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Of a regular hexagon,

X8R. A.

Of a regular 20n,

= R. A. 14×90°-120°


zx 36 R. A. = 3; R. A.= 1×90°=162° Of a regular 1000n,

TooX 1996 R. A.-1988 R. A.-1X90°-179,840°

156. Suppose a circle to be circumscribed about a regular polygon, and a straight line to be drawn from its centre to each angular point of the polygon; there will thus be formed about the centre as many equal angles as the polygon has sides.

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The greater the number of the sides of the inscribed. polygon, the greater will be its angles, but the less will be the centre angles. One centre angle, and one angle of the inscribed polygon are together equal to 2 R. A.; because they are equal to the three angles of a triangle. For example, each angle of a regular hexagon= 120°; each centre angle=60°; and 60° + 120° = 180°.

157. The angles made by producing the sides of a polygon are called exterior angles. Each exterior angle with its adjacent interior angle are together = 2 R. A. Therefore the sum of all the exterior and interior angles is equal to as many times 2 R. A. as the polygon has sides. But the sum of all the interior angles is equal to as many times 2 R. A. as the figure has sides, less 4 R.

A.; consequently, the sum of all the exterior angles of a polygon is equal to 4 R. A.

158. The size of the interior angles increases with the number of sides, consequently that of the exterior angles must decrease in the same proportion. The size of each exterior angle of a polygon may be found by dividing 4 R. A. by the number of sides in the polygon. each exterior angle of a triangle = R. A. = 120°

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345 SO


R. A.=120°

R. A.=90°

R. A.=72°

Number of Sum of the Interior the Sides. Angles.

Size of each Interior Angle.

Size of each Exterior Angle.

2 R. A.

3 R. A.=60°


4 R. A.

R. A.=90°


6 R. A.

§ R. A.=108°


8 R. A.

& R. A.=120°

R. A.=60°

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R. A.=45°

R. A.=40°

R. A.=36°

R. A.=32°
R. A.=30°
R. A.=4°

ʊ R. A.=1°

2100 R. A.=1′

1296000 2591996 R. A.81888 R. A.-179°. 59.59′′. 129000 R. A.=1"

Size of each Angle at the centre
of a circle circumscribed about
a regular figure.

R. A.=120°

R. A.=90°

R. A.=72°
R. A.=60°
+ R. A.=513°
R. A.=45°

R. A.=40°
R. A. =36°
R. A.=32°

R. A.=30°
R. A. -4°

30 R. A.=1°

Z1800 R. A.=1'

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In a regular polygon having an infinite number of sides, (and a circle may be supposed to be such a polygon,) each interior angle = 180°, and each exterior angle=0°.


159. A quantity is measured by comparing it with some known quantity of the same kind, (98.) The most simple of surfaces is a square, for in this figure the two dimensions are the same. The square, therefore,

has been adopted as the unit of measure for surfaces, or superficial unit; and its side is taken equal to some linear unit, as an inch, a foot, a mile, &c. To measure a surface we seek how often the unit of measure is contained in that surface, that is into how many squares, of equal magnitude with the unit, the surface to be measured can be divided. Area and surface are nearly synonymous terms. But area is more particularly a surface measured, and expressed in numbers of superficial units.

160. Here we have a right parallelogram ABCD, (fig. 94.) Let the side AB contain 5 linear units, and the side AD contain 3 of the same units; and let the sides be divided accordingly. From each division-point of AB draw a line parallel to AD; and from each division-point of AD draw a line parallel to AB. The surface of the parallelogram will thus be divided into 5 × 3 equal squares; each of the sides of which is equal to the linear unit. If the linear unit be 1 inch, then each square is 1 square inch, and the area of the parallelogram is 5 × 3 15 square inches. If the linear unit is a rod, then each square is a square rod, and the parallelogram contains 15 square rods.


Thus we learn that

the area of a right parallelogram is found by multiplying together two adjacent sides, that is, the base by the altitude.

But the oblique parallelogram is equal in surface to a

right parallelogram of equal base and equal altitude, (151 ;) consequently the area of any parallelogram may be found by multiplying its base by its altitude.

161. Every triangle is equivalent to the half of a parallelogram of equal base and equal altitude. Hence the area of a triangle is found by multiplying its base by its altitude, and dividing the product by 2. We shall come to the same result by multiplying the whole base by the altitude, or the whole altitude by the base. Thus the

area of triangle ABC (fig. 86) =

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= AB X

162. To find the area of polygons in general we must divide them into triangles, and then find the area of each triangle by itself; the sum of the areas of all the triangles will be the area of the polygon. In the pentagon ABCDE, (fig. 96,) draw the diagonals AD and AC. From the vertex of the angle B let fall upon AC the perpendicular BF. From the vertices of the angles C and E let fall upon opposite sides of AD the perpendiculars CG and EH. The area of triangle ABC ACX BF ADXCG 2 2 ADXEH


-; triangle ACD=


; triangle ADE=

; therefore the area of the pentagon ABCDE=

(AC × BF)+(AD × CG)+(AD × EH.)


163. The area of the trapezoid ABDC, (fig. 95) is equal to (CF × ♣ AB) + (BE × CD.) CF = BE, therefore the area of ABDC is equal to CF × (AB+ CD;) that is, the area of a trapezoid is found by multi

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