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Fig. 351.

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Fig. 332.

Let the cube AM and the pyramid C (fig. 351.) have the same base AD, and let the vertex of the pyramid be at the centre of the cube C; this pyramid is equal to a third part of the product of its height and base.

Conceive right lines drawn from the centre of the cube to its right angles A, B, D, F, N, G, L, M, the cube will be divided into six equal pyramids, each of which has one surface of the cube for its base, and half the height of the cube for its height; for example, the pyramid CABDF.

Three of these pyramids will therefore be equal to half the cube. Now the solid content of half the cube is (Prop. 99.) equal to the product of the base and half the height. Each pyramid, therefore, will be equal to one third part of the product of the base, and half the height of the cube; that is, the whole height of the pyramid.

991. Prop. CX. The solid content of a pyramid is equal to a third part of the product of its height and base.

Let RPS (fig. 352.) be a pyramid, its solid content is equal to a third part of the product of its height and its base RS.

Form a cube the height of which BL is double of the height of the pyramid RPS. A pyramid the base of which is that of this cube and the vertex of which is C, the centre of the cube, will be equal to a third part of the product of its base and height.

'I'he pyramids C and P have the same height; they are therefore (Corol. to Prop. 108.) to one another as their bases. If the base AFDB is double of the base RS, the pyramid C will therefore be double of the pyramid P.

But the pyramid C is equal to a third part of the product of its height and base. The pyramid P will therefore be equal to a third part of the product of the same height, and half the base AFDB, or, which is the same thing, the whole base RS.

992. Prop. CXI. The solid content of a cone is equal to the third part of the product of its height and base.

For the base of a cone may be considered as a polygon composed of exceedingly small sides, and consequently the cone may be considered as a pyramid having a great number of exceedingly small surfaces ; whence its solid contents will be cqual (Prop. 110.) to one third part of the product of its height and base.

993. Pror. CXII. The solid content of a cone is a third part of the solid content of a cylinder described about it.

Let the cone BAC and the cylinder BDFC (fig. 353.) have the same height and base, the cone is a third part of the cylinder.

For the cylinder is equal to the product of its height and base, and the cone is equal to a third part of this product. Therefore the cone is a third part of the cylinder.

994. Prop. CXIII. The solid content of a sphere is equal to a third part of the product of its radius and surface.

Two points not being sufficient to make a curve line, three points will not be sufficient to make a curve surface. If, therefore, all the physical

Fig. 353. points which compose the surface of the sphere C (fig. 354.) be taken three by three, the whole surface will be divided into exceedingly small plane surfaces; and radii being drawn to each of these points, the sphere will be divided into small pyramids, which have their vertex at the centre, and have plane bases.

The solid contents of all these small pyramids will be equal (Prop. 110.) to a third part of the product of the height and bases. Therefore the solid content of the whole sphere will be equal to a third part of the product of the height and all the bases, that is, of its radius and surface.

995. Prop. CXIV. The surface of a sphere is equal to four of its great circles.

Fig. 354. If a plane bisect a sphere, the section will pass through the centre, and it is called a great circle of the sphere.

Let ABCD (fig. 355.) be a square ; describe the fourth part of the circumference of a circle BLD; draw thé diagonal AC, through G, the right line FM, parallel to AD, and the right line AL.

In the triangle ABC, on account of the equal sides AB, BC, the angles A and C are (Prop. 4.) equal; therefore, since the angle B is a right angle, the angles A and C are each half a right angle. Again, in the triangle AFG, because the angle F is a right angle, and the angle A half a right angle, the angle G is also half a right angle; therefore (Prop. 26.) AF is Á equal to FG.

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The radius AL is equal to the radius AD: but AD is equal to FM; therefore AL is equal to FM.

In the rectangular triangle AFL the square of the hypothenuse AL is equal (Prop. 32.) to the two squares of AF and FL taken together. Instead of AL put its equal FM, and instead of AF put its equal FG; and the square of FM will be equal to the two squares of FG and FL taken together.

