MENSURATION OF SPECIFIC FIGURES (GEOMETRICAL)
22. Areas of Plane Rectilinear Figures.—The following are expressions for the areas of some simple figures; the expressions in (i) and (ii) are obtained arithmetically, while those in (iii)–(v) involve dissection and rearrangement.
(i) Square: side a. Area=a2.
(ii) Rectangle: sides a and b. Area = ab.
(iii) Rightangled triangle: sides a and b, enclosing the right an ie. Area =lab.
iv) Parallelogram: two opposite sides a and a, distance between them h. Area = ha.
(v) Triangle: one side a, distant h from the opposite angle. Area = ;ha.
if the data for any of these figures are other than those given above, trigonometrical ratios will usually be involved. If, for instance, the data for the triangle are sides a and b, enclosing an angle C, the area is lab sin C.
23. The figures considered in § 22 are particular cases of the trapezium, which is a quadrilateral with two parallel sides. If these sides are a and b, at distance h from one another, the area is h.; (a+ b). In the case of the triangle, for instance, b is zero, so that the area is j ha.
The trapezium is also sometimes called a " trapezoid," but it will be convenient to reserve this term for a different figure (§ 24).
The most important form of trapezium is that in which one of the two remaining sides of the figure is at right angles to the two parallel sides. The trapezium is then a right trapezium; the two parallel sides are called the sides, the side at right angles to them the base, and the fourth side the top.
By producing the two parallel sides of any trapezium (e.g. a paralellogram), and drawing a line at right angles to them, outside the figure, we see that it may be treated as the difference of two right trapezia.
It is, however, more simple to convert it into a single right trapezium. Let CABD (fig. I) be a trapezium, the sides CA and DB being parallel. Draw any straight line at right angles to CA and DB (produced if necessalry), meeting them in M and N. Along CA and DB, on the same side of MN, take MA' =CA, NB' = DB ; andjoin A'B'. Then MA'B'N is a right trapezium, whose area is equal to that of CABD; and it is related to the latter in such a way that, if any two lines parallel to AC and BD meet AB, CD, MN, A'B', in E, G, P, E', and F, H, Q, F', respectively,
the area of the piece PE'F'Q of the right trapezium ,B
is equal to the. area of the piece GEFH of the original S
, ,
trapezium. The right trapezium so constructed F' '
may be called the equivalent right trapezium. In
the case of a parallelogram, the equivalent right , trapezium is a rectangle; in the case of a triangle,
it is a rightangled triangle.
24. If we take a series of right trapezia, such that one side (§ 23) of the first is equal to one side of the second, the other side of the second is equal to one side of the third, and so on, and place them with their bases in a straight line and their equal sides adjoining each other, we get a figure such as MABCDEFS (fig. 2), which has two parallel sides MA and SF, a base MS at right angles to these, and the remainder of its boundary from A to F rectilinear, no part of the figure being outside the space between MA (produced) and SF (produced). A figure of this kind will be called a trapezoid.
(i) If from the.other angular points B, C, D, E, perpendiculars BN, CP, DQ, ER, are drawn to the base MS (fig. 2), the area is
MN.2(MA+NB)+NP. z (NB+ B
PC)+.. . .+RS.Z(RE+SF) = B 1(MN. MA + MP. NB +
NQ. PC+ . . + RS . SF). The lines MA, NB, PC, .. . are called the ordinates of the points A, B, C, . . . . from the base MS, and the portions MN, NP, PQ, . . . . of the base are
the projections of the sides AB, BC, CD, . . . . on the base.
(ii) A special case is that in which A coincides with M, and F with S. The figure then stands on a base MS, the remainder of its boundary being a broken line from M to S. The formula then becomes
area=1,(MP.NB+NQ . PC+... +QS . RE),
i.e. the area is half the sum of the products obtained by, multiplying each ordinate by the distance between the two adjacent ordinates. It would be possible to regard this form of the figure as the general one; the figure considered in (i) would then represent the special case in which the two endpieces of the broken line are at right angles to the base.
(iii) Another special case is that in which the distances MN, NP, PQ, . . . RS are all equal. If this distance is h, then
area =h(;MA+NB+PC+...+2SF).
