PROBLEM VIII. Having the difference of the two adjacent sides of a rectangular parallelogram, and the sum of their squares given, to find the sides. Rule-From the sum of their squares take the square of their difference, add the remainder to the sum of their squares, and the square root of this last sum will be the sum of the two numbers, then to half the sum add half the given difference and the sum will be the longest side, subtracted leaves the less. EXAMPLES. 1. The difference between the two adjacent sides of a rectangular parallelogram is 8 chains, and the sum of the squares of those sides is 194-What is the length and breadth of the parallelogram? Ans. 13 and 5 chains. 2. The difference between the two adjacent sides of a rectangular parallelogram is 11 inches, and the sum of the squares of those sides is 281-What is the length and breadth of the parallelogram? Ans. 16 and 5 inches. PROBLEM IX. Having the sum of the two adjacent sides of a rectangular parallelogram, and the sum of the squares of those sides given, to find the area. Rule-From the square of their sum take the sum of their squares, then from the sum of their squares take this remainder, and the square root of the difference will be the difference of the two numbers; to half their sum add half their difference and the sum will be the length of the longest side, subtracted leaves the shortest, then multiply the length by the breadth and the product will be the area. EXAMPLES. 1. The sum of the two adjacent sides of a rectangular parallelogram is 50 chains, and the sum of the squares of those sides is 1300-Required the area of the paral lelogram. Ans. 30 and 20 chains, the length of the sides required, and the area, 60 acres. 2. The sum of the squares of the two adjacent sides of a gentleman's garden, in the form of an oblong square or rectangular parallelogram, is 464 rods, and the length of the wall that encloses it, is 56 rods-Required the area of the parallelogram. Ans. 1 acre. PROBLEM X. Having the area of a parallelogram, and the length of the diagonal line given, to find the length of the sides. Rule-From the square of the diagonal, subtract twice the area, and the square root of the remainder will be the difference of the two adjacent sides. To the square of the diagonal add twice the area, and the square root of the sum will be the sum of the two sides; then to half the sum add half the difference, and the sum will be the length of the longest sidesubtracted, leaves the shortest. EXAMPLES. 1. The area of a rectangular parallelogram is 30 acres, and the length of the diagonal is 25 chainsRequired the sides of the parallelogram. Ans. 20 and 15 chains the length and breadth. 2. The area of a rectangular parallelogram is 6 acres and 12 rods, and the length of the diagonal is 45 rods What is the length of the sides of the parallelogram? Ans. 36 rods the length, and 27 the breadth. PROBLEM XI. Having the area of a rectangular parallelogram, and the difference of the squares of the two adjacent sides given, to determine the length of the sides of the parallelogram. Rule-To the square of the area add the square of half the difference of their squares, and, from the square root of that sum, subtract half the difference of their squares, and the remainder will be the square of the shortest side; extract the square root and that will be the length of the shortest side-divide the area by the shortest side, and the quotient will be the longest. EXAMPLES. 1. The area of a rectangular parallelogram is 4 acres, 3 roods, and 8 rods, and the difference of the squares of the adjacent sides is 28 chains-What is the length of the sides? Ans. 8 chains by 6. 2. The difference of the squares of the adjacent sides of a parallelogram is 128 chains, and the area 48 chains—What is the length of the sides? Ans. 12 chains by 4. OF THE RHOMBUS. A rhombus is a quadrilateral figure, whose sides are equal, but whose angles are not right angles. PROBLEM I. To find the area of a rhombus. Rule Multiply the length of one side by a perpendicular let fall fr: one of the angles to the opposite side. EXAMPLES. 1. The length of the sides of a rhombus are each 12,24 feet, and the perpendicular height 9,16 feetRequired its area. Ans. 112,1184 feet. 2. Required the area of a rhombus, whose length is 12 feet, 6 inches, and its height 9 feet, 3 inches. Ans. 115,625 feet. 'PROBLEM II. I The area of a rhombus, and the length of the side being given, to find the perpendicular height. RuleDivide the area by the length of the side, the quotient will be the length of the perpendicular; or, divide by the height, the quotient will be the length of the side. EXAMPLES. 1. The area of a rhombus is 24 rods, and the length of the side 5 rods-What is the length of the perpendicular? Ans. 4,8 rods the perpendicular. 2. The area of a rhombus is 125 feet, and the perpendicular height, 8 feet, 6 inches; required the length of the side? Ans. 14,7 + feet. OF THE RHOMBOID. A rhomboid, or rhomboides, is a quadrilateral figure, whose sides are parallel, and angles not right angles. PROBLÈM 1. To find the area of a rhomboid. Rule-Multiply the length by a perpendicular let fall from one of the angles to the opposite side. EXAMPLES. 1. The length of a rhomboid is 26 rods, and the perpendicular height 8 rods-Required the area. Ans. 1 acre, 1 rood, and 8 rods. 2. The length of a rhomboid is 18 feet, 9 inches, and the perpendicular height 12 feet, 3 inches-How many square yards does it contain? Ans. 25,5208+ square yards. 3. The length of a rhomboid is 36 feet, and the perpendicular height 24 feet, 9 inches-How many square rods does it contain? Ans. 31 square rods. PROBLEM II. When the area of a rhomboid, and the length are given, to find the perpendicular. Rule-divide the area by the length, the quotient will be the perpendicular; or, divide the area by the perpendicular, the quotient will be the length. EXAMPLES. 1. The area of a rhomboid is 4 acres, 3 roods, and 12 perches, and the length 9 chains and 20 links-What is the length of the perpendicular? Ans. 5,2445+chains. 2. The area of a rhomboid is 47 feet 9 inches, and the perpendicular is 4 feet, 8 inches-What is the length? Ans. 12,141 + feet the length. 3. The area of a rhomboid is one square rod, and the perpendicular is 7 feet, 2 inches-What is its length? Ans. 37,99-feet. OF RIGHT-ANGLED TRIANGLES, A right-angled triangle, is a figure bounded by three sides, and has one right angle: the longest side is called the hypotenuse, and the other two, the legs, otherwise known by the names of base, and perpendicular. PROBLEM I. To find the area of a right-angled triangle, the base and perpendicular being given. Rule-Multiply half the length of the base by the length of the perpendicular, the product will be the area; or half the perpendicular by the whole length of the base, the product will be the area. EXAMPLES. 1. The base of a right-angled triangle is 36 rods, and the perpendicular 12 rods-How many acres does it contain? Ans. 1 acre, 1 rood, and 16 rods. 2. The perpendicular of a right-angled triangle is 24 chains and 76 links, and the base 41 chains and 23 links-How many acres does it contain? Ans. 51 acres, 61% rods. PROBLEM II, Rule-di When the area and the length of either base or perpendicular are given, to find the other. vide twice the area by the base, the quotient will be the perpendicular; or, divide by the perpendicular, the quotient will be the base. EXAMPLES. 1. The area of a triangle is 16 feet 6 inches, and the |