Conceive the square ABCD to revolve about the line AB. In the revolution the square will describe a cylinder, the quadrant a hemisphere, and the triangle ABC an inverted cone the vertex whereof will be in A. Also the line FM will form a circular section of a cylinder, the line FL will form a circular section of a hemisphere, and the line FG a cir. cular section of a cone.

These circular sections, or circles, are to each other (Prop. 83.) as the squares of their radii ; therefore, since the square of the radius FM is equal to the squares of the radii FL and FG, the circular section of the cylinder will be equal to the circular sections of the hemisphere and cone.

In the same manner it may be demonstrated that all the other sections or circular surfaces whereof the cylinder is composed are equal to the corresponding sections or surfaces of the hemisphere and cone. Therefore the cylinder is equal to the hemisphere and cone taken together : but the cone (Prop. 112.) is By equal to a third part of the cylinder ; the hemisphere is therefore equal to the remaining two thirds of the cylinder; and consequently the hemisphere is double of the cone. The cone BSC (fig. 356.) is (Prop. 111.) equal to a third part of the product of the radius and base BC, which is a great circle of the sphere: the hemisphere ALD is therefore equal to a third part of the product of the radius and two of its great circles ; and consequently the whole sphere is equal to a third part of the product of the radius and four of its great circles.

Lastly, since the sphere is equal (Prop. 113.) to a third part of the product of the radius and surface of the sphere, and also to a third part of the product of the radius and four of its great circles, the surface of the sphere is equal to four of its great circles.

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Fig. 356.

Sect. II.

PRACTICAL GEOMETRY.

996. Practical Geometry is the art of accurately delineating on a plane surface any plane figure. It is the most simple species of geometrical drawing, and the most generally useful; for the surfaces of buildings and other objects are more frequently plane than curved, and they must be drawn with truth, and of the required proportions, before they can be properly executed, unless in cases where the extreme simplicity of the form renders it improbable that mistakes should arise. It has been defined as the art which directs the mechanical processes for finding the position of points, lines, surfaces, and planes, with the description of such figures on diagrams as can be intelligibly understood by de. finition, according to given dimensions and positions of lines, points, &c.

No part of a building or drawing can be laid down or understood without the assistance of practical geometry, nor can any mechanical employment in the building department be conducted without some assistance from this branch of the science. Cases frequently occur requiring a knowledge of very complex problems, as in masonry, carpentry, and joinery; but these will be given in other parts of this work.

The demonstration of most of the following problems will be found in the preceding section; we therefore refer the reader back to it for definitions, and for the proof of those enunciations which will follow,

PROBLEMS.

997. PROLEH I. To bisect a line AB; that is, to divide it into two equal parts.

From the two centres A and B (fig. 357.) with any equal radii describe arcs of circles intersecting each other in C and D, and draw the line CD. This will biscct the given line in the point E.

998. PROB. II. To bisect n angle BAC.

From the centre A (fig. 358.) with any radius describe an arc cutting off the equal laes AD, AE; and from the two centres D, E, with the same radius describe arcs in. tersecting in F, then draw AF, and it will bisect the angle A, as required. 999. PROB. III. At a given point C in a line AB to erect a perpendicular.

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From the given point C (fig. 359.) with any radius cut off any equal parts CD, CE of the given line; and from the two centres D and E with any one radius describe arcs intersecting in F. Then join CF, and it will be the perpendicular required.

Otherwise — When the given point C is near the end of the line.

From any point D (fig. 360.) assumed above the line as a centre, through the given point C describe a circle cutting

AD the given line at E, and through E and

Fig. 357.
Fig. 358.

Fig. 309.
the centre D draw the diameter EDF;
then join CF, and it will be the perpendicular required.

