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

EXAMPLE Ι.

Required the area of a triangle, its three fides being 20, 30, 40 Scots chains.

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METHOD II. By Logarithms.

RULE.

Add the logarithms of the three remainders and half fum together, and half their sum will be the logarithm of the area.

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AC: AB + BC :: AB-BC: AD-DC.

:: 10 : 12.5 diff. feg. base.

Now to find BD the perpen.
AB-AD=BD*, or

BC-DC = BD2

2

That is, 40: 50

To half base

20

Add half diff.

6.25

AB2 = 900

The greater seg. 26.25 AD

AD=689.0625

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2

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 logarithmic operation is the easiest. It were to be wished, that all land furveyors 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 fcales.

Ex. 2. Required the area of a triangle, whose three fides are 500, 300, and 400 links. Ans. 2 roods, 16 falls.

Ex. 3. Required the area of a triangle, whose sides are 80, Anf. 2400 feet.

60, 100, 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 fides Anf. 84 yards.

are 39, 42, and 45 feet ? Ex 7. Required the area of a triangle, whose fides are 90,84.

and 78 yards.

Anf. 3024 yards.

PRO

PROBLEM VII. Plate 6. fig. 75

Two fides of a right-angled triangle being given, to find the other fide.

RULE.

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.

EXAMPLE I.

The hypothenuse is 60, and the base AC 45; required the perpendicular.

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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.

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Ex. 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. 80.

Ex. 5. The hypothenuse of a right-angled triangle is 100, and

one of the legs 80: Required the other leg.

PROBLEM VIII. Plate 6. fig. 76.

To find the area of a trapezoid.

RULE.

Anf. 60.

Multiply one half of the sum of the parallel fides by the perpendicular distance between them, and the product will be the

area.

EXAMPLE I.

Required the area of a trapezoid, whose parallel fides are 15, 19+ chains, and their perpendicular distance 14 chains.

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