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Prob. 31. To find the Area of any regular Polygon.

Rule. Firft find the Center of the Polygon, (which may readily be done by Prob. 17) and draw Lines from the Center to each Angle of the Polygon; thus will the given Polygon be divided into as many equal Triangles as there are Sides: A Perpendicular, being let fall from the Center to the Middle of one of the Sides, will divide the Triangle into two rightangled Triangles, in which will be given the Angles and one Side to find the other Side, which may be done by Prob. 19 or 24; and then the Area may be found by Art. 56.

Exam. 1. The Side of the Hexagon ABCDEF being 10, to find the Area?

Divide 360, the Degrees in a
Circle, by the Number of Sides in
the given Polygon, and the Quo-
tient is the Angle at the Center of
each Triangle, which in this Ex- A
ample is 60°, the Half of which
is the Angle FOG 30°.
the right-angled Triangle F O G,
right-angled at G, whofe oppofite

In

Side FG is 5, being Half the Side

B

C

FGE

D

FE, to find the Perpendicular GO, oppofite the Angle at
F 60°, being the Complement of the Angle at O to 90°.

As the Sine of the Angle at Q, 30°,

Is to the Side F G, 5,

So is the Sine of the Angle at F, 60°,

To the Perpendicular G O, 8.66025,

Co. Ar. 0.301030

0.698970

9.937531

0.937531

Having found the Perpendicular GO, if that is multiplied by the Side FG, the Product will be double the Area of the Triangle FOG, by Art. 56, that is, it will be the Area of the Triangle FOE, which, being multiplied by 6, gives the Area of the given Hexagon.

8.66025 the Length of GO.

5 the Length of F G.

43.30125 Area of the Triangle FOE.
43.30125

259.80750 Area of the Hexagon.

Hh

This

This is the Area of an Hexagon whofe Side is 10; therefore, to find the Area of an Hexagon whofe Side is 1, the Proportion must be, As the Square of 10 is to the Square of 1, fo is the Area of the given Hexagon to the Area of an Hexagon whofe Side is ; fince all fimilar Figures are as the Squares of their Sides.

As roo is to I, fo is 259.8075.

100) 259.8075 (2.598075 the Multiplier for finding the Area of an Hexagon in the Table, Art. 60.

Thus the Area of any regular Figure, whofe Side is 1, being found, is a Multiplier, by which the Square of the Side of a like Figure being multiplied, the Product is the Area of the Figure fought.

Exam. 2. The Side of the Pentagon ABCDE being 10, to find the Area?

For the Angles at the Center, 360° being divided by 5, the Number of Sides, give 72°, the Half of which is the Angle DOF 36°, whofe Complement to go° is the E Angle O D F 54°; therefore, in the right-angled Triangle DOF, right-angled at F, are given the Side D F 5, and the Angle at D 54°, to find the Perpendicular FO, oppofite the Angleat D: Whence, As the Sine of the Angle at O, 36°, Is to the Side D E, 5,

A

B

D F

So is the Sine of the Angle at D, 54°,

To the Perpendicular FO, 6.88191,

Co. Ar. 6.230781

0.698970

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Which, being multiplied by the Side DF 5, gives the Area of the Triangle DO C; and that, being multiplied by 5, the Number of Triangles, gives the Area of the Pentagon :

6.88191 Length of F O.

5 Length of DF.

34.40955 Arca of the Triangle DOC.

172.04775 Area of the Pentagon.

Now,

Now, as 100, the Square of 10, the given Side, is to the Square of 1, fo is the Area of the given Pentagon to the Area of a Pentagon whose Side is 1.

As 100 is to 172.04775.

100) 172.04775(1.7204775 the Multiplier for the Pentagon in the Table, Art. 60.

In the next Place, I fhall fhew the Ufe of Trigonometry in measuring the Heights of Objects, or the Distance between them.

Prob. 32. To measure the Height of the Tower A B.

At any convenient Distance from the Tower A B, as at C, with a Quadrant take the Angle to the Top of the Tower, which let be 37°.30'; measure the Distance CD, which let be 39.7 Feet; and, when you have taken the Angle at C, measure how high the Center of the Quadrant was from the Ground, which let be 5.6 Feet.

