3. The diagonal of a trapezium is 24Ch. and the perpendiculars are 8.27 Ch. and 12.43 Ch.; what is the area? Ans. 24A. 3R. 14P.

PROBLEM VII.

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

Note. When three of the angles are given the fourth may be found by subtracting their sum from 360°.

RULE.

Consider one of the given sides and its adjacent angles, or their supplements when their sum exceeds 180°, as the side and adjacent angles of a triangle, and find its double area by prob. 4. Proceed in the same manner with the other given side and its adjacent angles: Half the difference of the areas thus found will be the area of the trapezium.**

EXAMPLES.

1. In a four-sided field ABCD, there are given the following bearings and distances, viz. AB, N. 24° E. dist. 6.90 Ch. BC, N. 64° 40′ E. CD, S. 35° 20′ E. dist. 11.50 Ch. and DA, S. 88° W.: required the area. Fig.

71.

* DEMONSTRATION. Let AB, CD, Fig. 71, be the given sides of the trapezium ABCD. Produce DA, CB, to meet in E; then 2 ABCD:

From the given bearings, the angles may be found as

=

Construction.

=

Make AB 6.90, and draw DA, CB, making the angle DAB = 64o, and ABC = 139o 20'; produce DA and make the angle EAF 56° 40′ the given angle ADC; lay off AF = 11.50 = the given side BD, and parallel to AD draw FC, meeting BC in C; lastly draw CD parallel to AF, meeting AD in D, then will ABCD be the trapezium.*

Calculation.

The angle E=180° the sum of the angles BCD, ADC23° 20'.

* DEMONSTRATION. By construction FC is parallel to AD and CD to AF, therefore (34.1.) CD=AF and (29.1.) the angle ADC EAF; hence it is evident that the sides AB, CD, and the angles of the trapezium ABCD are respectively equal to the given sides and angles.

N

ABCD 102.163 Ch. 10A. OR. 34.6P.

2. In a trapezium ABCD, the angles are, A = 65o, B=81o, C=120°, and consequently D = 94°; also the side AB=20 Ch. and CD = 11 Ch.: required the

area.

Ans. 22A. 2R. 27P.

3. Required the area of a four-sided piece of land, bounded as follows:

1. N. 12° 30′ E.

2. N. 81 00 E. dist. 23.20 Ch.

3. S. 36 00 W.

4. N. 89 00 W. dist. 12.90 Ch.

墨

PROBLEM VIII.

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

As radius,

RULE.

Is to the sine of one of the given angles;

So is the rectangle of the sides including this angle,
To a certain quantity.

As radius,

Is to the sine of the other given angle;

So is the rectangle of the sides including this other angle,

To a second quantity.

Take the difference between the sum of the given angles and 180°; Then,

As radius,

Is to the sine of this difference;

So is the rectangle of the opposite given sides,
To a third quantity.

If the sum of the given angles be less than 180°, subtract the third quantity from the sum of the other two, and half the difference will be the area of the trapezium. But if the sum of the given angles exceed 180°, add all the three quantities together and half the sum will be the area.*

* DEMONSTRATION. Let ABCD (Fig. 72 or 73) be the trapezium, having the given sides, AD, AB, BC, and given angles DAB, abc. Complete the parallelograms ABCE, ABFD, and join ED, CF; then because EC, DF, are each parallel and equal to AB, they are (30.1.), parallel and equal to each other, and (33.1.) ECFD is a parallelogram; therefore ABFD = ABHG + GHFD = (35.1.) ABCE + ECFD (34.1.) ABCE + 2 ECD; to the first and last of these equals add ABCE, then ABFD + ABCE = 2 ABCE + 2 ECD = 2 ABCDE.

=

But

EXAMPLES.

1. In a trapezium ABCD, there are given AD = 23.32 Ch., AB = 25.70 Ch., and BC 15.84 Ch., the angle DAB = 64° and ABC = 82°: required the area.

As rad.

Ar. Co. 0.00000

But Fig. 72, when the sum of the given angles DAB, ABC, is less than 180o, 2 ABCDE = 2 ABCD + 2 EAD; therefore in this case ABFD + ABCE 2 ABCD + 2 EAD; or ABFD + ABCE 2 EAD 2 ABCD.

And, Fig. 73, when the sum of the given angles DAB, ABC, exceeds 180o, 2 ABCDE = 2 ABCD. - 2 EAD; therefore ABFD + ABCE 2 ABCD 2 ABCD.

2 EAD; or ABFD + ABCE + 2 EAD =

But by prob. 3. one of the first two proportions gives 2 BAD (= ABFD), and the other gives 2 ABC (ABCE); also because the angle EAD is the difference between the sum of the given angles and 180°, and the side EA BC, the third proportion gives 2 EAD: