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EXAMPLES.

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

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12*V2=

43 4X4X4

4x4

16

16 X 2= 12

3

X V2=3XV2=3* 22.624 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 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 bnACX f De=An x gve=content of the solid.

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?

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X v2=94x V 2 = 21.333, &c. * / 2=21.333, &c.

64

3 x 1.414, &c.=30.16486=solidity required. 2. Required the solidity of an octaedron whose side is 8.

Ans. 241.3568.

PROBLEM III.

To find the solidity of a dodecaedron.

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 an' x jde = an? x fle=an? x }BA = Jan x An + Ac=fan? 2an'=fan 2. Q. E. D.

If i= linear side as before, then will 213=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 IV 10+2 3 5, and the angle aFD contains 36 degrees, D whose sine is £ 710—25, the radius in both cases being taken equal to 1.

5-75=aD

DF

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Therefore, by trigonometry, 1 10–2V5:V 10+ 75

5+15 1+5 ::ad: DF=ADV

=aD =*+175Xad.

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 1 V3; whence, by trigonometry, I v3: DF::

1 +15 : DC =

; and consequently ac=ad-D0

2v3 3-5 =aDV =aD1-775.

6 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.

ada ad' Whence ac : aD::a:

3+15

=aD ao ADV1-375

2

3+15 =diameter of the circumscribing sphere, and aD

4 =R=radius of the circumscribing sphere. Again, the angle Fon contains 72°, whose sine is }

1+5 710+2,5; and the angle oFn is 54°, whose sine is

4 whence by trigonometry, I 7,10+275:

-1+5 +

:: Fn (ad)

4 1+5

5+5 OF=aD

CAD =adv}+16 75.
✓10+25
P

10

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

D

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215+47

21 x 2.23606 +47 ✓ -x 27 x 5=V

x27x53 40

40 46.95726 +47

x 135=206.901 solidity required. 40 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 VR? - Of=

25+115 v( i3+15ad-175ad= aDV

40 v +18V5=radius of the inscribed sphere.

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

5aD each of whose bases are by Prob. VIII. vitv5; 60ad?

60ad?

ad therefore V1+375 x fr= 1+ 5 x 4

4

3 25 +115 47 +21 75 =5ad

= solidity of the dode 40

40 caedron. Q. E. D.

L 15L =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.

7+3 75 That is & SØ x ✓

solidity when S is=to the

linear side.

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A * Demon. Let A be a solid angle of the icosaedron, formed by 5 faces, or triangles, whose bases make the pentagon BCDEm.

Then, if a perpendicular be demitted from A upon the pentagonal plane BCDÉm, it will fall into the centre n, and Bn, by the demon

B stration of the last problem, will be = AB 5+v5

B and the radius of the circle 10 circumscribing one of the faces ABCABÍV3. 0

BA But the radius of the circumscribing sphere is R=

2AN BA

5+5 =ABV

found as in the last problem. 2 V AB2-BN

8 And, since R is the hypothenuse of a right angled triangle, one of whose legs is } AB V3, the radius of the circle circumscribing the face ABC, and the other r, the radius of the inscribed sphere, we shall

5 + 5 bave r= R-}(AB 3)= V

- JAB? 7 +35

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

20ABS whose bases is 3 by Problem VIII. ; therefore

AB2

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= AB

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