37°30'

B

32.7 D

A

To conftruct the Figure: Draw the Line CD at Pleafure, and fet off the measured Diftance 39.7 from your Station at C to the Tower; at C make the given Angle 37°. 30', by Prob. 8, and draw CB; at D erect the Perpendicular DB, 'till it meets the Hypothenufe in B, and continue it downward to A; make D A 5.6 Feet the Height of the Center of the Quadrant; draw Ce parallel and equal to D A, and draw A equal to CD, 'till it meets BD continued in A; thus will A B be the Height of the Tower.

A

Now, in the right-angled Triangle CBD, there is given, befides the right Angle, the Bafe CD 39.7 Feet, the Angle at C 37°. 30', and the Angle at B, the Complement of 37.30 to 90°, viz. 52°. 30', to find the Side BD, oppofite the Angle at A; which may be found by the following Proportion:

I

Hh 2

As

As the Sine of the Angle at B, 52°. 30',
Is to the Side CD, 39.7,

So is the Sine of the Angle at A, 37°. 30',

To the Side BD, 30.46,

Co. Ar. 0.100 533

1.598790 -9.784447

1.483770

To which adding the Height of the Center of the Quadrant, 5.6, it gives 36.06 Feet for the Height of the Tower A B from the Ground:

30.46 B D, by Calculation.

5.6 Cc, the measured Height of the Quadrant,

36.06 Height of the Tower.

equal to D A.

Prob. 33. Having the Height of any given Object, A B 59.4, Feet, to measure how far it is to a Tree at C, growing on the fame Level with the Bafe of the given Object.

Keep the Center of the Quadrant even with the Top of the Object, and measure the Angle B 51°. 45', and the Angle C, being the Complement of it to go, will be 38°.15; fo that we have given the Angle C, oppofite the known Side A B, and the Angle B, to find the Side C A.

C

B

A

Conftruction. Draw the Bafe-Line CA at Pleasure, on which, at any Point, as A, erect A B perpendicular to CA; lay 59.4 from A to B, by the general Rule, Prob. 19; at B make an Angle of 51.45, by Prob. 8, and draw BC, which completes the right-angled Triangle CA B. Then,

As the Sine of the Angle at C, 38°. 15',
Is to its oppofite Side A B, 59.4,
So is the Sine of the Angle at B, 51°.45',

Co. Ar. 0.208244

1.773786

9.895045

To the Side CA, or Distance of the Object, 75.35, 1.877075

Prob. 34. To measure the Height of an inacceffible Object, AB; that is, where the Distance between the Station and the Bafe of the Object cannot be measured.

At

At any convenient Distance, as at C, on the fame Plain with the Object, measure the Angle at C to the Top of the Object, which let be 31° 25'; then measure the Height of the Center of the Quadrant from the Ground, which fuppofe 4.9 Feet; and then meafure, in a right Line from C to

D

B

wards the Object, any convenient Diftance, as CD, which fuppofe 36.5 Feet; here, holding the Quadrant the fame Height as before, meafure the Angle D to the Top of the Object 52°. 10.

Conftruction. Draw the Bafe-Line CE at Pleasure, on which lay 36.5 from C to D; at C, by Prob. 8. make an Angle of 31°. 25', and draw CB at Pleafure; at D make an Angle of 52°. 10', and draw D B, meeting CB in B; from B let fall the Perpendicular BE; make Cc perpendicular to CE, on which lay 4.9; then draw c A parallel to CE, and continue BE 'till it meets A, and the Figure is conftructed.

Now, in the oblique-angled Triangle CBD, are known the Side CD 36.5 Feet, and the Angle at C 31°. 25', also the obtufe Angle CDB 127°. 50', being the Supplement of the Angle BDE, or 52°. 10', to 180°; therefore the Angle CBD is found by deducting the Sum of the other two from 180°:

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Having the Angle at B 20°. 45' oppofite the known Side CD 36.5, and the Angle at C 31°.25' oppofite the Side DB; hence,

As the Sine of the Angle at B, 20°. 45',
Is to its oppofite Side CD, 36.5,
So is the Sine of the Angle at C, 31°.25',

To its oppofite Side B D, 53.7,

Co. Ar. 0.450640

1.562293

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9.717053

1.729986

Now

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