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METHOD II. By Logarithms.
RULE. Add the logarithms of the three remainders and half sum together, and half their fum will be the logarithm of the area.
10 : 12.5 diff. feg, base.
Now to find BD the perpen, To half base
ABP-AD2=BD’, or Add half diff. 6.25
AB? = 900
To find ang. A.
To find the area. As AD = 26,25 1,41913 | As radius
90 10,00006 is to rad. : 0° 10,00000
is to fine 28° 57' 9,68489 So is AB
30 1,47712 So is ABX AC 600 2,77815
To fec.ang.A 28° 57' 10,05799 | To area 290,4738 = 2.46304
Which is 29 ac. or. 7 falls, 21 ells, as before.
From these four different varieties, it appears, that the lo
garithmic operation is the easiest. It were to be wished, that all land surveyors would take the trouble of computing their measurements by logarithms; then would they agree in their calculations, and depend less upon the accuracy of their scales.
Ex. 2. Required the area of a triangle, whose three dides are 500, 300, and 400 links.
Anf. 2 roods, 16 falls. Ex. 3. Required the area of a triangle, whose sides are 80, 60, 100, feet.
Anf. 2400 feet. Ex 4. How many square yards are in a triangular court, whose three fides are 36, 24, and 30 feet?
Anf. 39 yards, 6.17 feet. Ex. 5. How many square yards are in a triangle, whose three fides are 63, 123,5 and 148 yards ? Anf. 4168 yards.
Ex. 6. How many square yards are in a triangle, whose sides are 39, 42, and
Anf. 84 yards. Ex 7. Required the area of a triangle, whose fides are 90,84. and 78 yards.
Anf. 3024 pards.
PROBLEM VII. Plate 6. fig. 75.
Two sides of a right-angled triangle being given, to find the other side.
To find the hypothenuse, add the square of both the legs, and the square root of the sum is the hypothenuse.
To find one of the legs, subtract the square of the given leg from the square of the hypothenuse, and the square root of the remainder is the leg required.
The hypothenufe is 60, and the base AC 45; required the perpendicular. 45
Ex. 2. Required the length of a ladder, to reach the top of a tower 56 feet high, the foot of the ladder being 48 feet from the wall.
Ans: 73 feet 9.072 inches.
3. The hypothenuse is 600, and one of the legs 360 : Required the other leg.
Anf. 480. Ex. 4. The legs of a right-angled triangle are 64, and 48: Required the hypothenuse.
Anf. 8o. Ex.
5• The hypothenuse of a right-angled triangle is 100, and one of the legs 80: Required the other leg.
PROBLEM VIIL Plate 6. fig. 76.
To find the area of a trapezoid.
Multiply one half of the sum of the parallel fides by the perpendicular distance between them, and the product will be the
Required the area of a trapezoid, whose parallel sides are 15, 19+ chains, and their perpendicular distance 14 chains.
Ex. 2. Required the area of trapezoid, whose sides are 12,185 feet, and the perpendicular distance between 7 feet.
Anf. 106 square feet 2 inches. Ex. 3. Required the area of a trapezoid, the parallel fides being 180 and 200 yards, and their perpendicular distance 100 yards.
Anf. 19000 square yards. Ex. 4 How many square yards are in a trapezoid, whose parallel fides are go and 100 feet, and breadth 50 feet?
Anf. 527 yards 7 feet. Ex. 5. Required the area of a trapezoid, whose parallel fides are 3, 4 feet, and perpendicular breadth 3 feet.
Anf. 101 feet. Ex. 6. How many square feet are in a plank, 13 inches broad at one end, and 15 at the other, the length being 16 feet 5 inches?
Anf 19 feet i inch 10 parts. Ex. 2. Required the expence of causewaying a bridge 150 feet long and 30 broad, at is. 6d. per square yard.
Anf. 37h 10s.
PROBLEM IX. Plate 6. fig. 77.
To find the area of a trapezium.
Resolve the trapezium into triangles; compute the area of each of the triangles separately, and the sum will be the area of the trapezium.
Required the area of a trapezium ABCD, the diagonal AC 60, BF 50, and DE 40 feet.