B 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 irregular polygon. By draw C K H ing the diagonal AD and dropping perpendiculars upon it from the other corners, the figure is divided into two trapezoids BGH C, KFEL, and four triangles, AB G, CHD, 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 :-A G-8, GH =24, HD=7, GB=15, HC=24, A K=12, KL=18, LD 9, KF-9, and L E-12. Area of triangle A G B Area of trapezoid B G H C Area of triangle C HD Ex. ×8×15-60 sq. ch. (Art. 66). =1×39×24 468 sq. ch. (Art. 74). XXXIV. (1) Find the area of the field ABCDE, when the following dimensions are given in feet:A F-13.8, BF=3.2, GE=2.5, and the portion AC FD is a square. [The figure consists of a square and two triangles, of which one is rightangled.] E D F B (2) A B C D E is a fivesided figure; the following lengths are given in feet:A C-18, A D=16, E a=8, E Db=10, Bc=6; find the area in feet. [The figure consists of three triangles, in each of which the base and perpendicular height are given.] (3) A B C D E is a five-sided field, and the angles at B, C, and D are right angles. A B=20 ft., B C 18 ft., CD=32 ft., and DE=13 ft. Find the length of the remaining side, and the area of the field. [From the area of the rectangle, take the area of the right-angled triangle AFE.] lks. The portion A B D E is a rectangle. G F [The figure consists of a rectangle and two triangles.] (5) Find the area of the irregular polygon ABCDEFG, when the following measurements are given in yards :AD=375, Ga=75, Fb=125, Ec=130, Bd=150, C e=125, Ad 90, Ae=280, A a=70, Ab =185, Ac=300. [The figure is divided into three trapezoids and four right-angled triangles.] A B с A B (7) In the annexed figure there are given in links, A F=250, Gb =100, Ba=75, B F=225, Ec= 150, CE 300, Bd=75, De=75; find the area in roods and poles. [The figure is divided into five triangles, in each of which the base and perpendicular height are given. Ba is perpendicular to A F, E c is perpendicular to BF, and B d is perpendicu lar to CE.] (8) In the irregular polygon ABCDE represented in the annexed diagram, the following measurements were taken in yards :-FA=8, E F=6, BG =5, G C=12, F G=30, E C= 32, HD=8. The angles at FE and G are right angles; find the area. F [From the area of the trapezoid E F G C take the sum of the areas of the triangles EFA, B G C, the remainder will be the area of the figure Е А В С.] (9) Required the area of the field A B C D E when the following measurements are given:-A C= 1125 lks., E F-425 lks., D G=550 lks., A F=375 lks., C G=225 lks., BH=750 lks. [The figure consists of a trapezoid and three triangles, of which two are rightangled. To find the length of FG, take the sum of AF and G C from A C. Now A C=1125, and the sum of AF and G C= 375+225 600 lks. Hence F G=1125-600 =525 lks.] (10) ABCDEFG is an irregular polygon; find its area from the following dimensions:-FD-12 ch. 50 lks., E a 5 ch. 14 lks., GD =15 ch. 10 lks., Fb-8 ch. 75 lks., G C-15 ch., Dc= 10 ch., Ad=6 ch. 25 lks., A C 8 ch. 50 lks., Be=2 ch. 75 lks. [Fb is perpendicular to GD, De to G C, and Ad to G C.] F 6 A E D с (11) The sides of a five-sided rectilineal figure ABCDE taken in order, A B, B C, 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, BC=65, CD=75, DE=14, and E A=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 ABCDE there are given the following measurements in links, viz.:-A B=20, B C=65, CD=39, DE=52, E A=70; also these two diagonals, viz. :— CA-75, CE=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 E sides. 88. An octagon has eight sides. 89. A nonagon has nine sides. 90. A decagon has ten sides. O F 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 perpendicular 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 obvious to beginners by taking the triangles into which the regular pentagon ABCDE is divided, and placing them as in the annexed diagram, so that their bases may be in a straight line. ABCDE A', the perimeter of the polygon, will be the base of a rectangle A A' GH, and OF, 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 A' and OF, 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 perimeter and the perpendicular distance of the centre from the side. 93. To find the area of a regular polygon when the side and perpendicular upon it are given. RULE. 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 ABCDE, each side of which is 24 ft., and the perpendicular O F 16·5144 ft. Perimeter of polygon=24x5=120 ft. Area of polygon ×120×16·5144-990.864 sq. ft. |