Measurement of Area and Volume 27 



Example : 



(1) In the triangle MNO (Fig. 8) the perpendicular 

 MP (usually called the " height ") from M (the 

 " apex ") on NO represents 3'60 cm., while NO (the 

 "base") represents 4'00 cm. Hence the area of the 

 triangle is (J x 3'60 x 4*00) sq. cm. = 7'20 sq. cm. 



(2) In the triangle M'N'O' the height M'P' repre- 

 sents 3-60 cm. and the length of the base N'O' represents 

 2 -20 cm. Hence the area of the triangle M'N'O' is 

 (i x 3-60 x 2-20) sq. cm. = 3*96 sq. cm. 



There are other special rules for finding by measure- 

 ment and calculation the areas of certain other recti- 

 linear figures, but, generally speaking, it is best to 

 find the area of any such figure by dividing it into 

 simpler figures, e.g. parallelograms or triangles, by 

 diagonally drawn lines, finding separately the area of 

 each of these component figures, and adding the areas 

 thus obtained. 



Thus the area of the regular hexagon RSTUVW 

 (Fig. 8) is evidently six times that of the triangle RSX, 

 and that of the irregular figure 12345678 is the sum 

 of the areas of the smaller figures into which it is divided 

 by the dotted lines. 



For practice in the above rules, the student should do 



EXPERIMENT 6. Find by actual measurement and 

 calculation the areas in square inches and in square 

 centimetres of the regular hexagon and of the irregular 

 figure shown in Fig. 8, (5) and (6). 



From the results of the determinations made in 

 each case, calculate the number of square centimetres 

 in one square inch, and the number of square inches 

 in one square centimetre. 



18. In all cases already considered the sides of 