25. To find the area of any rectilinear figure, various methods are available.
(i) The figure may be divided into triangles. The quadrilateral, for instance, consists of two triangles, and its area is the product of half the length of one diagonal by the sum of the perpendiculars drawn to this diagonal from the other two angular points.
For figures of more than four sides this method is not usually convenient, except for such special cases as that of a regular polygon, which can be divided into triangles
by radii drawn from its centre. D
(ii) Suppose that two angular points, A and E, are joined (fig. 3) so as to form a diagonal AE, and that the whole cf the figure lies between lines through A and E at right angles to AE. Then the figure is (usually) the sum of two trapezoids on base AE, and its area can be G
calculated as in § 24. If BN, CP, FIG. 3. DQ, . . . FS, GT are the perpen
diculars to AE from the angular points, the ordinates NB, PC, ... . are called the offsets from the diagonal to the angular points.
The area of the polygon in fig. 3 is given by the expression
2(AP . NB+NQ . PC+PE . QD+ET . SF+SA . TG).
It should be noticed (a) that AP , NQ. SA are taken in tilts cyclical order of the points ABC . . . GA, and (b) that in fig. 3, if AN and NB are regarded as positive, then SF, TG, ET and SA are negative, but the products ET . SF and SA . TG are positive. Negative products will arise if in moving from A to E along the perimeter of either side of the figure the projection of the moving point does not always move in the direction AE.
(iii) Take any straight line intersecting or not intersecting the figure, and draw perpendiculars Aa, Bb, Cc, Dd, . . . Ff, Gg to this line. Then, with proper attention to signs,
area=2(gb . aA+ac'. bB+bd . cC+... +fa . gG).
(iv) The figure may be replaced by an equivalent trapezoid, on the system explained in § 23. Take any base X'X, and draw lines at right angles to this base through all the angular points of the figure.
F
Let the lines through B, G, C, D and F (fig. g) cut the boundary of the figure again in B', G', C', D' and F , and meet the base X'X in K, L, M, N and P; the points A and E being at the extremities of the figure, and the lines through them meeting the base in a and e. Then, if we take ordinates Kb, Lg, Mc, Nd, Pf, equal to B'B, GG', C'C, D'D, FF', the figure abecdfe will be the equivalent trapezoid, and any ordinate drawn from the base to the LM N P e X top of this trapezoid will be
ordinate (produced) which falls within the original figure.
26. Volumes of Solids with Plane Faces.—The following are expressions for the volumes of some simple solid figures.
(i) Cube: side a. Volume =a3.
(ii) Rectangular parallelepiped : sides a, b, c. Volume =abc.
(iii) Right prism. Volume =length of edge X area of end.
(iv) Oblique prism. Volume =height X area of end =length of edge X area of crosssection; the " height " being the perpendicular distance between the two ends.
The parallelepiped is a particular case.
(v) Pyramid with rectilinear base. Volume = height X } . area of base.
The tetrahedron is a particular case.
(vi) Wedge: parallel edges a, b, c; area of crosssection S. Volume = i(a+b+c)S.
This formula holds for the general case in which the base is a trapezium; the wedge being thus formed by cutting a triangular prism by any two planes.
(vii) Frustum of pyramid with rectilinear base: height h; areas of ends (i.e. base and top) A and B. Volume=h. j(A+ fAB+B).
27. The figures considered in § 26 are particular cases of the prismoid (or prismatoid), which may be defined as a solid figure with two parallel plane rectilinear ends, each of the other (i.e. the lateral) faces being a triangle with an angular point in one end of the figure and its opposite side in the other. Two adjoining faces in the same plane may together make a trapezium. More briefly, the figure may be defined as a polyhedron with two parallel faces containing all the vertices.
If R and S are the ends of a prismoid, A and B their areas, h the Perpendicular distance between them, and C the area of a section by a plane parallel to R and S and midway between them, the volume of the prismoid is
§h(A+4C+B).
This is known as the prismoidal formula.
The formula is a deduction from a general formula, considered later (§ 58), and may be verified in various ways. The most instructive is to regard the prismoid as built up (by addition or subtraction) of simpler figures, which are particular cases of it.
(i) Let R and S be the vertex and the base of a pyramid. Then A O, C = ;B, and volume = 4hB = eh(A 1 4C + B). The tetrahedron is a particular case.
