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.. the whole arca P ̧ P ̧ PÅ N1 N2 =

4 2

(4 Y + Y2 +Y4)

2

4

2 2

2

Hence the area of the figure PA BQ (fig. 18), which equals P A N2 P2 + P2 N P4 N4 + P4 N4 P6 No +

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reckoning that there are 2n + 1 (an odd number) ordinates drawn between the two extreme ones PA, QB.

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=

x (sum of extreme sides + 4 sum of odd ordinates + 2 sum of even ordinates.) Hence the Rule:- -At equal distances along the base AB draw an odd number of ordinates parallel to the parallel sides of the figure; then take the sum of the parallel sides, of four times the odd ordinates, and twice the even ordinates, and multiply the third part of this sum by the common distance between the ordinates, and this product is the area of the figure.

Besides the above areas of plane surfaces, there are areas of certain solids which will be best given here, viz., the areas of a prism, a cylinder, a pyramid, a cone, and a sphere.

DEF. 1.-A prism is a solid bounded by two equal and similar rectilinear figures in parallel planes and by parallelograms.

Thus ABCDE, abcde, is a prism bounded by two equal and similar pentagons whose planes are parallel (ABCDE, and abcde), and by parallelograms AaBb, BbcC, &c. A prism is called a right prism when the planes of the parallelograms are perpendicular to the plane of the base (i.e. abcde in the accompanying figure).

DEF. 2.-A pyramid is a solid bounded by any plane rectilinear figure, and by triangles having a common vertex, and for bases the sides of the rectilinear figure respectively. Thus (Fig. 21) PABCD is a pyramid, on a quadrilateral base, ABCD.

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B

E

Fig. 20.

DEF. 3.-A cylinder is a solid whose surface is traced out by a straight line which always moves parallel to its first position and

whose extremity is guided by a given curve.

Thus (Fig. 22) if acb is a circle, and Aa a straight line perpendicular to the plane of the circle, then ABC abc is a cylinder which is traced out by a line Cc, that is always parallel and equal to Aa. This cylinder is strictly defined as a "right cylinder with a circular base" it is, however, in elementary treatises generally called "a cylinder."

D

Fig. 21.

DEF. 4.-A cone is a solid, the surface of which is traced out by a straight line,

one end of which passes through a fixed point, and the other end through a given plane curve, called its base.

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If BCD (Fig. 23) is a circle, and a a fixed point, the surface ABCD is called a cone on a circular base. If O is the centre of the circle, join A; then if straight lines, draw from A to different points in circumference of circle, make equal angles with AO; or if (which is the same thing) AO is perpendicular to plane of circle, ABCD is strictly defined as "a right cone with a circular base;" or as it is more generally called in elementary treatises, a "right cone."

N.B. It is manifest that if a regular polygon is inscribed in the circle BCD, and its angular points are joined with A, that the resulting solid will be a pyramid inscribed in the cone; and also, that if we increase the number of sides in the polygon, the inscribed pyramid will approach more nearly to the cone; and, since the circle is the limit of the inscribed polygon, the cone will be the limit of the inscribed pyramid. Similarly, the cylinder will be the limit of the inscribed prism.

(15.) To find the Area of a right Prism.

In figure (20) AEea is a rectangle. Since the planes Ab, Ae being perpendicular to the base, their cone of intersection Aa is also perpendicular to the base, and therefore angle Aae is a right angle. Hence the area of AEea is Aa × AE, similarly of all the other parallelograms. Hence the area of the parallelograms is Aa (AE+ED + DC + ...) AE (the perimeter of the base.) Hence the whole area = height x perimeter of base areas of the two ends.

(16.) To find the Area of the surface of a right Cylinder.

=

Since the surface of the cylinder is the limit of the surface of the inscribed prism, and the area of the prism height x perimeter of base + 2 x base, whatever be the number of sides to the base, this will be true in the limit when base is a circle. .. Area of a cylinder height perimeter of base + 2 x base.

Hence, if h

=

height, and a = radius of base.

Area of cylinder = 2 παh + 2 πα = 2πa (a + h).

(17.) To find the Arca of the curved surface of a right Cone.

