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Required the area of the irregular figure ABCDEF, of which the fide AB is 40, AC 5C, AD 55, AE 69, AF 36 Scots chains; and the angles are as follow. Plate 6. fig. 80.

BAC 40°.
CAD 43°
DAE 40° 30':
EAF 48° 20'.

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METHOD II. By finding the perpendiculars.

To find Bb.

To find Cc. As rad. 90° 10.00000 As rad. = 90=.

10.00000 is to AB 40

1,60206 is to AC = 50 = 1,69897 So is fine 40° 9,80807 | So is fine 43° 9,83378

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To Bb 25.71

1,41013 | To Cc 34.1 =

1.53275

To find Dd.

To find Ff.
As rad. 90 =

10,00000
As rad. 90°

10.00000 is to AD = 55 = 1,74036

is to AF = 36 = 1.55630 So is fine 40° 30' 9,81254 So is fine 48° 20' - 9,87334

To Dd 35,72

1,55290 | To Ff 26.89 =

1,42964

Now, to find the area by bases and perpendiculars.

25.71 X 50 = 1285.5
34,1 x 55 = 1875-5
35.72 x 69 = 2464.68
26.89 x 69 = 1855.41

2)7481,09 twice the area.

10)3740.545

374.0545

Ans. 374.0545 acresa

Lx. 3. Required the area of the following polygon, where of the sides are as follow, viz. AF 31.5, ,FE 33.5. ED 25.5, DC 38.5, CB 43.5, BA 34.5, AE 60.5, AD 81.7, BD 74.3 English chains.

Anf. 277 acres 3 ro. 12 perches.

PRO

PROBLEM XII.

To find the angles of any regular polygont.

By cor. ist, I. 32. Euclid. All the anterior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has fides. Hence the following rule.

RULE.

From double the number of fides fubtract 4, and the remainder is the number of right angles contained by all the fides of the polygon. Multiply the remainder by go, and divide the product by the number of fides, the quot gives the degrees in any of the angles.

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5)540

108 degrees in each angie. Ex. 2. Required the angle of a heptagon. Anf. 128° 347

of a hexagon. Anf. 120.

of a decagon. An. 144. Ex. 5.

of an octagon. Anf. 135. Ta

PRO

Ex. 3. Ex. 4.

PROBLEM XIII.

To find the area of a regular polygon.

RULE.

Find the area of a triangle, constructed on one of the fides of the polygon, and whose vertex is in the centre ; then mul. tiply this area by the number of sides, and the product will be the area of the polygon. Or,

Multiply the perimeter by the radius of the inscribed circle, and half the product is the area of the polygon.

EXAMPLE I.

As rad. 90 =

Required the area of a pentagon, whose fide is 10. 11, To find the angle.

To find the rad. of the infcribd. 5

10,00000 2

is to EG 5 0,69897

So is tang. 54 = · 10,13874 JO

To FG 6,882 = 4

0,83771

6 90

5)540

Angle 103

The perpen. 6,882
the base

5

34.410 No, fides

5

Area

172.050 Ex, 2. Required the area of a hexagon, whose side is 30.

Anf. 2338.2

!

Ex. 3. Required the area of a heptagon, whose fide is i.

Ans: 3.633912. Ex. 4. Required the radius of the inscribed circle, area; internal angles, and angles at the centre, for the following polygons : viz. The trigon, tetragon, pentagon, hexagon, heptágon, octagon, enneagon, decagon, undecagon, and dodecagon, the side of each being sy

72

No of
Names.

Rad. inf. circ. Areas.
Sides.

Int. Ang. Ang. t cent.
Trigon 3 0.288674 0.433013 60° 0

120° Tetragon 4 0.5

I.
go

go
Pentagon 5 0.688190 1.720475 108
Hexagon
0.866024 2.598072120

60 Heptagon 7 1.038260 3.633912 128 34;

51 25 Octagon 8 1.207106 4.828427135 45 Enneagon 9 1.373738 6.181824 140 0

40 Decagon 10 1.538841 7.694205 144 0 Undecagonu 1.7028401 9.365620 | 147 164 32 4317 Dodecagon I2 1.866024|11.1961441150 0 30

36 ò

Regular polygons of the like number of sides are similar, and similar surfaces are to one another in the duplicate ratio of their homologous fides; but the sides of the polygons in the foregoing table are each of them I ; therefore, as the square of i is to the tabular area, so is the square of the side of any given polygon to the area required : Hence the following

RULE.
Multiply the squate of the side of any given polygon into
the tabular area of the like polygon, and the product will be
the area of the polygon.
Ex. 5. Required the area of a pentagon whose fide is 20 feet.

Pentagonal tablular area 1.720475
Sq. of the given Pol. fide

400

20 20

400

688. 190000 fq. feet.

Ex. 6.

T2

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