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Similarly area CK= GFX CM= GFX BK, and so on.
But S = sum of areas of lateral faces, BH, CK, etc.
.. S = DG X BK + GFX BK + FO X BK + etc.

= (DG + GF+FO + etc.) BK
= P x E.

Q.E.D.

548. Cor. The lateral area of a right prism equals the perimeter of the base multiplied by the altitude.

Ex. 1081. Find the lateral area of a right prism if its altitude is 15 inches, and its base is a triangle the sides of which are 8 inches, 10 inches, and 11 inches, respectively.

Ex. 1082. Find the lateral area of a right prism if its altitude is 18 inches, and its base is a quadrilateral the sides of which are 7 inches, 8 inches, 11 inches, and 12 inches, respectively.

Ex. 1083. Find the altitude of a right prism if its base is an equilateral triangle inscribed in a circle of radius 5 inches, and its lateral area is 135 square inches.

Ex. 1084. Find the radius of the circle circumscribing an equilateral triangle which forms the base of a right prism whose altitude is 9, and whose lateral area is 540 V3.

PROPOSITION III. THEOREM 549. If two prisms have the three faces including a triedral angle of one equal respectively to the three faces including a triedral angle of the other, and the faces similarly placed, they are equal.

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To prove

Hyp. The faces AD, AH, AO, of prism AM= the faces A'D', A'H', A'O' of prism A'M' respectively, and are similarly placed.

AM= A'M'.
Proof. 2 BAE = 2 B'A'E', Z BAG= 2 B'A'G',
LEAG= LE'AG'.

(Why ?) .. triedral angle A= triedral angle A'. (526) Place triedral Z A on its equal Z A'.

Then face AD coincides with face A'D', face AH with face A'H', and face AO with face A'O'. The point C falls at C", and D at D'.

(Why ?) CK will fall on C'K', and DM on D'M'. (Since the lateral edges of a prism are parallel.)

The points G, H, and ( coincide respectively with G', H', and O'.

(Why ?) . planes GM and G'M' coincide. (454, 2) Hence point K coincides with K', and M with M'. (Why?)

:: prisms coincide throughout and are equal.

Q.E.D.

550. Cor. 1. Two right prisms having equal bases and equal altitudes are equal.

551. Cor. 2. Two truncated prisms are equal when they fulfil the hypothesis of the above theorem.

552. DEF. The volume of any solid is the number of times it contains the unit of volume. The unit of volume is a cube whose edges are equal to the linear unit.

553. DEF. Equivalent solids are solids whose volumes are equal.

PROPOSITION IV. THEOREM

554. An oblique prism is equivalent to a right prism whose base is a right section of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism.

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Hyp. GM is a right section of oblique prism AD', and GM a right prism whose altitude is equal to a lateral edge of AD'.

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Proof. The lateral edges of GM'=lateral edges of AD'.

(Why ?) .. AH=A'H', BK=B'K', CM=C'M', etc. (Why?) Face AK= face A'K', face BM=face B'M', and face BE.= face B'E',

(Why?) .. truncated prism AM= truncated prism A'M'. (551) Hence AM' – AM= AM' A'M',

(Ax. 3) AD' = GM'.

or

Q.E.D.

PROPOSITION V. THEOREM

555. The opposite lateral faces of a parallelopiped are equal and parallel.

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To prove

Hyp. AH and DK are opposite faces of parallelopiped AK.

AH = DK and AH || to DK.
Proof.
AB = and Il to DC.

(Why ?)
AE = and Il to DM.
.:: ZEAB= _ MDC.

(468) .:. AH =

(142) and AH || DK.

(486) Similarly faces AM and BK' are equal and parallel.

DK,

556. COR. Any two opposite faces of a parallelopiped may be taken as bases.

Ex. 1085. Find the total area of a right prism if its altitude is 12 inches, and its base is a triangle the sides of which are 9 inches, 6 inches, and 5 inches respectively.

Ex. 1086. Find the total area of a right prism if its altitude is 9 inches, and its base is a rhombus of side 4 inches and having an acute angle of 60°.

PROPOSITION VI. THEOREM

557. The plane passed through two diagonally opposite edges of a parallelopiped divides it into two equivalent triangular prisms.

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To prove

Hyp. DOLB is a plane passing through edges DO and BL.

prism ABD-K = prism BCD-M. Proof. Draw right section EFGH, intersecting DOLB in HF. AL || DM, AO || BM.

(Why?) .:. EF || HG.

(478) And

EH is II to FG.
.:. EFGH is parallelogram.
A EFH=AFGH.

(Why?) The prism ABD-K is equivalent to a right prism whose base is A EFH and altitude AK, and prism BDC-M is equivalent to a right prism whose base is AFGH and altitude AK. (554)

.. prism ABD-K = prism BDC-M.

Q.E.D

Ex. 1087. The diagonals of a parallelopiped bisect each other.

Ex. 1088. Find the edge of a cube having a surface equal to the sum of the surfaces of two cubes whose edges are 120 inches and 209 inches.

Ex. 1089. The dimensions of a rectangular solid are proportional to 3, 4, and 5. Find them if the entire surface contains 2350 square inches.

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