(ii) Let R be one edge of a wedge with parallel ends, and S the face containing the other two edges. Then A =0, C = iB, and volume =;hB = sh(A+4C+B)•
(iii) Let R and S be two opposite edges of a tetrahedron. Then the tetrahedron may be regarded as the difference of a wedge with parallel ends, one of the edges beingR, and a pyramid whose base is a parallelogram, one side of the parallelogram being S (see fig. 9, § 58). Hence, by (i) and (ii), the formula holds for this figure.
(iv) For the prismoid in general let ABCD . . . be one end, and abcd . . . the other. Take any point P in the latter, and form. triangles by joining P to each of the sides AB, BC, . . . ab, bc, .. . of the ends, and also to each of the edges. Then the prismoid is divided into a pyramid with vertex P and base ABCD ..., and a series of tetrahedra, such as PABa or PAab. By (i) and (iii), the formula holds for each of these figures; and therefore it holds for the prismoid as a whole.
Another method of verifying the formula is to take a point Q in the midsection, and divide up the prismoid into two pyramids with vertex Q and bases ABCD . . . and abcd . . . respectively, and a series of tetrahedra having Q as one vertex.
28. The Circle and Allied Figures.—The p mensttration of the circle is founded on the property that the areas of different circles are proportional to the squares on their diameters. Denoting the constant ratio by 4r, the area of a circle is ira2, where a is the radius, and r=•14159 approximately. The expression zra for the length of the circumference can be deduced by considering the limit of the area cut off from a circle of radius a by a concentric circle of radius a—a, when a becomes indefinitely small; this is an elementary case of differentiation.
The lengths of arcs of the same circle being proportional to theangles subtended by them at the centre, we get the idea of circulai measure.
Let 0 be the common centre of two circles, of radii a and b, and let radii enclosing an angle B (circular measure) cut their circumferences in A, B and C, D respectively (fig. 5). Then the area of ABDC is
Zb28—za20=(b—a) 4(b+a)B.
If we bisect AB and CD in P and Q respectively, and describe the arc PQ of a circle with centre 0, the length of this arc is z (b+a)e; and b —a =AB. Hence area ABDC =AB X arc PQ. The figure ABDC is a sector of an annulus, which is the portion of a circle left after cutting out a concentric circle.
29. By considering the circle as the limit of a polygon, it follows that the formulae (iii) and (v) of § 26 hold for a right circular cylinder and a right circular cone; i.e.
volume of right circular cylinder =length X area of base;
volume of right circular cone =height X a area of base. These formulae also hold for any right cylinder and any cone.
30. The curved surfaces of the cylinder and of the cone are developable surfaces; i.e. they can be unrolled on a plane. The curved surface of any right cylinder (whether circular or not) becomes a rectangle, and therefore its area =length .X perimeter of base. The curved surface of a right circular cone becomes a sector of a circle, and its area = slant height X perimeter of base.
31. If a is the radius of a sphere, then
(i) volume of sphere= ire;
(ii) surface of sphere=4ra2=curved surface of circumscribing cylinder.
The first of these is a particular case of the prismoidal formula (§ 58). To obtain (i) and (ii) together, we show that the volume of a sphere is proportional to the volume of the cube whose edge is the diameter; denoting. the constant ratio by 8X, the volume of the sphere is Xaa, and thence, by taking two concentric spheres (cf. § 28), the area of the surface is 3X¢2. This surface may be split up into elements, each of which is equal to a corresponding element of the curved surface of the circumscribing cylinder, so that 3Xa2=curved surface of cylinder=2a. 2ra=4ra2. Hence a=tr.
The total surface of the cylinder is 4ra2+ra2+ra2=62ra2, and its volume is 2a.ir¢2=22raa. Hence
volume of sphere = volume of circumscribing cylinder; surface of sphere = i surface of circumscribing cylinder.
These latter formulae are due to Archimedes.
32. Moments and Centroids.—For every material body there is a point, fixed with regard to the body, such that the moment of the body with regard to any plane is the same as if the whole mass were collected at that point; the moment being the sum of the products of each element of mass of the body by its distance from the plane. This point is the centroid of the body.
