Multiply

PROBLEM I.

To find the solidity of a tetraedron.

RULE.*

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

*Demon. From one angle c of the tetraedron ABCn, let fall the perpendicular ce, upon the opposite side, and draw Ae.

Then Ac2-Ae2=ce2; and since the point e is equally distant from the three angles A, B, and n, † ac2= ( AB2) Ae as is shown in the demonstration of the rule for regular polygons, page 61. Consequently Ac2- AC2=}AC2 = ce2, or ce AC √ }. But the area of the triangle ANB=AB2√3= Ac2√3; and therefore Ac✓ (ce) × ‡ac2√3(aanb)=√AC3√2. Q. E. D.

If L be put = length of the linear edge, then will Ľ2√3 =whole surface of the tetraedron.

The rule for the hexaedron, or cube, has been given before.

EXAMPLES.

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

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

PROBLEM II.

Ans. 25.452.

Multiply

To find the solidity of an octaedron.

RULE.*

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 let fall the perpendicular de.

Then since the solid is composed of two equal square pyramids, each of whose bases BRAC are equal to the square of the linear side AG or AD, we shall have BnAC×

De=.

= An2

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

EXAMPLES.

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

G

× √2=21.333, &c. × √/2=21.333, &c.

=

x 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 = An2× Be=An2 × BA = an2 × √An2+Ac2=an2 √2an2=}an3/2. 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. Also join the points D, c and c, F.

Then the angle DaF contains 108 degrees, whose sine is 1 √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+2√5

Again, since c is the centre of DBF, 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:: } 1+ √5 ;

DF :DC=αD

3-5

6

2/3 and consequently ac=√aD2-DC2

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.

Whence ac: aD::a:

=

ac

=diameter of the circumscribing sphere, and aD

=R=radius of the circumscribing sphere.

√3+ √15

2

√3+15

4

Again, the angle Fon contains 72°, whose sine is √10+2√5; and the angle oFn is 54°, whose sine is. 1+ √5

whence by trigonometry, &√/10+2√5:

EXAMPLES.

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

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 √R2 · 012 25+11√5

`√(√3+√15αD)—+1%√5аD2 = AD √

✓ +1√5=radius of the inscribed sphere.

40

=

And because the solid is composed of 12 équal pentagonal pyramids,

each of whose bases are by Prob. VIII. =

If L be put for the linear side, then will 15Ľ2√

5+2√5

5

surface of the dodecaedron.