Page images

in feet, or yards, the remainder should be divided by the number of square feet, or square yards in an acre, then, the remainder by the number of square feet or square yards in a rood, and the remainder again by the number of square feet or square yards in a perch.

As precisely the same remarks are applicable to all future examples, they need not be repeated.


[ocr errors][ocr errors]

112. To find the area of a triangular piece of ground.

1. If the base and perpendicular are known, multiply them together, and divide the product by 2, the quotient is the area of the triangle. (176)*.

2. If two sides and the included angle are known, add together the logarithms of the sides, and the logarithmic sine of the contained angle; from this sum subtract 10, which is the logarithm of the radius, and the remainder will be the logarithm of double the area of the triangle. Find, from the tables, the number answering to this logarithm, and divide it by 2, the quotient is the required area.

Example 1.—Let AB=57.65 chains, AC=125.81 chains, and the angle CAB=57° 25', and 2Q=twice the required area. +log. AB=log. 57.65


2.099715 +log. AC=log. 125.81 Then log. 2Q= +log. Sin. A=log. Sin. 57° 25' 9.925626

-10 Log. 2Q=


[ocr errors]

-log. R

* Let ABC (Pl. I. Fig. 4) be the triangle, having the sides AB, AC, and the contained angle BAC given.

Let BD be drawn perpendicular to the base AC, and Q put equal to the area of the triangle.

AB X Sin. A
Then R: AB :: Sin. A:BD (43); or BD=

But, Q=

(176)*.; hence by substituting for BD its value,

or 2Q=

Taking the logarithms, log. ?Q=log. AC+log. AB+log. Sin. A--!og. R,

[ocr errors]

And 2Q=6111.4 square chains, or Q, the area of the triangle, equal to 3055.7 square chains, equal to 305 acres, 2 roods, and 11.2 perches.

Example 2.-What is the area of a triangle whose sides are 30 and 40, and their contained angle 28° 57' ? Ans. 290.4276.

Example 3.-What is the number of square yards contained in a triangle, of which the sides are 25 and 21.25 feet, and the included angle 45°? Ans. 20.8694.

3. If the three sides are known, add them together and take half their sum ; from this half sum subtract each of the sides severally; then multiply the half sum and the three remainders together; the product is the square of the area of the triangles: then extract the square root for the required

area. *

Or, after having obtained the three remainders, add together the logarithm of the half sum and the logarithms of the respective remainders, and divide the sum by two: the quotient will be the logarithm of the area.

* Demonstration from Hutton's Mensuration. Let ABC (pl. 1, fig. 19,) be the given triangle. Draw the parallels BD, AE meeting the two sides AC, and CB produced in the points D and E, and making CD=CB, and CE=CA. Also draw CFG bisecting DB and AE perpendicularly in F and G. Draw FHI parallel to AB, meeting CA in H, and AE produced, in I. Lastly, with the centre H, and radius HF, describe a circle, meeting AC produced in K: this circle will pass through I, because AI=FB=FD, hence HF=HI=AB; and it will also pass through the point G, because CGA is a right angle.

Hence HA or HD is half the difference of the sides AC, CB; and HC= half their sum=}AC+CB; also KH=HI= IF=AB; consequently, CK =AC+*CB+AB=half the sum of the sides of the triangle ABC, equal }S, calling S the sum of those sides. Again HK=HI=IF=fAB, or KL= AB; therefore, CL=CK – KL={S-AB, and AK=CK-CA=}S-AC, and AL=DK=CK-CB= S-CB.

Now AGXCG=the area of the triangle ACE, and AGXFG=ABE; therefore AGXCF=ACB. Also by the parallels AG:CG::DF: CF, or AI :CF; therefore, AGXCF=triangle ABC=CGXAI=CGXDF; consequently, AGXCF XCGXDF=the square of the area ABC.

But CGXCF=CKXCL='SX(}S:- AB); and AGXDF=AKXAL= (18-AC)X(IS-BC), therefore, AGX CFXCGXDF=}S(JS-AB)x (S- AC)*(S - BC)=the square of the area of the triangle ABC.

[ocr errors]
[ocr errors]

Example 1.—To find the area of a triangle whose three
sides are 20, 30, and 40.

45 half sum.

25 1st rem.
15 2d rem.

5 3d rem. 2390

45 half sum.

Then 45 x 25 x 15 x5=84375.
The square root of which is 290.4737, the required area.

Example 2.—How many acres are contained in the triangle whose sides are 2569, 4900, 5035 links? Ans. 61 acres, 2 roods, 8 perches.

[ocr errors]

PROBLEM. 113. To find the area of a piece of ground in the form of a trapezoid.

Multiply the half sum of the parallel sides by the perpendicular distance between them, the product is the required area. (178.*)

Example 1.-Required the area of the trapezoid, of which the parallel sides are, respectively, 30 and 49 rods, and their perpendicular distance 61.6.

30+49=79, and, dividing by 2, gives 39.5
Multiply by


Gives for the area in square rods 2433.20 Or 15 acres, 33.2 perches.

Example 2.—To find the area when the parallel sides are respectively 20 and 32 rods, and their perpendicular distance 26. Ans. 4 acres, 0 roods, and 36 perches.


114. To find the area of a quadrilateral.

Draw a diagonal dividing it into two triangles; find the areas of the triangles: the sum of these areas is the area re


To measure long irregular figures having a right line for a base.

115. At the two extremities of the base, measure in perpendiculars to the base, the breadth of the figure. Then divide the base into any convenient number of equal parts, and measure the breadth of the figure at the points of division. Add together the intermediate perpendiculars and half the sum of the extreme ones, and multiply this sum by one of the equal parts of the base line; the product is the required area very nearly; or, multiply by the whole base, and divide the product by the number of equal parts.*

If it is not convenient to erect the perpendiculars at equal distances from each other, the areas of the trapezoids into which the whole figure will be divided, must be computed separately, and the sum taken.

Example 1.—The breadths of an irregular figure at five equidistant places, being 8.2, 7.4, 9.2, 10.2, 8.6, and the whole length 40, required the area.



* Let ABEeda (pl. 1, fig. 20.) be an irregular figure ; AE its base, h, i, 1, n, m, the perpendiculars drawn at equal distances from each other. The area of the trapezoid ABba, is equal to




[ocr errors]


The area of the trapezoid BCch=


1+n The area of CDder


2 The area of DEed

netim XDE

2 The sum of these trapezoids, which is equal to the area of the figure, is

it Itn, ntm equal to


2 equal.




+ + ' * + stk)x AB, since the divisions on AE But this latter sum is evidently equal to (+m+ i +1+)*AB,

C++ n


[blocks in formation]

35.2 sum. Example 2.-The length of an irregular figure being 84, and the breadths, at six equidistant places, 17.4, 20.6, 14.2, 16.5, 20.1, 24.4 rods; required the area. Ans.-9 acres, 2 roods, 30.64 perches.

[ocr errors]



116. To find the area of a circle.

Square the radius and multiply the number so obtained by 3.1416, the product is the required area. (291.*)



117. To find the area of an ellipsis.

Multiply the axes together, and their product by 7854, the last product will be the area of the ellipsis.

Example. Required the area of an ellipsis, whose two axes are 70 and 50 poles. Ans.--17 acres, 28.9 perches.

118. With respect to the content of land, it may be remarked, that irregular pieces must, in general, be divided into triangles, rectangles, parallelograms, or trapezoids. The manner of making such divisions, depends on the lines and angles which are known. They should always, however, be so made as to simplify the calculations, and so that a sufficient number of known parts may fall into each figure, to

« PreviousContinue »