisecting two sides of a triangle at right angles, and from the point of their intersection a perpendicular be drawn to the third side, it will bisect the third side.

3. If two angles of a triangle be bisected, and from the point where the bisecting lines cut one another, a straight line be drawn to the third angle, it will bisect the third angle.

4. The difference of any two sides of a triangle is less than the third side.

5. The sum of two sides of a triangle is greater than twice the straight line drawn from the vertex to the middle of the base.

6. If the opposite sides of a quadrilateral figure be equal, the figure is a parallelogram.

7. If the opposite angles of a quadrilateral figure be equal, the figure is a parallelogram.

8. If a straight line bisect the diagonal of a parallelogram, it will bisect the parallelogram.

9. The diagonals of a rhombus bisect one another at right angles. 10. If a straight line bisect two sides of a triangle it will be parallel to the third side, and equal to the half of it, and will cut off a triangle equal to one-fourth part of the original triangle.

11. The diagonals of a right angled parallelogram are equal.

12. From a given point between two indefinite straight lines given in position, but not parallel, to draw a line which shall be terminated by the given lines, and bisected in the given point.

13. If the sides of a quadrilateral figure be bisected, and the adjacent points of bisection joined, the figure so formed will be a pa rallelogram, equal to half of the quadrilateral figure.

14. If a point be taken either within or without a rectangle, and straight lines drawn from it to the angular points, the sum of the squares of those drawn to the extremities of one diagonal will be equal to the sum of the squares of those drawn to the extremities of the other.

15. In any quadrilateral figure, the sum of the squares of the diagonals, together with four times the square of the line joining their middle points, is equal to the sum of the squares of the sides.

16. If the vertical angle of a triangle be two-thirds of two right angles, the square of the base will be equal to the sum of the squares of the side, increased by the rectangle contained by the sides; and if the vertical angle be two-thirds of one right angle, the square of the base will be equal to the sum of the squares of the sides, diminished by the rectangle contained by the sides.

17. To bisect a triangle by a line drawn from a given point in one of its sides.

18. A perpendicular drawn from an angle of an equilateral triangle to the opposite side, is equal to three times the radius of the inscribed circle.

19. If perpendiculars be drawn from the extremities of a diameter to any chord in the circle, they will cut off equal segments.

20. If, from the extremities of any chord in a circle, straight lines be drawn to any point in the diameter to which it is parallel, the sum of their squares will be equal to the sum of the squares of the segments of the diameter.

21. If, from two angular points of any triangle, straight lines be drawn, to bisect the opposite sides, they will divide each other into segments, having the ratio of two to one, and, if a line be drawn through the third angle and their point of intersection, it will bisect the third side, and divide the triangle into six equal triangles.

22. If two triangles have two angles equal, and other two angles together equal to two right angles, the sides about the remaining angles will be proportional.

23. If a circle be described on the radius of another circle, any straight line drawn from the point where they meet to the outer circumference is bisected by the interior one.

24. The rectangle contained by two sides of a triangle, is equal to the rectangle contained by the perpendicular from the contained angle upon the third side, and the diameter of the circumscribing circle.

25. If a straight line be drawn bisecting the vertical angle of a triangle, the rectangle contained by the two sides will be equal to the square of the bisecting line, together with the rectangle contained by the segments of the base.

26. A straight line drawn from the vertex of a triangle to meet the base, divides a parallel to the base in the same ratio as the base.

27. If the three sides of one triangle be perpendicular to the three sides of another, the two triangles are similar.

28. If from any point in the diameter of a circle produced, a tangent be drawn, a perpendicular from the point of contact to the diameter will divide it into segments, which have the same ratio which the distances of the point without the circle from each extremity of the diameter have to each other.

29. A straight line drawn from the vertex of an equilateral triangle, inscribed in a circle to any point in the opposite circumference, is equal to the two lines together which are drawn from the extremities of the base to the same point.

30. If the interior and exterior vertical angles of a triangle be bisected by straight lines, which cut the base, and the base produced, they will divide it into three segments, such that the rectangle contained by the whole base thus produced, and the middle segment, shall be equal to the rectangle contained by the two extreme segments. (A line divided in this manner is said to be divided harmonically.)

31. The side of a regular decagon inscribed in a circle, is equal to the greater segment of the radius divided medially; the side of a regular hexagon is equal to the radius; and the side square of a regular pentagon is equal to the side square of a regular hexagon, together with the side square of a regular decagon.

L

PRACTICAL GEOMETRY.

PROBLEM I.

To divide a given straight line AB into two equal parts.

From the centres A and B, with the same radius half of AB, describe arcs intersecting in D and E, and draw the DCE, it will bisect AB in the point C.

PROBLEM II.

To divide a given angle ABC into two equal parts.

From the centre B with any radius, describe the arc AC, and from the centres A and C with the same radius, describe arcs intersecting in D, and join BD; the angle ABC will be bisected by the BD.

PROBLEM III.

To trisect a right angle ABC'; that is, to divide it into three equal parts.

From the centre B with any radius, describe the arc AC; and from the centre C with the same radius, cut the arc in D, and from the centre A with the same radius, cut the arc in E, and join BD and BE, and they will trisect the angle as required.

PROBLEM IV.

A

To erect a perpendicular from a given point A, in a given line AC.

CASE I. When the point is near the middle of the line.

On each side of the point A take any two equal distances, Am, An. From the centres m, n, with any radius greater than Am or An, describe two arcs intersect

B

m

ing in D. Through A and D draw the straight line AD, and it will be the perpendicular required.

CASE II. When the point is near the end of the line. From the centre A with any ra

the centre m with the same radius,
turn the compasses twice over on
the arc at n and r. Again, from
the centres n and r with the same B-
radius, describe arcs intersecting in

DX

D. Then draw AD, and it will be the perpendicular required.

Another Method.

From any point n as a centre, with a radius=nA, describe an arc, not less than a semicircle, cutting the line in m and A. Through m and n draw the diameter mnr, cutting the arc in r, and Bjoin Ar, and it will be the perpendicular required.

m

Cor. If the point from which the perpendicular is to be drawn were the extremity of the line C, it would only be necessary to take nC as a radius instead of nA, and the other parts of the construction would be the same.

With a marquois square, which is a right-angled triangle cut out of ivory or wood, apply the right angle to the point A, and make one side coincide with AB, a line drawn along the other side will be the perpendicular required.

PROBLEM V.

From a given point C, without a straight line AB, to let fall a perpendicular.