54 



MENSUEATION. 



Memuu- hence the figure CH B h is a rectangle, and B A=C = 

 tloo f CD , llgo BH=C/.= C d. (26.1.) 



Join BE, B e; and because E e is parallel to BC, 

 the triangle BAG is equal to each of the triangles BEC, 

 B e C, (0 4.) But the triangle BEC is equal to EC x 

 BH, (2.4) that is to ECxC d; and in like manner 

 the triangle B e C is equal to e C x B h, that is, to 

 ieCxCD; therefore the triangle ABC is equal to 

 , and also to \ eCxCD. 



Fig. S. 



Now since CD : C d : : CE xCD : CE xC d (3.4.) 

 and also CD : C d : : CexCD : C exC d; 



2)5.24293 

 The area =418.28 sq. eh. 2.62146 



PROBLEM III. 



To find the area of a trapezoid. 



Nnte. A trapezoid is a quadrilateral, of which two 

 opposite sides are parallel but not equal. 



RULE. Multiply the sum of the parallel sides by the 

 perpendicular distance between them, and half the 

 product is the area. 



In the trapezoid ABCD, draw the diagonal AC, 

 and from its extremities draw AE, CF at right angles 

 to the parallel sides DC, AB. The figure is made up 



Plane PI 



cures. 



therefore CE x CD : CExCrf: : CexCD : 



that is, because CExCD = FCxCG, (29.4.) and 



C exCrf=/CxCff, FCxCG : CExC d : : C ex 



CD-.fCxCg. 



From this proportion, by taking one-fourth of each 

 term, and putting the triangle ABC for its equivalent 

 values | CE xC d, and \ C exCD, we also have 

 ^FCxiCG: trian. ABC:: trian. ABC : ^Cx^Cg. 



Instead of FC, ^ CG, ^/C, C g, substitute their 

 values found above, and the proportion becomes 

 pX (p b) : trian. ABC : : trian. ABC : (p a) (p e.) 



Hence it appears that the triangle is a mean propor- 

 tional between two rectangles, one contained by half the 

 perimeter, and the excess of half the perimeter above one 

 of the sides, and tlie other contained by the excesses of half 

 the perimeter above the other two sides. 



The rule is got from this theorem, by considering 

 that the mean of three proportionals is the square root 

 of the product of the extremes. 



NOTE. This rule is particularly well adapted to 

 logarithmic calculation. 



Ex. 1. The sides of a triangle are 24, 36, and 48, 

 chains. Find the area. 



a=24 

 6=36 

 o=48 



2*= 108 



*=54 



s n=30 



s i=18 



s e= 6 



* X (s a) X (sb) X U c)=54 X 30 X 18 X 6= 



174960, 



V/174960=4I8.282 square chains the area. 

 Or, by logarithms, 



s=54- 1.73239 



* a=30 1.47712 



s 6=18 1.25527 



s c= 6 0.77815 



of the triangles ACB, CAD ; the area of the former Mensum 

 is ,J ABxCF, and that of the latter CDxAE, or Hen of 

 CD xCF, because AE=CF: therefore the area of 

 the trapezoid is ^ABxCF+4CDxCF=J(AB+CD) 



X A 



Ex. Let AB and CD, the parallel sides of a trape- 

 zoid, be 7-5 and 12.25 chains respectively, and CF 

 their perpendicular distance 154 chains. What is the 

 area ? 



7 5+12.25=19-75 the sum of the par. sides. 

 19.75x15.4 _ 152075 sq ch _ 15 fl 



area. 



PROBLEM IV. 



To find the area of a trapezium. 



CASE 1. When a diagonal, and perpendiculars on it, 

 from the opposite angles are given. 



RULE. Multiply the diagonal by the sum of the 

 perpendiculars, if they are on opposite sides of the dia- 

 gonal, or their difference, if they are on the same side, 

 and half the product is the area. 



For the trapezium ABCD is the sum of the triangles Fig. G. 

 ABC, ADC, the areas of which are by Prob. 2. Rule 1. 



In this figure, the perpendiculars are on opposite sides. 

 of the diagonal ; but the truth of the rule may bo 

 shewn in the same way when they are on the same 

 side. 



Ex. The diagonal of a trapezium is 20 feet, and the 

 perpendiculars on opposite sides of it are 4.2 feet and 

 3.8 ; feet find the area, 



4.2 + 3.8=8 the sum of the perpendiculars 



8 X 20 



=80 sq. feet the area. 



CASE II. When the two diagonals and the angle they 

 make with each other are given. 



RULE. Radius is to the sine of the angle contained 

 by the diagonals as their product to double of the area. 



The trapezium ABCD is made up of the triangles Fig. 7. 

 BEC, CED, DEA, AEB. The doubles of the areas ofthe 

 two first, by Prob. 2. Case 2. are BE xEC x sin. E 

 and DE x EC x sin. E ( supposing rad. = 1 ), and twice 

 their sum is BD X EC X sin. E. In like manner twice 

 the sum of the triangles BEA, AED is BDxAEx 

 sin. E, therefore twice the whole area is 



BDxECx sin. E + BDxAEx sin. E= 

 BDxACxsin.E. 



Ex. The diagonals of a trapezium are 326.8 and 

 269.2 feet, and they contain an angle of 54^, how 

 many square yards are in the area ? 



rad ....... 10.0000 



sin. 54*30' 9.91069 



269.2 2.43008 



326.8 2.51438 



twice area 71621.5 4.85505 



area 35810.7 sq. feet. 



PROBLEM V. 



To find the area of a regular polygon. 



Note. A polygon is said to be regular when its sides 

 are equal, also its angles. 



RULE. Radius is to the tangent of half the angle 

 contained by two adjacent sides of the polygon, as half 

 the side to the radius of the inscribed circle ; and the 

 area of the polygon is equal to the rectangle contained 

 by half the perimeter and the radius of the inscribed 

 circle. 



