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50 40

90
60

2)5400

2700 Anf. Ex. 2. In the trapezium ABCD, AB is 64, BC 46, CD 60, DA 66, and the diagonal AC 72, English chains. To find ABC.

To find ACD. 64 90 91 91

72 99 99 99 64 46 72

60

72 60 66 72

66 27 45 19

27 39 33 2) 182

2)198

40

91 = 1.95904
27 = 1.43136
45 = 1.65321
19 = 1.27875

99 = 1.99563
27 = 1.43136
39 = 1.59106
33 = 1.51851

2)6.32236

2)6.53656 1449.5 = 3.16118* 1855 = 3.26828

BAC 1449.5

1855 ACD=

10)3304.5

330.45

1.8
40

32.00

Ans. 330 as I rood 32 perches

Ex. 3. Required the area of the trapezium ABCD, whose diagonal AC is 20, and perpendiculars BE and DE, 8 and 10 Scots chains.

Anf. 18 acres. Ex. 4. In the trapezium ABCD, the sides AB is 45, BC 39, CD 42, DA 36, and the diagonal AC 48: Required the area.

Anf. 1552, 7223. Ex. 5. Required the area of the trapezium ABCD, whereof the fide AB is 10.25, BC 35, CD 50, DA 30, and the diagonal AC 40 chains. Anf. 76 acres 2 roods 19 falls 6 ells,

Ex. 6. How many square yards paving are in a trapezium, whose diagonal is 20, and perpendiculars 104 and 6 feet?

Anf. 19 yards 2 feet. Ex. 7. How many acres are in a field ABCD, of which thę fide AB is 8000, AD 6000, and AC the diagonal 9560 links of the Seots chain : Also the angles * BAC, CAÐ are each of them 30°?

Ans: 334 acres 2 roods 16 fallsa

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To find the area of a trapezium, its two diagonals and the included

angle being given.

RULE.
As radius,
Is to the fine of the included angle
So is į product of the diagonals,

to the area.

EXAMPLE

It will be worth the learner's while to observe, that when one of the angles of a right angled triangle is 30°, the leg opposite to it will be exactly one half of the hypothenuse. Hence the perpendiculajs BF and DE are 40co, 3000 thro halves of the fides AB, AD.,

EXAMPLE I.

Required the area of a trapezium, whose diagonals are 100, 80 feet, and the included angle 60°. As radius 90°

10.00000 is to sine 60°

9.93753 So is prod. 4000

3.60206 To the area 3464 3.53959 Ex. 2. Required the area of a trapezium, whose diagonals are 120, and 140 yards, and the included angle 30°.

Ans 4200 square yards. Ex. 3. What is the area of a trapezium, of which the diagonals are 80 and 60 Scots chains, and the included angle 60°?

Anf. 207 acres 3 roods 8 falls. If the trapezium be inscribed in a circle, its area may be found by the following rule.

Add all the four fides together; from half their sum fubtract the sides severally; then multiply the remainders continually into each other, and the square root of the last product will be the area.

EXAMPLE. Required the area of a trapezium, whose fides are 12, 13, 14, 15

27 27 27 27 13

13 14 14 15 15 14 13 12

12

12

15

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2754

27

15*14*13*12=32760 and 32760=180,997 Anf.

PROBLEM

PROBLEM XI.

To find the area of an irregular polygon.

RULE.

Resolve the polygon into triangles by diagonals ; find the area of each triangle separately, and their sum will be the area of the whole polygon.

EXAMPLE I.

Required the area of the following figure, ABCDEF, whose perpendiculars and diagonals are given. Fig. 78. plate 6.

AC = 10507
Bb = 320
Ff = 420
FD =

links.

960 Cc = 600

Ee = 200 To find the area of ABC.

To find ACE.

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24.20000

Anf. 7 acres 3 roods 8 perches 245 yards. The above example may be rendered more simple, by reducing the figure to trapezias. Bb = 320

1050 Ff = 420

370

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Anf. 7 ac 3 ra 8 per. 247 yds, 7.80500 as before.

Required

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