CHAMBERS'S INFORMATION FOR THE PEOPLE. 



PROP. XII. To find the area of any rectilineal 

 figure. 



Thus, let it be required 

 to find the area of the 

 figure 

 divide 



Again, 



ABCDE. We 



this figure into 

 triangles by drawing the 

 lines AC, AD, and then 

 we find the area of these 

 triangles separately, and take their sum. 

 we may sometimes very 

 conveniently find the 

 area of the figure by 

 drawing the longer diag- 

 onal, and dropping per- 

 pendiculars on this diag- 

 onal from the opposite 

 angle ; then we can find 

 the areas of the triangles and trapezoids into 

 which the figure is thus divided. 



Similar Triangles. 



PROP. XIII. // has been proved that similar 

 triangles are to one another in area as the squares 

 upon their like sides. 



That is, if ABC and abc are similar, we have 

 ABC = AB 3 BC 3 AC 3 

 abc " ab* 



thus, for example, if AB = twice ab, then ABC 

 =four times abc. 



And we have seen that the same relation holds 

 good for similar figures of any number of sides ; 

 thus if ABCDE, abcde be two similar figures, we 

 have 



ABCDE = AB 3 

 abcde ~ afr ' 



Then it follows that if we know the area of 

 either of the two figures, and the ratio of the 

 sides, we can tell at once the area of the other 

 figure. For example, if the area of abcde be i 

 acre, and AB 5 times the length of ab ; then the 

 area of ABCD will be 25 acres. Or again, sup- 

 pose a field to contain i acre, and the plan of this 

 field is laid down to a scale of say I inch to 

 20 feet, we can at once tell how much paper the 

 plan will cover; or conversely, if we know the 

 size of the plan, we can as readily tell the size of 

 the field. 



636 



Area of a Circle. 



PROP. XIV. To find the area of a regui 

 polygon of any number 

 of sides. 



Let ABCD ... H be a 

 regular polygon of any 

 number of sides ; to find 

 its area. Joining O, the 

 centre of the figure with M 

 each of the angles, we 

 thus divide the figure 

 into triangles, which are 

 all equal to each other ; 

 then area of 



AR 



ABO = X OI, 



2 



BOC = ^- X OI, 

 COD = CD - X OI, 



2 



DF 

 DOE = -_ - X OI, 



HOA = ^~ X OI j 



.*. area of whole figure 



= (AB + CD + DE + ... HA) X OI 

 = $ perimeter X OI 



= perimeter X 



= perimeter X radius of inscribed circle. 



Now if we consider the number of the sides of 

 the polygon to be inde- 

 finitely increased, it follows 

 that the area of the polygon 

 will approach nearer and 

 nearer to the area of the 

 circle, and will ultimately 

 become equal to it ; and 

 the perimeter of the poly- 

 gon will be the circumfer- 

 ence of the circle, and the 

 radius of the inscribed 



circle will become the radius of the given circle ; 

 hence it follows that what has been proved for 

 the polygon namely, 



area of polygon = perimeter X radius, 

 becomes for the circle, 



area of circle = circumference X radius ; 



that is, we find the area of the circle by multiplj 

 ing half the circumference by the radius, or 



area = irr 2 ; 



that is, the area is found 

 by multiplying the square 

 of the radius by 37. 



It also follows from this 

 that the area of a sector of a 

 circle OABCDE is found 

 by multiplying half the arc 

 by the radius. 