1000. PROB. IV. From a given point A to let fall a perpendicular on a line BC.

From the given point A (fig. 361.) as a centre with any convenient radius describe an arc cutting the given line at two points D and E; and from the two centres D and E with any radius describe two arcs intersecting at F; then draw AF, and it will be the perpendicular to BC required.

Otherwise · When the given point is nearly opposite the end of the line.

From any point D in the given line BC (fig. 362.) as a centre, describe the arc of a circle through the given point A cutting BC in E; and from the centre E with the

Fig. 360.

Fig 361. radius EA describe another arc cutting the former in F; then draw AGF, which will be the perpendicular to BC required.

1001. PROB. V. At a given point A, in a line AB, to make an angle equal to a given angle C.

From the centres A and C (fig. 363.) with any radius describe the arcs DE, FG; then with F as a centre, and radius DE, de- OD

BC scribe an arc cutting FG in G; through G draw the line AG, which will form the angle required.

FB 1002. PROB. VI. Through a given point

Fig. 362.

Fig. 363. C to draw a linc parallel to a given line AB.

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Case II.
When the parallel is to be drawn at a given distance from AB.

From any two points c and d in the line AB, with a radius equal to the given distance describe the arcs e and f; draw the line CB to touch those arcs without cutting them, and it will be parallel to AB, as required.

1003. PROB. VII. To divide a line AB into any proposed number of equal parts.

Draw any other line AC (fig. 365.), forming any angle with the given line AB; on the latter set off as many of any equal parts AD, DE, EF, FC as those into which the line AB is to be divided; join BC, and parallel thereto draw the other lines FG, EH, DI; then these will divide AB, as required.

1004. Prob. VIII, To find a third proportional to two other Á lines AB, AC.

Let the two given lines be placed to form any angle at A (fig. 366.), and in AB take AD equal to AC; join BC, and draw DE parallel to it; then AE will be the third proportional sought.

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1005. PROB, IX.

To find a fourth proportional to three lines AB, AC, AD. Let two of the lines AB, AC (fig. 367.), be so placed as to form any angle at A, and set out AD on AB; join BC, and parallel to it draw DE; then AE will be the fourth proportional required.

1006. PROB. X. To find a mean proportional between tuo lines AB, BC. Place AB, BC (fig. 368.)

Fig. 566.

Fig. 567. joined together in one straight line AC, wbich bisect in the point 0; then with the centre (and radius OA cr OC describe the semicircle ADC, to meet which erect the perpendicular BD, which will be the mean proportional between AB and BC sought.

1007. ProB. XI. To find the centre of a circle.

Fig. 368.

Fig. 369.

Fig. 370, Draw any chord AB (fig. 369.), and bisect perpendicularly with the line CD, which bisected in 0 will be the centre required.

1008. Prob. XII. To describe the circumference of a circle through three points A, B, C.

From the middle point B (fig. 370.) draw the chords BA, BC to the two other points, and bisect these chords perpendicularly by lines meeting in 0, which will be the centre; from the centre 0, with the distance of any one of the points, as 0 A, describe a circle, and it will pass through the two other points B B, C, as required.

1009. ProB. XIII. To draw a tangent to a circle through a given point A.

When the given point A (fig. 371.) is in the circumference of the circle, join A and the centre 0, and perpendicular thereto draw BAC, which will be the tangent required.

If the given point A (fig. 372.) be out of the circle, draw A to the centre 0, on

Fig. 371.

Fig. 372. which, as a diameter, describe a semicircle cutting the given circumference in D, through which draw BADC, which will be the tangent required.

1010. Prob. XIV. To draw an equilateral triangle on a given line AB.

From the centres A and B (fig. 373.) with the distance AB describe arcs inter. secting in C; draw AC, BC, and ABC will be the equilateral triangle.

1011. PROB. XV. To make a triangle with three giren lines AB, AC, BC. With the centre A and distance AC (fig.

Fig. 373.

Fig. 371. 374.) describe an are; with the centre B and distance BC describe another arc cutting the former in C; draw AC, BC, and ABC will be the triangle required.

