of its base by one-third of its altitude, that is, by one third the radius of the sphere. Hence, the sum of their solidities will be equal to the sum of their bases multiplied by one-third of the radius, or onesixth of the diameter.

That is, the solidity of a sphere is equal, etc.

Cor. Since the surface of a sphere is equal to 3.14159×D2, it follows that its solidity is equal to D

3.14159XD2X- which, by reduction, becomes .5236×

" 6

D3. That is, the solidity of a sphere is equal to the cube of its diameter multiplied by .5236.

SUPPLEMENT.

SEC. XVIII.-MISCELLANEOUS EXAMPLES.

1. What is the side of a square inscribed in a circle whose diameter is 5 feet?

2. Prove that the side of a square is to its diagonal as 1 to the square root of 2.

3. Prove that the area of a square described about a circle is double the area of a square inscribed in the same circle.

4. What is the altitude of an equilateral triangle whose side is 12 feet?

5. If the altitude of a cone is 9 feet, and the diameter of its base 5 feet, 4 inches, what is its convex surface? Its solidity? 78159 12,56

6. What is the surface of a sphere whose diameter is 7 feet? Its solidity? 53.957 1958 r

7. If the altitude of a prism is 5 feet, and the area of its base 18 square feet, what is its solidity???

8. What is the solidity of a cube whose edge is 6 feet? Its surface? Level 144

9. What is the lateral surface of a regular pyramid whose slant hight is 15 feet, and the base 30 feet square?

Evans' Geometry:-9

(97)

10. What is the solidity of a pyramid whose altitude is 72, and the sides of whose base are 24, 24, and 40?

Its

11. If the altitude of a cylinder is 7, and the 'radius of its base 2, what is its whole surface? solidity? 21.972

(87.964521

12. Prove that a circle described on the hypotenuse of a right-angled triangle as diameter, is equivalent to the sum of the circles described on the other two sides.

13. Prove that if a perpendicular be let fall from the right angle of a right-angled triangle on the hypotenuse, either of the two sides containing the right angle will be a mean proportional between the hypotenuse and the adjacent part of the hypotenuse.

14. If from the top of a mountain 2.006 miles high a vessel can be seen in the horizon 126 miles off, what is the diameter of the earth? (Theo. XIII, B. II).

15. If the earth's diameter be 7912 miles, what is its circumference? Its surface? Its solidity? 662 16 16. If the circumference of the moon be 6783 miles, 217916 17 4 4 6 G what is its diameter? 216.40

17. Prove that the solidities, S and s, of any two spheres are to each other as the cubes of their diameters, D and d.

18. The diameter of the sun is 112 times that of the earth; what are the relative solidities of the two bodies?

SEC. XIX.-APPLICATIONS OF ALGEBRA.

FOR those acquainted with the elements of both branches, a few examples of the application of Algebra to Geometry are subjoined. The method is specially adapted to the solution of that class of problems where certain parts of a figure are given, from which to determine others; but it may also be used for abbreviating long demonstrations of theorems.

1. If a perpendicular be let fall from the vertex of a triangle upon its base, the sum of the parts of the base is to the sum of the other two sides as the difference of the latter is to the difference of the former.

Proof. Let BD be the perpendicular. Now, BC2-CD?= BD2 (Cor. 2, Theo. XIX, B. I).

B

Transposing, BC2-AB2=CD2—AD2.

Factoring,

(BC+AB) (BC—AB)=(CD+AD) (CD—AD). Hence, resolving into a proportion, we have CD+AD: CB+AB :: CB—AB : CD—AD.

2. The square of the side opposite to any acute angle in a triangle is equal to the sum of the squares of the other sides, minus twice the product of the base by the distance from the acute angle to the foot of the perpendicular let fall from the vertex on the base, or the base produced.

Proof. For the case where the perpendicular falls on the base produced, take the annexed figure; for the other case, take the preceding one.

B

D

A

Now, AB2=BD2+AD2 (Theo. XIX, B. I).

But,

Hence,

AD=CD-AC (or, AC-CD).

AD2=CD2+AC2—2 CDX AC.

Substituting this in the first equation, we have

AB2=BD2+CD2+AC2-2 CDXAC.

But BD2+CD2=BC2; whence,

AB2=BC2+AC2-2 CDXAC.

3. The square of the side opposite an obtuse angle in a triangle is equal to the sum of the squares of the other sides, plus twice the product of the base by the distance from the obtuse angle to the foot of the perpendicular, let fall from the vertex on the base. produced.

For proof use the last figure.

4. Given the base and the sum of the two other sides of a right-angled triangle, to find the hypotenuse and perpendicular.

Solution. Represent the base by b and the sum of the other sides by s. Also, represent the perpendicular by x, then the hypotenuse will be s-x. Now, b2x2=(8-x)2 (Theo. XIX, B. I).

Hence, by reduction, we have,

x=

2