The ideas of moment and of centroid are extended to geometrical figures, whether solid, superficial or linear. The moment of a figure with regard to a plane is found by dividing the figure into elements of volume, area or length, multiplying each element by its distance from the plane, and adding the products. In the case of a plane area or a plane continuous line the moment with regard to a straight line in the plane is the same as the moment with regard to a perpendicular plane through this line; i.e. it is the sum of the products of each element of area or length by its distance from the straight line. The centroid of a figure is a point fixed with regard to the figure, and such that its moment with regard to any plane (or, in the case of a plane area or line, with regard to any line in the plane) is the same as if the whole volume, area or length were concentrated at this point. The centroid is sometimes called the centre of volume, centre of area, or centre of arc. The proof of the existence of the centroid of a figure is the same as the proof of the existence of the centre of gravity of a body. (See MECHANICS.)
The moment as described above is sometimes called the first moment. The second moment, third moment, . . . of a plane or solid figure are found in the same way by multiplying each element by the square, cube, ... of its distance from the line or plane with regard to which the moments are being taken.
If we divide the first, second, third, . . . moments by the total volume, area or length of the figure, we get the mean distance, mean square of distance, mean cube of distance. . . . of the figure from the line or plane. The mean distance of a plane figure from a line in its plane, or of any figure from a plane, is therefore the same as the' distance of the centroid of the figure from the line or plane.
We sometimes require the moments with regard to a line or plane through the centroid. If No is the area of a plane figure, and NI, N2, . . . are its moments with regard to a line in its plane, the moments MI, M2, . . . with regard to a parallel line through the centroid are given by
Mi = NZ = xNo = o,
M2=N2—2xNi+x2No=N2x2No, 
MQ=NQ—gxNQ_i+q(q2( 2(—)4'igxQ iNi+
—)4xNo;
where x = the distance between the two lines = N1/No. These formulae also hold for converting moments of a solid figure with regard to a plane into moments. with regard to a parallel plane through the centroid ; x being the distance between the two planes. A line through the centroid of a plane figure (drawn in the plane of the figure) is a central line, and a plane through the centroid of a solid figure is a central plane, of the figure.
The centroid of a rectangle is its centre, i.e. the point of intersection of its diagonals. The first moment of a plane figure with regard to a line in its plane may be regarded as obtained by dividing the area into elementary strips by a series of parallel lines indefinitely close together, and concentrating the area of each strip at its centre. Similarly the first moment of a solid figure may be regarded as obtained by dividing the figure into elementary prisms by two sets of parallel planes, and concentrating the volume of each prism at its centre. This also holds for higher moments, provided that the edges of the elementary strips or prisms are parallel to the line or plane with regard to which the moments are taken.
33. Solids and Surfaces of Revolution.—The solid or surface generated by the revolution of a plane closed figure or a plane continuous line about a straight line in its plane, not intersecting it, is a solid of revolution or surface of revolution, the straight line being its axis. The revolution need not be complete, but may be through any angle.
The section of a solid of revolution by a plane at right angles to the axis is an annulus or a sector of an annulus (fig. 5), or is composed of two or more such figures. If the solid is divided into elements by a series of such planes, and if h is the distance between two consecutive planes making sections such as ABDC in fig. 5, the volume of the element between these planes, when h is very small, is approximately h X AB X arc PQ = h. AB. OP.0. The corresponding element of the revolving figure is approximately a rectangle of area h. AB, and OP is the distance of the middle point of either side of the rectangle from the axis. Hence the total volume of the solid is M.0, where M is the sum of the quantities h.AB.OP, i.e. is the moment of the figure with regard to the axis. The volume is therefore equal to S. y.0, where S is the area of the revolving figure, and Si is the distance of its centroid from the axis.
Similarly a surface of revolution can be divided by planes at right angles to the axis into elements, each of which is approximately a section of the surface of a right circular cone. By unrolling each such element (§ 30) into a sector of a circular annulus, it will be found that the total area of the surface is M'.9=L.a.0, where M' is the moment of the original curve with regard to the axis, L is the total length of the original curve, and 2 is the distance of the centroid of the curve from the axis. These two theorems may be stated as follows:
(i) If any plane figure revolves about an external axis in its plane, the volume of the solid generated by the revolution is equal to the product of the area of the figure and the distance travelled by the centroid of the figure.
(ii) If any line in a plane revolves about an external axis in the plane, the area of the curved surface generated by the revolution is equal to the product of the length of the line and the distance travelled by the centroid of the line.
