EXAMPLES.

1. The linear side of a tetraedron ABCn is 4: what.' the solidity?

1.414=

7.5413-solidity required. 3

2. Required the solidity of a tetraedron whose side is 6.

Ans. 25.452

PROBLEM II.

To find the solidity of an octaedron.

RULE.*

Multiply of the cube of the linear side by the square root of 2, and the product will be the solidity.

*Demon. From the angle D of the octaedron DBGA lei fall the perpendicular De.

Then since the solid is composed of two equal square pyramids, each of whose bases BnAC are equal to the square of the linear side AG or AD, we shall have BNAC× De=An' Xue=content of the solid.

But be evidently bisects the diagonal BA, and is equal to

EXAMPLES.

1. What is the solidity of the octaedron BGAD, whose linear side is 4?

43

64

3x √2=4× √2=21.833, &c. × √/ 2=21.333, &c.

× 1.414, &c.=30.16486=solidity required.

2. Required the solidity of an octaedron whose side is 8.

Ans. 241.3568.

PROBLEM III.

RULE.*

To 21 times the square root of 5 add 47, and divide the sum by 40: then the square root of the quotient being

=

Be; therefore An2x De An2x Be An2 X BA = JAN2X √ an2+ac2={an2√2an2=}an√2. Q. E. D.

If L=

linear side as before, then will 2123-surface of the octaedron.

*Demon. Let a be a solid angle of the dodecaedron, and ac a

multiplied by 5 times the cube of the linear side will give the solidity required.

perpendicular falling on the equilateral

plane, BDF. and c F.

Also join the points D, c

Then the angle DaF contains 108 degrees, whose sine is √10+2√✓5, and the angle aFD contains 36 degrees, whose sine is√10-2/5, the radius in both cases being taken equal to 1.

D

Therefore, by trigonometry, √10—2√5: † √10+ √5 5+√5 1+√5 5-5

:: aD: DF-αD √

2.

Again, since c is the centre of BDF, the angles cDF and cFD are each 30°, and the angle DcF-120°; but the sine of 30° is; and the sine of 120° is√3; whence, by trigonometry, √3: DF:: DF 1+ √5

; and consequently ac=aD-DO

But a perpendicular from a upon the plane BDF must pass through the centre of the circumscribing sphere, and ac will be the versed sine of an arc whose chord is aD, and radius equal to that of the said sphere.

=R=radius of the circumscribing sphere.

Again, the angle Fon contains 72°, whose sine is

10+25; and the angle oFn is 54°, whose sine is

1+√5

4

EXAMPLES.

1. The linear side of the dodecaedron ABCDE is 3, what is the solidity?

46.95726+47

40

x 135 206.901 solidity required.

2. The linear side of a dodecaedron is 1; what is the solidity? Ans. 7.6631.

But since the radius of the circumscribing sphere is the hypothenuse of a right angled triangle, whose legs are oF and the radius of the inscribed sphere, we shall have

√(‡√3+√15aD)2—}+7% √ 5αD2 = AD√

✔+1√5=radius of the inscribed sphere.

--

√R2 — or2 25+11/5

40

=

=

And because the solid is composed of 12 equal pentagonal pyramids.

each of whose bases are by Prob. VIII.

5AD

√1+ √5;

caedron. Q. E. D.

If L be put

for the linear side, then will 1512√5+2√5

surface of the dodecaedron.

PROBLEM IV.

To find the solidity of an icosaedron.

RULE.*

To 3 times the square root of 5 add 7, and divide the sum by 2; then the square root of this quotient being multiplied by of the cube of the linear side will give the solidity required.

But the radius of the circumscribing sphere is R=;

BA

2An

And, since R is the hypothenuse of a right angled triangle, one of whose legs is AB 3, the radius of the circle circumscribing the face ABC, and the other r, the radius of the inscribed sphere, we shall

But the solid is composed of 20 equal triangular pyramids, each of

AB2

20AB3 4

whose bases is- ✓ 3 by Problem VIII.; therefore4