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mainders continually together, and the square root of the last product will be the area. *
* DEMONSTRATION. Let ABC, Fig. 69, be the triangle. Bisect any two of the angles, BAC, ABC, by the straight lines AG, BG, meeting in G; let fall on the three sides of the triangle, the perpendiculars GD, GF, GE, and join GC; also produce AB, AC, and bisect one of the exterior angles, HBC, by the line BK, meeting AG* produced in K, join KC, and let fall the perpendiculars KH, KM, and KL. Then (26.1) AD is equal to AE and DG to GE; also BD is equal to BF and DG to GF; hence GF and GE are equal, and consequently (47.1) CF is equal to CE. In like manner it may be proved that AH is equal to AL, BH to BM, and CM to CL; as likewise that KH, KM, and KL are equal to each other. Now since BH is equal to BM and CL to CM, it is manifest that AH and AL together, are equal to the sum of the three sides AB, AC, and BC; hence AH or AL is equal to the semiperimeter of the triangle ABC. But since twice AD, twice BD, and twice CF are the sum of the sides of the triangle, or twice AH, it is obvious that AD, BD and CF together, are equal to AH; consequently CF is equal to BH or BM; hence CM or CL is equal to BF or BD; and therefore DH and BC are equal.
If now from the semiperimeter AH, the three sides AB, AC and BC be severally taken, the remainders will be BH, CL, (or BD) and AD respectively.
Again, since the angles DBF and DGF are together equal to two right angles, as likewise DBF and FBH together equal to two right angles, it is manifest that the angle DGF is equal to the angle HBF; and the angle DGB to the angle HBK; the triangles DBG and HKB are therefore similar. Hence BD: DG::KH: HB; also in the simi. lar triangles ADG, AHK, AD: DG :: AH : HK; therefore (C.6) AD X BD : DG2 : : AH :HB:: AH: AH X HB.
If therefore we take between AD and BD, and between AH and HB, the mean proportionals M and N respectively, the foregoing analogy will become M2: DG? :: AH? : No; hence (22.6) M: DG:: AH:N; consequently the rectangle Mx N is equal to the rectangle AH X DG; therefore ABC ABG + BCG + ACG AH X DG
MXN=V (AD X BD) XV (AH X HB)=V (AB X BD X HB X AH).
The angle BAC is less than the angle HBC (16.1); consequently BAG is less than HBK, and BAG, KBA, are together less than HBK, KBA; but HBK, KBA, are together equal to two right angles; hence BAG, KBA, are less than two right angles; therefore (cor. 29.1) the line BK will meet the line AG produced.
1. Required the area of a triangular tract of land, whose three sides are 49.00, 50.25 and 25.69 chains.
61.5 Acres = 61A. 2R. 2. What is the area of a triangular field whose sides measure 10.64, 12.28 and 9.00 chains?
Ans. 4A. 2R. 26P.
3. What quantity of land is contained in a triangle, the sides of which are 20, 30 and 40 chains?
Ans. 29A. OR. 7P.
To find the area of a trapezium, when one of the diagonals
and the two perpendiculars let fall on it from the opposite angles, are given.
Multiply the sum of the perpendiculars by the diagonal, and half the product will be the area.*
AC X BF
Note. When all the sides and one of the diagonals are given, the trapezium will be divided into two triangles, the area of each of which may be found by the last problem. The sum of these areas will be the area of the trapezium.
1. In a field ABCD in the form of a trapezium, the diagonal AC measures 20.64 chains, the perpendicular BF 6.96 chains, and DE 5.92 chains: required the area. Fig. 70.
Ch. 6.96 5.92
5152 7728 2576
132.9216Ch. = 13A. IR. 6P.
2. Required the area of a trapezium, whose diagonal measures 16.10Ch. and the perpendiculars 6.80Ch. and 3.40Ch. Ans. 8A. OR. 33 P.
AC X DE and the area of the triangle ADC
; therefore the sum of 2
AC X BF these areas, or the area of the trapezium ABCD
*2 ACX DE BF + DE
X AC. Fig. 70. 2
3. The diagonal of a trapezium is 24Ch. and the perpendiculars are 8.27 Ch. and 12.43 Ch.; what is the area?
Ans. 24A. 3R. 14P.
PROBLEM VII. To find the area of a trapezium, when all the angles and
two opposite sides are given.
Note. When three of the angles are given the fourth may
be found by subtracting their sum from 360°.
Consider one of the given sides and its adjacent angles, or their supplements when their sum exceeds 180°, as the side and adjacent angles of a triangle, and find its double area by prob. 4. Proceed in the same manner with the other given side and its adjacent angles: Half the difference of the areas thus found will be the area of theftrapezium. *
1. In a four-sided field ABCD, there are given the following bearings and distances, viz. AB, N. 24° E. dist. 6.90 Ch. BC, N. 64° 40' E. CD, S. 35° 20' E. dist. 11.50 Ch. and DA, S. 88° W.: required the area. Fig. 71.
* DEMONSTRATION. Let AB, CD, Fig. 71, be the given sides of the trapezium ABCD. Produce DA, CB, to meet in E; then 2 ABCD
2 EDC -2 EAB. 2 EDC -2 EAB or ABCD =
Hence the truth
From the given bearings, the angles may be found as follows:
Construction. Make AB = 6.90, and draw DA, CB, making the angle DAB = 64°, and ABC = 139° 20'; produce DA and make the angle EAF = 56° 40' = the given angle ADC; lay off AF = 11.50 = the given side BD, and parallel to AD draw FC, meeting BC in C; lastly draw CD parallel to AF, meeting AD in D, then will ABCD be the trapezium.*
* DEMONSTRATION. By construction FC is parallel to AD and CD to AF, therefore (34.1.) CD=AF and (29.1.) the angle ADC = EAF; hence it is evident that the sides AB, CD, and the angles of the trapezium ABCD are respectively equal to the given sides and angles.