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RULE. Divide the figure into convenient parts, forming triangles, trapezoids, etc., by drawing diagonals and dropping perpendiculars; find the area of each part separ- A ately, and the sum of the areas will be the area of the polygon.

ABCDEF is an ir. regular polygon. By drawing the diagonal AD and dropping perpendiculars upon it from the other corners, the figure is divided into two trapezoids BGHC, KFEL, and four triangles, ABG, CH D, LED, AKF, whose areas may be found as already directed, and their sum will be the area of the polygon.

Ex. Find the area of the field ABCDEF, when the following measurements are given in chains :-Å G=8, G H =24, HD=7, G B =15, HC=24, A K=12, KL=18, LD -9, KF=9, and L E=12. Area of triangle A GB 1 x8x15= 60 sq. ch. (Art.

66). Area of trapezoid B GHC = 1 x 39 x 24=468 sq. ch. (Art.

74). Area of triangle CHD =}x7x24=84 sq. ch. Area of triangle A KF =* x 12 x 9= 54 sq. ch. Area of trapezoid KFEL =} x 21 x 18=189 sq. ch. Area of triangle LED = x 9 x12=54 sq. ch. Area of polygon ABCDEF=60+468+84+54+189+54

· =909 sq. ch. = 90.9 ac.

=90 ac. 3 ro. 24 po.

Ex. XXXIV. (1) Find the area of the field A B C D E, when the following dimensions are given in feet: A F-13.8, BF=3.2, GE=2.5, and the portion A CFD is a square.

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(11) The sides of a five-sided rectilineal figure ABCDE taken in order, AB, BC, etc., are 5, 12, 14, 9, 12 ft. respectively; the diagonal A D is 15 ft., and the diagonal AC 13 ft. Find the area in feet.

[Make a rough sketch of the figure. In each triangle the three sides are given. See Art. 69.]

(12) In a five-sided yard I note the following measurements: A B = 11, BG=65, CD=75, D E=14, and EA=15, all in chains; the diagonal A D=13 chains, and the diagonal BD = 20 chains. Required the area in acres.

(13) In a five-sided field A B C D E there are given the following measurements in links, viz. :-AB=20, BC= 65, CD=39, DE=52, E A-70; also these two diagonals, viz. :CA = 75, C E=65. Required the area of the field in square links.

THE REGULAR POLYGON. 84. A regular polygon is one which has all its sides and all its angles equal.

85. A pentagon has five sides.

86. A hexagon has six sides.

87. A heptagon has seven sides.

88. An octagon has eight sides.

89. A nonagon has nine sides.

90. A decagon has ten sides.

91. The perimeter of a polygon is the sum of all its sides.

The perimeter of a regular polygon is therefore the length of each side multiplied by the number of sides.

O is the centre of the polygon, and OF the perpendicular drawn from it to the middle of one of the sides A B.

92. If straight lines be drawn from the centre to each angle, the polygon will be divided into as many

equal triangles as the figure has sides, and the height of each triangle will be the length of the perpendi. cular from the centre. The area of one triangle is equal to half the rectangle having the side for base, and the perpendicular distance from the centre for height, and therefore the area of the whole is equal to half the rectangle having the perimeter for base and the perpendicular for height. The truth of this theorem may be rendered more

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F B obvious to beginners by taking the triangles into which the regular pentagon A BODE is divided, and placing them as in the annexed diagram, so that their bases may be in a straight line. ĀBODE A', the perimeter of the polygon, will be the base of a rectangle A A'G H, and O F, the perpendicular from the centre of the polygon, the breadth of the rectangle. The area of the rectangle is equal to the product of A Å and O F, and the sum of the areas of the triangles to half this product. The same will be found true of the hexagon, heptagon, etc.; and hence, whatever the number of sides in a regular polygon, its area is measured by half the product of the peri. meter and the perpendicular distance of the centro from the side.

93. To find the area of a regular polygon when the side and perpendicular upon it are given.

ROLE. Multiply the perimeter of the polygon by the perpendicular from the centre, and half the product will be the area.

Ex. 1. Find the area of a regular pentagon A B C D E, each side of which is 24 ft., and the perpendicular 0 F 16-5144 ft.

Perimeter of polygon=24x5=120 ft.
Area of polygon = } x 120 x 16.5144=990·864 sq. ft.

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