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2. Required the area of a triangle, two sides of which are 49.2 and 40.8 perches, and their contained angle 144 degrees

Answer, 3A. 2R. 22P.

3. What quantity of ground is inclosed in an equilateral triangle, each side of which is 100 perches, either angle being 60 degrees?

Answer, 27A. 10P.

PROB. X.

To find the area of a trapezoid, viz. a figure bounded by four right lines, two of which are parallel, but unequal.

RULE.*

Multiply the sum of the parallel sides by their perpendicular distance, and take half the product for the area.

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EXAMPLES.

1. Required the area of a trapezoid, of which the parallel sides are, respectively, 30 and 49 perches, and their perpendicular distance 61.6?

61.6

30+49=79.

2)4866.4

Multiply.

Answer, 2433.2-15A. 33.2P.

* Demonstration. The trapezoid ABCD (Pl. 14. Fig. 8.) is equivalent to the rectangle contained by its altitude and half the sum of the parallel_sides BC and AD. For draw CE parallel to AB (Prob. 8.), bisect ED in F, and draw FG parallel to AB, meeting the production of BC in G. Because BC is equal to AE, BC and AD are together equal to AE and AD, or to twice AE with ED, or to twice AE and twice EF, that is to 2 AF; consequently AF(BC+AD). Wherefore the rectangle contained by the altitude of the trapezoid and half the sum of its parallel sides, is equivalent to the rhomboid BF; but the rhomboid EG is equivalent to the triangle ECD (Theo. 12. Cor. 2), add to each the rhomboid BE, and the rhom boid BF is equivalent to the trapezoid ABCD.

Note. On this proposition is founded the method of off-sets, which enters so largely into the practice of land-surveying. In measuring a field of a very irregular shape, the principal points only are connected by straight lines, forming sides of the component triangles, and the distance of each remarkable flexure of the extreme boundary is taken from these rectilineal traces. The exterior border of the polygon is therefore considered as a collection of trapezoids, which are measured by multiplying the mean of each pair of off sets or perpendiculars into their base or intermediate distance, which is one of the other sides, because the parallel sides are perpendicular to it.

PL. 9. fig. 10.

2. In the trapezoid ABCD the parallel sides are, AD, 20 perches, BC, 32, and their perpendicular distance, AB, 26; required the content? Answer, 4A. 36P.

PROB. XI.

To find the Content of a Trapezium,
RULE I.*

Multiply the diagonal, or line joining the remotest opposite angles, by the sum of the two perpendiculars falling from the other angles to that diagonal, and half the product will be the

area.

EXAMPLE.

PL. 7. fig. 3.

Let ABCD be a field in form of a trapezium, the diagonal AC 64.4 perches, the perpendicular Bb 13.6 and Dd 27.2, required the content?

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* Demonstration. For the trapezium ABDC=the triangles ABC+ADC ACX Bb, ACxDd_Bb+Dd

2

+

·XAC. Q. E. D.

2

NOTE. The method of multiplying together the half-sums of the opposite sides of a trapezium for the content is erroneous, and the more so the more oblique its angles are.

To draw the map set off Ab 28 perches, and Ad 34.4, and there make the perpendiculars to their proper lengths, and join their extremities to those of the diagonal.

NOTE. When one of the diagonals and the four sides of a trapezium are given, it is divided into two triangles, whose sides are given; the area of each triangle may be found (by prob. 8.), and their sum will give the area of the trapezium.

RULE II.*

If there be drawn two diagonals cutting each other, the product of the diagonals, multiplied by the natural sine of the angle of intersection of the diagonals, will be double the area. And this rule is common to a square, rhombus, rhomboides, &c. as well as to all other quadrilateral AC × BD ×x Nat. S R figures. That is,

2

= the

area. Pl. 14. Fig. 9. Or, as radius: S. R ACX BD: the area.

1

1

* Demonstration. Pl. 14. Fig. 9. For the trapez. the four A's ARB, BRC, CRD, DRA = ·(AR×RB+BR>RC+CR×RD+DR×RA) X + S. 2R = (AR + RC × BR+ CR+RA × DR) × † S. L R = AR+RC× BR+RD× 3 S. LR=AC× BD× à S. LR. Q. E. D.

NOTE. Because the diagonals of a square and rhombus intersect at a right angle, whose sine is 1, therefore half the product of their diagonals is the area.

EXAMPLE.

Let the two diagonals be 40 and 30 chains, and at their intersection one of the less angles is 48°, the area is required?

Then, since the natural sine of 48° is .7431448, 40 x 30 x .7431448

the area =

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600 x .7431448

=445.88688 square chains = 44 A. 2R. 14.19P.

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To find the area of a trapezium, when three sides and the two included angles are given.

EXAMPLE.

In a quadrangular field the south side is 23.4, the east side 19.75, and the north side 20.5 chains; also the south-east and north-east angles are 73° and 87° 30': what is the area?

First, (by rule 2. prob.6.).

.9990482×19.75x20.5 2

.4995241 x 19.75 x 20.5 = 202.24482, the area of the north-east triangle BCD. Pl. 14. Fig. 10.

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