76 



THE POPULAR EDUCATOR. 



LESSONS IN MENSURATION. II. 



FOLLOWING up our subject from the point at which we left it 

 in the former lesson, we subjoin a few examples in the mea- 

 surement of sides of right-angled triangles, and then pass on to 

 the consideration of triangles which do not contain a right angle. 



EXAMPLE 1. A wall is 30 feet high, and it is required to 

 know what length a ladder must be which shall reach to its top, 

 the foot of the ladder not being able to stand nearer the wall 

 than 14 feet. 



The 47th Proposition of the First Book of Euclid gives us 

 at once the means of solving the question. The right angle is 

 formed by the wall and the ground ; the ladder is therefore the 

 hypothenuse (see Definitions in "Geometry"), and this is equal 

 to the square root of the sum of the squares of the base and 

 perpendicular, or VW + 14 2 = V900 + 196 = V 1096 = 

 885*5 * eet > approximately. 



EXAMPLE 2. A ladder is 45 feet long, and when its foot rests 

 upon the edge of the footpath, which is 12 feet wide, its top 

 just reaches the cave of the roof. "What height is this eave 

 from the ground ? 



In this case the hypothenuse and base are known, and the 

 height of the perpendicular is required. This, by the before-men- 

 tioned Proposition, is equal to the square root of the difference 



of the squares of the base and hypothenuse, or V 45 2 12 2 

 = V~!2025^~i44 = -v / "l881 = 43^ feet, approximately. 



EXAMPLE 3. The side of a square is 9'774 feet ; what is 

 the length of the diagonal ? 



It is necessary here to observe that being a square, the sides 

 are all equal ; the length of the diagonal which is, of course, 

 the same thing as the hypothenuse of either of the triangles 

 formed by the bisection of the square by the diagonal will, 

 therefore, be the square root of twice the square of one side. 

 This will be in this case nearly 13'823 feet. 



EXAMPLE 4. The side of an isosceles triangle (see Definitions 

 in " Geometry ") is 65 feet, and the base 50 feet ; what is the 

 altitude ? 



The Btudent must here remember that the sides of the isos- 

 celes triangle being equal, the perpendicular bisects the base ; 

 hence we have two right-angled triangles formed, in both of 

 which the hypothenuse and base are equal, each to each ; the 

 base of each being one-half of that of the isosceles triangle. 

 The rule for right-angled triangles will then apply, and the 

 altitude or perpendicular will be found to be CO feet. 



EXERCISE 1. 



1. The base of a right-angled triangle is 4 ft. 6 in. (54 in.), 

 and the hypothenuse 7 ft. 5 in. (89 in.). What is the height of 

 the perpendicular ? 



2. The base being 513, and the perpendicular 684 ft., what is 

 the hypothenuse ? 



3. The hypothenuse is 2 ft. 10 in. and the base 2 ft. 6 in. 

 What is the perpendicular ? 



4. What is the side of a square whose diagonal is 8 ft. 5 in. ? 



5. A ladder 50 feet long, being placed in a street, reached a 

 window 40 feet from the ground, on one side of the street ; but 

 -when the ladder rested against the house upon the other side 

 of the street, the position of its foot not being altered, it reached 

 a window 48 feet high. What was the breadth of the street ? 



6. What is the height of an equilateral triangle whose side 

 isl? 



We subjoin a few examples having reference to the proportion 

 which exists between the homologous sides of similar right- 

 angled triangles, as explained in Lesson I. 



EXAMPLE 1. Two poles stand upright on level ground; the 

 height of one is 10 feet, and its shadow projected upon the ground 

 by the sun is 15 feet. The shadow of the other pole measures 

 150 feet. What is its height ? Ans. 100 feet. 



EXAMPLE 2. I wish to draw an oblong or right-angled 

 parallelogram, similar to one whose length is 200 feet and 

 oreadth 20, but have only room on my paper to make the length 

 15 inches ; what must be its breadth ? Ans. 1^ inches. 