1012. Prob. XVI. To make a square on a given line AB.

Raise AD, BC (fig. 375.) each perpendicular and equal to AB, and join DC; then ABCD will be the square sought.

1013. Pro, XVII. To inscribe a circle in a given triangle ABC. Bisect the angles at A and B with the two Fig. 375,

Fig. 376. lines AD, BD (fig. 376.); from the intersection D, which will be the centre of the circle, draw the perpendiculars DE, DF, DG, and they will be the radii of the circle required.

101 4. Prob. XVIII. To describe a circle about a given triangle ABC.

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Bisect any two sides with two of the perpendiculars DE, DF, DG (fig. 377.), and J will be the centre of the circle.

1015. PROB. XIX. To inscribe an equilateral triangle in a given circle.

Through the centre C draw any diameter AB (fig. 378.); from the point B as a centre, with the radius BC of the given circle, describe an arc DCE; join AD, AE, DE, and ADE is the equilateral triangle sought.

1016. PROB. XX. To inscribe a square in a given circle. (Half AB, BC, &c. forms an octagon.)

Fig. 377. Draw two diameters AC, BD (fig. 379.) crossing at right angles in the centre E; then join the four extremities A, B, C, D with right lines, and these will form the inscribed square ABCD.

1017. Prob. XXI. To describe a square about a giren circle.

Draw two diameters AC, BD crossing at right angles in the centre E (fig. 380.); then through the four extremities of these draw FG, IH parallel to AC, and FI, GH parallel to BD, and

Fig. 379.

Fiz..99 they will form the square FGHI.

1018. PROB XXII. To inscribe a circle in a given squarc.

Bisect the two sides FG, FI in the points B and A (see fig. 380.); then through these two points draw AC parallel to FG or IH, and BD parallel to FÍ or GH. Then the point of intersection E will be the centre, and the four lines EA, EB, EC, ED radii of the inscribed circle,

1019. PROB. XXIII. To cut a given line in extreme and mean ratio.

Let AB be the given line to be divided in extreme and mean ratio (fig. 381.); that is, so that the whole line may be to the greater part as the greater part is to the less part.

Draw BC perpendicular to AB, and equal to half AB; join AC, and with the centre C and distance CB describe the circle BDF; then with A as a centre and distance AD desc:ibe the arc DE. Then AB will be divided in E in extreme and mean ratio, or so that AB is to AE as AE is 1o EB.

1020. PROB. XXIV. To inscribe an isosceles triangle in a given circle that shall have cach of the

Fig. 381.

Fig. 382. angles at the base double the angle at the verter.

Draw any diameter AB of the given circle (fig. 382.), and divide the radius CB in the point D in extreme and mcan ratio (by the last problem); from the point B apply the chords BE, BF, each equal to the greater part CD; then join AE, AF, EF; and A EF will be triangle required.

1021. Proe. XXV. To inscribe a regular pen- D tugon in a giren circle. (Half AD, &c. is a decagon.)

Inscribe the isosceles triangle AB (fig. 383.) having each of the angles ABC, ACB double the angle BAC (Prob. 24.); then bisect the two ares ADB, AEC, in the points D, E; and draw the chords AD, DB, AÈ, EC; then ADBCE will Fig. 383.

Fig. 354. be the inscribed regular pentagon required.

1022. PROB. XXVI. To inscribe a regular hexacou in a circle. (Holf AB, &c. foras a dodecagon.)

Apply the radius of the given circle AO as a chord (fig. 384.) quite round the circumference, and it will form the points thereon of the regular hexagon ABCDEF.

1023. ProB. XXVII. To describe a re. gular pentagon or hexagon about a circle.

In the given circle inscribe a regular polygon of the same name or number of sides as ABCDE (fig. 385.) by one of the foregoing problems; then to all its angular points draw (Prob. 13.) tangents, and these will by their intersections form the

Fis. 385.

Fig. 126, circumscribing polygon required.

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