Imágenes de páginas
PDF
EPUB
[blocks in formation]

24.20000

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

1050
370

Ff = 420

[blocks in formation]

Anfo 7 ac. 3 ra 8 per. 24; yds, 7.80500 as before.

Required

[ocr errors]

Required the area of the irregular figure ABCDEF, of which the side 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 430
DAE 40° 30':
EAF 48° 20'.

[merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

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

[ocr errors]

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 * 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

Anf. 374.0545 acres,

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.

PROPROBLEM 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.

[blocks in formation]

Ex. 3. Ex. 4.

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. 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

[ocr errors]

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

« AnteriorContinuar »