3. How many acres does a triangle contain, whose base is 40, and its perpendicular height 30 rods? Ans. 3.75 acres. To solve this example by the sliding rule, first find the side of an equal square, as directed under the rectangle, which in this example we find to be 24.5 rods nearly; then shut the slider, and over 24.5 rods, found on the line D, we find 3.75 acres for the answer. To find the altitude of an equilateral triangle, its side being given : From the square of the given side subtract one-fourth of the square of the given side, and extract the square root of the remainder. 4. What is the altitude, and what is the area of an equilateral triangle whose side is 1? The altitude is .86602540378444, and its area is .43301270189222. Trigons, or equilateral triangles, are to each other as the squares of their sides. Therefore, to find the area of an equilateral triangle, its side being given : Multiply the square of the given side by the decimal .4330127. 5. What is the area of an equilateral triangle whose side is Ans. 500.5, nearly. 34? To find the side of an equilateral triangle whose area shall be 1: Divide unity, or 1, by the decimal .4330127, and the quotient is the area of a square drawn on one side of the triangle, [viz. 2.3094, very nearly, the square root of which is the side of the triangle, viz. 1.51967, nearly. By the above rule, the side of any equilateral triangle may be found, its area being given; or, multiply the side of a square by 1.51967, and the product will be the side of a trigon of equal area. 6. What is the side of an equilateral triangle whose area is 144 square inches? Ans. 18.236 inches. 7. What is the side of an equilateral triangle whose area is 1 acre? Ans. 19.225 rods. Since the areas of trigons are as the squares of their sides, and since the numbers on the line C are the squares of those on the line D, if we place 1 on C over 18.236 on D, over the side of any equilateral triangle on D, (its side being given in inches,) will be found the number of feet on C. Or, set 1 on C over 19.225 on D, and over the side of any equilateral triangle given in rods, will be found its area in acres. 8. The side of a trigon is 31 inches; its area in feet is required. Ans. 2.9 feet, nearly. 9. What is the area of an equilateral triangle whose side is 38 rods? Ans. 3.9 acres, nearly. 10. What is the area of a trigon whose side is 45 rods? When the sides of a scalene triangle are given, to find its area: From half the sum of the three sides subtract each side severally; then multiply together the half sum of the three sides, and the three remainders, and extract the square root of the product for the area. 11. What is the area of a triangle whose sides are 20 and 30 and 40 rods? Ans. 1.8159 acre. The half sum of the three sides is 45, and the three remainders 25 and 15 and 5; and 45×25×15×5=84375, the square root of which is 290.473 rods=to 1 acre, 3 roods, and 10 rods. 12. If the sides of a triangle are 134 and 108 and 80 rods, what is its area? Ans. 4319.9 square rods. Having the three sides of any triangle given, to find the point in the base, or longest side of the triangle, where a perpendicular let fall from the opposite angle will meet the base, or longest side, say :-As the longest side is to the sum of the other two sides, so is their difference to the difference of the segments formed on the base by the perpendicular drawn from the opposite angle; then add half of this difference to half the sum of the two segments, (that is, to half of the longest side,) and it will give the greater segment; and subtract the said half difference from the said half sum, and it will give the less segment. It is not necessary that the longest side should always be taken for the base of the triangle, but the rule may be applied in any case, when the perpendicular drawn from the angle opposite the side chosen for a base, does not fall without the triangle, or beyond the base. 13. Suppose the lower ends of two rafters are 42 feet apart, and that one of the rafters is 36 feet in length, and the other 32; the height of the ridge, above the plates on which the rafters rest, is required. As 42:68:: 4 to the difference of the segments, viz., 6.4762, and 6.4762 42 er segment, [as cg, see the figure,] and -=24.2381 the great 42 6.4762 =17.7619, the less segment. Having found the segments, if we square the length of either of the rafters, and subtract from its square the square of the segment under that rafter, the remainder will be the square of the height of the ridge, the square root of which will give the required height. Thus, (36×36)-(24.2381 ×24.2381)=708.51450839, the square of the height of the ridge, the square root of which is 26.61773 feet, the required height.-A roof is said to have a true pitch when the length of the rafters is the breadth of the building. 14. The sides of a triangle are 20, 28, and 31 rods. What is the length of a perpendicular falling from the opposite angle upon the longest side? Ans. 17.703 rods, very nearly. Similar triangles (that is, triangles whose angles are equal) are to each other as the squares of their corresponding sides, or as the squares of their altitudes; and this rule, or statement, is true in respect to all similar superficies. 15. If the sides of a triangle are 42, 36, and 32 rods, at what distance from the apex of the triangle must a line be drawn parallel to the longest side, so as to cut off of the area? And what will be the length of the base of the part cut off? Answers, 15.36788 rods; and base, 24.24871 rods. The square root of of the square of the altitude of the given triangle will be the altitude of the part cut off; and the square root of of the square of the given base will be the base of the part cut off. To find the area of a triangle when two of its sides and the contained angle are given : Multiply the NATURAL SINE of the contained angle by the product of the two sides into each other, and half this product will be the area. (See the table, 73.) Required the number of square yards in a triangle, two of whose sides are 50 feet and 42 feet 6 inches, and the contained angle 45 degrees ? Ans. 83.47788 yards. How many square yards are contained in a triangle, two of whose sides are 42 and 75 yards, and the included angle 50 degrees ? Ans. 1221 yards. A trapezium is a figure bounded by four right lines, having no two sides equal, and no two angles which contain the same number of degrees. To find the area of the trapezium : Divide it into two triangles by a diagonal line, (as ab in the figure,) to which draw perpendiculars from the opposite angles; measure the length of the diagonal and the two perpendiculars, and multiply the diagonal by the sum of the perpendiculars, and half the product is the area. Or, Find the areas of the two triangles as directed in ¶ 19, and the sum of their areas will be the area of the trapezium. EXAMPLES. 1. What is the area of a trapezium, whose diagonal is 42, and the perpendiculars falling upon it from the opposite angles 16 and 18? Ans. 714. 2. How many square yards of paving are there in a trapezium whose diagonal is 65, and the perpendiculars falling upon it 28 and 32.5 feet? Ans. 222.08 yards. When a trapezium can be circumscribed by a circle, and its four angles made to touch the circumference, its area may be found by the following rule. From the sum of the four sides subtract each side, severally; then multiply together the four remainders and extract the square root of their product. 3. The sides of a trapezium inscribed in a circle, are 10, 12, 15, and 21 rods; its area is required. Ans. 190.2 rods. 4. The four sides of a trapezium in a circle, are 600, 650, 700, and 750 links; its area is required. Ans. 4 acres, 2 roods, 4 rods. POLYGONS. Any figure bounded by more than four right lines, is called a polygon; and if the sides are of equal length, it is called a regular polygon; but if the sides are of unequal length, it is called an irregular polygon. To find the area of an irregular polygon : Divide it into triangles; measure each of the triangles sepa |