These theorems were discovered by Pappus of Alexandria (c. A.D. 300), and were made generally known by Guldinus (c. A.D. 1640). They are sometimes known as Guldinus's Theorems, but are more properly described as the Theorems of Pappus. The theorems are of use, not only for finding the volumes or areas of solids or surfaces of revolution, but also, conversely, for finding centroids or centres of gravity. They may be applied, for instance, to finding the centroid of a semicircle or of the arc of a semicircle.
34. Segment of Parabola.—The parabola affords a simple example of the use of infinitesimals. Let AB (fig. 6) be any arc of a parabola; and suppose we require the area of the figure bounded by this arc and the chord AB.
Draw the tangents at A and B,
meeting at T; draw TV parallel to
the axis of the parabola, meeting the
arc in C and the chord in V; and
M draw the tangent at C, meeting AT
and BT in a and b. Then (see
PARABOLA) TC=CV, AV=VB, and K ab is parallel to AB, so that aC=Cb.
Hence area of triangle ACB = twice
area of triangle aTb. Repeating the process with the arcs AC and CB, and continuing the repetition indefinitely, we divide up the required area and the remainder of the triangle ATB into corresponding elements, each element of the former being double the corresponding elements of the latter. Hence the required area is double the area of the remainder of the triangle, and therefore it is twothirds of the area of the triangle.
The line TCV is parallel to the axis of the parabola. If we draw a line at right angles to TCV, meeting TCV produced in M and parallels through A and B in K and L, the area of the triangle ATB is iKL. TV = KL .CV; and therefore the area of the figure bounded by AK, BL, KL and the arc AB, is
i(L }(AK+BL)+IKL1CM—4(AK+BL)} .=}KL(AK+4CM+BL).
Similarly, for a corresponding figure K'L'BA outside the parabola, the area is
eK'L'(K'A+4M'C+LB).
35. The Ellipse and the Ellipsoid.—For elementary mensuration the ellipse is to be regarded as obtained by projection of the circle, and the ellipsoid by projection of the sphere. Hence the area of an ellipse whose axes are 2a and 2b is Trab ; and the volume of an ellipsoid whose axes are 2a, 2b and 2c is tirabc. The area of a strip of an ellipse between two lines parallel to an axis, or the volume of the portion (frustum) of an ellipsoid between two planes parallel to a principal section, may be found in the same way.
36. Examples of Applications.—The formulae of § 24 for the area of a trapezoid are of special importance in landsurveying. The measurements of a polygonal field or other area are usually taken as in § 25 (ii) ; a diagonal AE is taken as the baseline, and for the
points B, C, D, . there are entered the distances AN, AP,
AQ, along the baseline, and the lengths and directions of the offsets NB, PC, QD, . . . The area is then given by the formula of § 25 (ii).
37. The mensuration of earthwork involves consideration of quadrilaterals whose dimensions are given by special data, and of prismoids whose sections are
such quadrilaterals. In the ordinary case three of the four lateral surfaces of the prismoid are at right angles to the two ends. In special cases two of these three lateral surfaces are equally inclined to the third.
(i) In fig. 7 let base BC=2a, and let h be the distance, measured at right angles to BC, from the middle point of BC to AD. Also, let angle ABC =a0, angle BCD=r4', angle between BC and AD =0. Then (as the difference of two triangles)
area ABCD= (h cot 4.+a)2 (h cot 4'—a)2
2(cot 4'—cot 4') 2(COt ¢+cot e)'
(ii) If 4'=0, this becomes
tang
area =tan, B—tang 4'(h +a tan 0)2 — a2 tan 0.
(iii) If ¢ =o, so that AD is parallel to BC, it becomes
area = 2ah+ z (cot 0 + cot ¢)h2.
(iv) To find the volume of a prismoidal cutting with vertical ends, and with sides equally inclined to the vertical, so that 0=0, let the values of h, 4' for the two ends be and ha, 4'a, and write
ot cot 01
m,— cot 4" cot o (a + h, cot B), n,= cot ,p, + cote (a + hi cot 0), m cot ,k
cot
2 cot B (a + h2 cot B), n2FTe c°t 02
+ cot B (a + ha cot 0). cot 02 Then volume of prismoid =length X slmini +man,+
1 (mina + man,) 3a2} tan 0.
End of Article: MENSURATION OF SPECIFIC FIGURES 

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