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Р

Let PACDB be the cone, ACDB its circular base, O the centre of the circle, then PO is at right angles to plane of the circle. In the circle inscribe any regular polygon, ACD. and join PA, PC, PD... Bisect CD in n, join Pn. Then Pn is perpendicular to CD, and.. the area of triangle PCD = CD, Pn. Now the line joining P with the bisection of any other side of the polygon is equal to Pn.

Hence area of pyramid inscribed in cone

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slant

polygon may have, and hence is true in the limit; when perimeter of polygon circumference of circle, and Pn is drawn to a point in circumference, or is side of cone,

slant side x circumference of base.

.. Area of curved surface of cone = COR.-If we suppose that the surface of a cone is capable of being unwrapt, it is plain that its surface will be a sector of a circle whose radius is the slant side of cone, and base of the same length as the circumference. It is plain (Mensuration, Art. 9,)

1 that the area of this sector is, as it should be, the same as that of the surface of cone.

DEF. The frustum of a cone or pyramid is the portion cut off by a plane parallel to the base; thus, ABba is a frustum of the cone CAB.

C

a

b

A

(18.) To find the Area of the Frustum of a Cone.

Fig. 25.

P

P

Suppose a sector, OPQ, (Fig. 26) of a circle is described with radius OP = CA (Fig. 25), and if its base PQ = circumference of AB, we have seen that the area of POQ is the same as that of the cone CAB. Take Oq Ca, and describe the arc O¶p. Then as before, area of opq = area of Cab, and.. the area of frustum

=

qQPp.

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.. Area of frustum rectangle between Qq and tt' or between slant side of cone and circumference of mean section of frustum.

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B

P

A
M

n

(19.) To find the Area of a portion of the surface of a Sphere.

Let AB be a quadrant of a circle whose ra lits

is AB or OB. If we suppose the quadrant to p.24

revolve round AO it will describe a hemisphere,

and if we suppose a number of equal chords, P.3

AP2, P.' P2, P2, P3, &c. to be drawn from point to point of AB, these chords in the revolution will describe frustums of cones; now the are

AR is the limit of the chords Apı + P P2 + N

P2 P3t... and hence the area of the portion
of sphere described by AR will be the limit of
the sum of the frustums of cones described by

AP,P.P2 P2P3 :
Fig. 27.

Let PQ (Fig. 28) be one of these chords :

draw PM, QN perpendicular to A0. Draw O perpendicular to PQ, and in perpendicular to AO. Now tn is the radius

of the mean section of the frustum of cone described
by PQ, and therefore area of that frustum

= 2 * tn x PQ
Now tn = Ot sin. tOA.
and PQ sin. PQN = NM.
or since PQN = OA.
PQ sin. tOA = NM

i. tn. PQ = 0t. NM.
And area of frustum = 2Ot. NM.
In (fig. 27) draw P, N, P, N, P, ng perpendicularly
AO. Then, since Ap, Pa P2 P2 P3 are all equals, the

perpendiculars on them are equal, and therefore the Fig. 28.

sum of areas of frustum of cone
== 2Ot (An, + ng nafn, ng t....)

= 27 Ot x AN. if we only consider the portion of sphere described by AR. This is true, however great the number of chords, and is therefore true in the limit; but in the limit Ot = radius of sphere :: Area of portion of sphere, whose height is AN

= 2. OA. AN. If we take the whole sphere. AN = 20A

.. Area of sphere 41. (OA).
Cor.-In the sphere 27 * OA = circumference of a great circle.

Area of portion of a sphere = height of portion x circumference of a great circle.

If we imagine a right cylinder to be described about a sphere, its curved area = circumference of a great circle x diameter of sphere = area of sphere.

Hence, " area of sphere = area of circumscribing cylinder.”

N

B

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Exam: How many square miles of sea are visible from the top of a mast 80 feet above the surface?

B

Let O be the centre of earth, draw OA. OC., so that AC is perpendicular to OC. draw CN perpendicular to OA. Then, if BN = x. W OBr. The area visible from A will be the area of the part of the sphere whose depth is BN

i.e. will =2π rx.

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(1.) If two solid Angles are each contained by three Plane Angles, that are equal each to each, then the inclination of the Planes will be equal, each to each.

Let O, o, be two solid angles contained by the plane angles AOB, BOC, COA, and

C

B

a

Fig. 30.

aob, boc, coa, respectively, Then the plane AOB is same angle that aob is inclined to aoc. For, take OA

inclined to plane COA at the oa and let the plane CAB be

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