We shall now briefly glance at the relations which exist 

 between the sides and angles of triangles which are not right- 

 angled, that is, neither of whose angles is a right angle. 



In the case of similar triangles, that is, of those in which any 



two of the corresponding angles are equal, no matter how great 

 the disparity of the triangles as to area, the corresponding sides 

 are all respectively proportional. This we have already noticed 

 in our first lesson with respect to right-angled triangles, and the 

 rule holds good in all similar triangles. Let ABC, a 6 c (Fig. 9) 

 be two similar triangles, having the two angles at A and B 

 equal to the two angles at a and b; then since the three interior 

 angles of every tri- 

 angle are equal to 

 two right angles 

 (Euc. I. 32), the 

 angle at c must be 

 equal to the angle 

 at c, and the simi- 



lar sides are proportional in each triangle ; that is to eay, A B 

 is to A c as a 6 is to ac ; hence if two sides of one triangle, 

 and a similar side of another and similar triangle be known, the 

 other similar side of the second triangle is found by proportion. 

 We may here observe that this simple and useful rule is equally 

 applicable with respect to the similar lines of alt similar figures, 

 whether plane or solid. 



In calculating the length of the third side of a triangle, not 

 right-angled two being known it is obvious that the rule of 

 the squares (Euo. 1. 47) cannot apply, and for this reason : when 

 the angle formed by the sides ia 

 variable, the sides which contain 

 that angle may remain the same as 

 to length, whilst the hypothenuse 

 may alter. Let ABC (Fig. 10) be 

 a triangle, of which A c is the hy- 

 pothenuse ; then if the angle ABC 

 be not right-angled, it must be 

 either less than a right angle, as 



Fig. 10. 



A' B c, or greater than one, as A" B c, and the lines A' B, A" B may 

 remain equal as to length, whilst the hypothenuse A' c will be 

 very unequal to the hypothenuse A" c ; hence if A c 2 = A B* + B c*, 

 it is obvious that A' c s and A" c 2 cannot equal the same or equal 

 quantities ; a different mode of treatment must in thia case be 

 adopted, which we shall introduce subsequently. In order to 

 find the area of a triangle it is desirable to ascertain the height 

 of the perpendicular, that is, of the line falling vertically upon 

 the base from th opposite angle ; and we will here premise that 

 the base of a triangle means its longest side, except in the case 

 of a right-angled triangle, and that the greatest angle of every 

 triangle is always opposite the 

 greater side. (Euc. I. 18.) 



Given the three sides of a tri- 

 angle, it is required to find the 

 height of its perpendicular. Let 

 ABC (Fig. 11) be a triangle of 

 which AC is the base, it is required 

 to find the length of the perpen- 

 dicular B D. Let B c be greater than B A, then the segment D c 

 will be greater than the segment D A. (Euc. 1. 18.) 



DC - DA 



Then AC : BC + BA : :BC-BA : DC-DA, and ---- ~ 



+ ^-= the length of the greater segment D c, which being 

 a 



subtracted from A c gives the lesser segment D A. 



We have thus ascertained the position of the point D. Then 



in either of the right-angled triangles A D B, c D B we have the 



two sides A D, A B, and c D, c B, from which, by Euc. I. 47, 



we find the height of D B. 



Next, having given the length of the base, and the height 

 of the perpendicular of a triangle, to find its 

 area. The rule is of the very simplest kind : 

 Multiply the base by half the perpendicular, 

 and the result is the area of the triangle. The 

 reason of this we will prove : Let ABC (Fig. 

 12) be a right-angled triangle, right-angled at 

 B. Complete the parallelogram A B c D ; then 

 A c bisects it (Euc. I. 34). Now the area of a 

 square or of a right-angled parallelogram is the 



Fig. 12. 



product of any two adjacent sides. Hence A B X B c is the area of 

 the parallelogram A B c D, but this is double the area of the tri- 

 angle ABC (Euc. I. 41). Hence if A B X B c = area of parallelo- 

 gram A B c D. A B X 5- = area of triangle ABC. 



