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2. What is the area of a triangle two sides of which measure 15.36 chains and 11.46 chains respectively, and their included angle 47° 30? Ans. 6A. 1R. 38P.

3. One side of a triangular field bears N. 12° E. distance 18.23 chains, and at the same station the other adjacent side bears N. 78° 30 E. distance 13.84 chains: required the area. Ans. 11A. 2R. 11P.

4. Required the area of a triangular piece of ground, one side of which bears N. 82o 30'W. dist. 19.74 chains and at the same station, the other adjacent side S. 24° 15'E. dist. 17.34 chains. Ans. 14A. 2R. 8P.

PROBLEM IV.

To find the area of a triangle when one side and the two adjacent angles are given.

RULE.

Subtract the sum of the two given angles from 180°, the remainder will be the angle opposite the given side. Then,

As the rectangle of radius and the sine of the angle opposite the given side,

Is to the rectangle of the sines of the other angles,
So is the square of the given side,

To double the area.*

* DEMONSTRATION. Let AB, Fig. 49, be the given side of the triangle ABC, and A and B the given angles; also let CD be perpendicular on AB: Then by trig.

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Therefore (C.6) rad. × sin. ACB: sin. A × sin. B:: AB × AC: CD X AC (cor. 1.6) AB: CD:: AB2: AB x CD; but AB x CD is equal to double the area of the triangle ABC; therefore (11.5) rad. × sin. ACB: sin. A × sin. B : : AB2: double the area of the triangle

EXAMPLES.

1. In a triangular field ABC, the side AB measures 76 perches, the angle A 60°, and the angle B 50°: required the area. Fig. 47.

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2. One side of a triangle measures 24.32 chains, and the adjacent angles are 63° and 74°; required the area. Ans. 37A. OR. 22P.

3. What is the area of a triangular field, one side of which is 17.36 chains, and the adjacent angles 37° 30′ and 48° 15'? Ans. 6A. 3R. 18P.

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To find the area of a triangle, when the three sides are

given.

RULE.

From half the sum of the three sides subtract each

mainders continually together, and the square root of the last product will be the area.*

* DEMONSTRATION. Let ABC, Fig. 69, be the triangle. Bisect 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 similar triangles ADG, AHK, AD : DG :: AH: HK; therefore (C.6) AD × BD : DG2 :: AH : HB : : AH2 : AH × 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: DG2 :: AH2 : N2; hence (22.6) M : DG :: AH: N; consequently the rectangle M x N is equal to the rectangle AH × DG; therefore ABC = ABG + BCG + ACG = AH × DG =M × N = √(AD × BD) × √ (AH × HB) = √(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;

1. Required the area of a triangular tract of land, whose three sides are 49.00, 50.25 and 25.69 chains.

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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.

PROBLEM VI.

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.

RULE.

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.

EXAMPLES.

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

12.88

20.64

5152

7728

2576

2)265.8432

132.9216Ch. 13A. 1R. 6P.

2. Required the area of a trapezium, whose diagonal measures 16.10Ch. and the perpendiculars 6.80Ch. and 3.40Ch. Ans. 8A. OR. 333P.

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 =

AC X DE BF + DE

+

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