2. What is the area of a four-sided lot of ground, three sides of which, taken in order, measure 6.15, 8.46 and 7.00 chains, respectively, the angle contained by the first and second sides 56°, and that contained by the second and third sides 98° 30'? Ans. 4A. OR. 25P.

3. One side of a quadrilateral piece of land bears S. 7 E. dist. 17.53 Ch., the second, N. 87 E. dist. 10.80 Ch. and the third, N. 25 E. dist. 12.92 Ch.: what is the area? Ans. 21A. 3R. 2P.

Note. As in triangles any three parts, except the three angles, being given, the area may be found, so in trapeziums, unless two sides are parallel, any five parts except the four angles and one side, being given, the area may be found. Several other problems might therefore be introduced for finding the areas of triangles and trapeziums, depending on the different parts, sufficient to limit them, that may

be given; but as they seldom occur in practice, and when they do may readily be solved by means of trigonometry and the preceding problems, they are omitted.

Multiply the sum of the parallel sides by their perpendicular distance, and half the product will be the

area.*

EXAMPLES.

1. Required the area of a trapezoid ABCD, of which the parallel sides AD, BC measure 6.14 and 9.48 chains respectively, and their perpendicular distance BF or DE, 7.80 chains.

Ch.

6.14

9.48

15.62

7.80

124960

10934

2)121.8-60

60.9180 Ch. 6A. OR. 15P.

* DEMONSTRATION. The trapezoid ABCD, Fig. 74, the triangle

AD X BF BC X DE,

ABD + BDC (by prob. 2),

+

2

AD X BF BC X BF

BF = DE,)

+

2

AD + BCX BF

= (because

2. The parallel sides of a trapezoid are 12.41 and 8.22 chains, and their perpendicular distance 5.15 chains: required the area. Ans. 5A. 1R. 10P.

3. Required the area of a trapezoid whose parallel sides are 11.34 and 18.46 chains, and their perpendicular distances 13.25 chains. Ans. 19A. 2R. 39P.

PROBLEM X.

To find the area of a circle, or of an ellipsis.*

RULE.

Multiply the square of the circle's diameter, or the product of the two diameters of the ellipsis, by .7854, for the area.†

Note. 1. If the diameter of a circle be multiplied by 3.1416, the product will be the circumference; also if the circumference be divided by 3.1416, the quotient will be the diameter.

2. If the area of a circle be divided by .7854, the square root of the quotient will be the diameter.

* If two pins be set upright in a plane, and a thread, the length of which is greater than twice the distance between the pins, having the ends tied together be put about the pins; and if the point of a pin or pencil, applied to the thread, and held so as to keep it uniformly tense, be moved round, till it return to the place from which the motion began; then the point of the pin or pencil will have described on the plane, a curved line called an Ellipsis.

†The demonstration of this rule is too abstruse to admit of a place in this work. The student who wishes to see a demonstration is re

EXAMPLES.

1. How many acres are in a circle a mile in diameter?

1 mile = 80 Ch.

80

6400

.7854

3141600

47124

5026.5600 Sq. Ch. 502A. 2R. 25P. nearly.

2. Required the area of an ellipsis, the longer diameter of which measures 5.36 Ch. and the shorter 3.28 Ch.

Ch. 5.36

3.28

4288

. 1072

1608

17.5808

.7854

703232

879040

1406464

1230656

3. Required the area of a circular park, the diameter of which is 100 perches. Ans. 49A. OR. 14P.

4. Required the area of an elliptical fish pond, the longer diameter of which is 10 perches, and the shorter 5 perches. Ans. 39.27 Sq. Per.

PROBLEM XI.

To protract a Survey, and to find its area by dividing it into triangles and trapeziums.

The method of doing this will be best understood by an example. Thus,

Suppose the following field-notes to be given, it is required to protract the survey and find its area.