But AB2 is the area of the square, so that we see that to find the area we simply take half the square of the diagonal. EXERCISES (C). 1. How many square yards of paper will be required for papering a wall 23 ft. long and 14 ft. 6 in. high? 2. A table is 73 ft. long and 3 ft. broad; what is its area? 3. A rectangle is 9 in. wide and contains exactly a square yard; find its length. 4. Find the length of a rectangle whose area is 540 sq. ft. and breadth 4 yds. 5. The diagonal of a square is 92 ft.; find its area. 6. What is the area of a rectangular table 5 ft. 4 in. long and 4 ft. 6 in. broad? 7. A rectangle contains sq. ft. and is 10 inches broad; find its length. 8. A room is 23 ft. 10 in. long and 25 ft. broad; how much will it cost to cover it with carpet, at 4s. 10d. a square yard? 9. Each side of a square field is 693 ft.; find its area. 10. Find the side of a square field which contains 64 acres. 11. The diagonals of two square court yards are respectively 110 yds. and 1112 yds.; find the area of each. 12. Find the area of a garden 20 ft. long and 15 ft. broad. 13. A certain road is 14 miles long and 75 ft. broad; what was the cost of its construction at one halfpenny a sq. foot? 14. A square field contains exactly 1 acre, find the length of a side. 15. The sides of a rectangle are 30 yds. 2 ft. 8 in. and 9 yds. O ft. 9 in.; find its area. 16. Find the area of a square whose side measures 871 links. LESSON VI.-THE RHOMBUS AND RHOMBOID. In the preceding lesson we dealt only with rectangular quadrangles; we have now to consider two figures which differ from the square and rectangle in one respect, viz., that their angles are not right angles. They are, indeed, like squares and rectangles that have been forcibly wrenched (like a loose slate-frame), from their true form, the rhombus (I) being a distorted square, and the rhomboid (II) a distorted rectangle. These figures, because their opposite sides are parallel, are frequently called parallelograms, although that name really includes the square and rectangle. A square has all its sides equal, and all its angles right angles. A rectangle (or oblong) has its opposite sides equal, and all its angles right angles. A rhombus has all its sides equal, but its angles are not right angles. A rhomboid has its opposite sides equal, but its angles are not right angles. To find the area of a rhombus or rhomboid, multiply the base by the perpendicular height. The In each of the above figures the area is AB × BE. reason of this rule is manifest; for in each case the parallelogram ABCD is equal in area to the square or rectangle, ABEF, since they are upon the same base AB, and contained between the same parallels. (Euclid I. 35). But the area of the latter is in each case AB BE, and, therefore, the area of the rhombus or the rhomboid is AB X BE. EXERCISES (D). 1. The base of a rhomboid measures 2345 links, and the perpendicular height is 2464 links; find the area. 2. The base of a rhomboid measures 385 yds., and its area is 21 ac. 1 r. 30 p.; find the perpendicular height. 3. Find the area of a rhomboid whose length is 40 ft. 6 in. and breadth 12 ft. 9 in. 4. A field in the form of a rhomboid contains 3 ac., and its perpendicular breadth is 198 ft; find its length. 5. What is the rent of a field, at 30s. per acre, which is of rhomboidal form, and 6 chains 20 links long, with perpendicular width 5 chains 14 links? 6. A rhombus is 12.4 ft. long, and its perpendicular breadth is 10.9 ft.; find the area. LESSON VII.-TRIANGLES. D J F CASE I.-The easiest way by which to find the area of a triangle is by measuring the base and perpendicular height. In the accompanying figure, let us suppose that we know the length of BC and AE. E Now if we multiply these together their product gives the area of the enclosing rectangle DBCF, which we know to be exactly double of the triangle ABC. (Euc. I. 41). The area of which can, therefore, readily be found by dividing by 2. To find the area of a triangle, multiply half the base by the perpendicular height. CASE II.—The area of a triangle may, however, be determined from its three sides by the following rule:-- From half the sum of the three sides, subtract each side separately; then multiply the half sum and the three remainders together, and the square root of the last product will be the area of the triangle. For practical purposes this rule may be more conveniently stated in the form of the following formula, where s represents half the sum of the sides: Area = ss(s-a) (s—b) (s-c) Note. If the triangle is equilateral, all that is necessary is evidently to measure one side. The area may then be calculated from this by the following short and useful rule, multiply the square of the side by 433, and the product will be the area. The proof of this rule may be exhibited thus: EXERCISES (E). Find the area of each of the following triangles, the base and per pendicular respectively being and 9ft. 4 in. 3178 links. 18.92 ft. 9 in. 1. 22 ft. 5 in. 2. 3568 links 3. 21.68 ft. 4 ft. 9 in. Find the area of each of the triangles, whose sides are respectively as follows: 17. 16 chains 25 links; 15 chains 70 links; 12 chains 45 links. 18. 1500 links; 1200 links; 1450 links. 19. 404 yds.; 495 yds.; 217 yds. 20. 1432 links; 1264 links; 1346 links. 21. 4 ft. 7 in.; 15 ft. 3 in.; 11 ft. 4 in. 22. 10.75 chains; 7.5 chains; 8.25 chains. 26. 164 yds.; 225 yds.; 349 yds.. LESSON VIII.—THE TRAPEZOID. A trapezoid is a four-sided figure having two opposite sides parallel. To find the area of a trapezoid multiply half the sum of the parallel sides by the perpendicular distance between them. The reasonableness of this rule may be demonstrated without any strictly mathematical investigation, by remembering what has been said previously-that we calculate all areas by reference to the rectangle; for if we multiply the longer side CD by the perpendicular distance, the product will obviously be greater than the area of the trapezoid; while if we multiply the perpendicular distance by the shorter side AB, the product will be too little; hence we find a side of mean or average length by dividing the sum of AB and CD by 2, and the product of this and the perpendicular distance is the exact area. But the rule may be proved mathematically thus: bisect BD in F; join AF, and produce it to meet CD produced in G. Now it is evident that the triangle ABF is similar and equal to FDG, and that AB is equal to DG; so that the trapezoid is exactly equal in area to the triangle ACG. area of triangle ACG CG. AE. =(CD+DG) AE. LESSON IX. THE TRAPEZIUM. A trapezium is a four-sided figure, whose opposite sides are not parallel. This is an irregular figure, and its area is usually found by dividing it as in the figure, into two triangles. Now, the area of the A triangle ABC is AC. BF, and area of triangle ACD is AC. DE; and, therefore, the area of the trapezium (.e., of the two triangles together) is AC. BF+ AC. DE =AC (BF+DE). |