Let S represent half the sum of the three sides of the triangle; then a+b+c= 2 S; a+c―b=2 (S — b); b+ca=2 (S— a); a+b―c=2 (S — c) ; hence = ABC 16 S (Sa) X (S-b) x (Sc), which, being reduced, gives as the area of the triangle, as given above, ✔S EXAMPLES. 1. What is the area of a triangle, ABC, whose sides, AB, BC, CA, are 40, 30, and 50 feet? 30+40502 = 60, half the sum of the three sides. 603030, first remainder. 604020, second remainder. 605010, third remainder. 60 X 30 X 20 x 10 = 360,000; 360,000 square feet, the area required. = 600, 2. How many square feet in a triangular floor, whose sides are 15, 16, and 21 feet? 3. Required the area of a triangular field whose sides. are 834, 658, and 423 links. Ans. 1 A. 1 R. 20 P. 4 yd. 1.6 ft. 4. Required the area of an equilateral triangle, of which each side is 15 yards. 5. What is the area of a garden in the form of a parallelogram, whose sides are 432 and 263 feet, and a diagonal 342 feet? Ans. 2 A. 10 P. 11.46 yd. 6. Required the area of an isosceles triangle, whose base is 25 and each of its equal sides 40 rods. 7. What is the area of a rhomboidal field, whose sides are 57 and 83 rods, and the diagonal 127 rods? Ans. 22 A. 3 R. 21 P. 26 yd. 5 ft. PROBLEM X. 623. Any two sides of a RIGHT-ANGLED TRIANGLE being given, to find the third side. To the square of the base add the square of the perpendicular; and the square root of the sum will give the hypothenuse (Prop. XI. Bk. IV.). From the square of the hypothenuse subtract the square of the given side, and the square root of the difference will be the side required (Prop. XI. Cor. 1, Bk. IV.). EXAMPLES. 1. The base, AB, of the triangle ABC is 48 feet, and the perpendicular, B C, 36 feet; what is the hypothenuse? 482362 = 3600; 3600 60 feet, [the hypothenuse required. A C B 2. The hypothenuse of a triangle is 53 feet, and the perpendicular 28 feet; what is the base ? 3. Two ships sail from the same port, one due west 50 miles, and the other due south 120 miles; how far are they apart? Ans. 130 miles. 4. A rectangular common is 25 rods long and 20 rods wide; what is the distance across it diagonally? 5. If a house is 40 feet long and 25 feet wide, with a pyramidal-shaped roof 10 feet in height, how long is a rafter which reaches from the vertex of the roof to a corner of the building? 6. There is a park in the form of a square containing 10 acres; how many rods less is the distance from the centre to each corner, than the length of the side of the square? Ans. 11.716 rods. PROBLEM XI. 624. The sum of the hypothenuse and perpendicular and the base of a RIGHT-ANGLED TRIANGLE being given, to find the hypothenuse and the perpendicular. To the square of the sum add the square of the base, and divide the amount by twice the sum of the hypothenuse and perpendicular, and the quotient will be the hypothenuse. From the sum of the hypothenuse and perpendicular subtract the hypothenuse, and the remainder will be the perpendicular. 625. Scholium. This problem may be regarded as equivalent to the sum of two numbers and the difference of their squares being given, to find the numbers (National Arithmetic, Art. 553). NOTE. The learner should be required to give a geometrical demonstration of the problem, as an exercise in the application of principles. EXAMPLES. 1. The sum of the hypothenuse and the perpendicular of a right-angled triangle is 160 feet, and the base 80 feet; required the hypothenuse and the perpendicular. Ans. Hypothenuse, 100 ft.; perpendicular, 60 ft. 160280232,000; 32,000 (160 × 2) = 100; 60. 2. Two ships leave the same anchorage; the one, sailing due north, enters a port 50 miles from the place of departure, and the other, sailing due east, also enters a port, but by sailing thence in a direct course enters the port of the first; now, allowing that the second passed over, in all, 90 miles, how far apart are the two ports? 3. A tree 100 feet high, standing perpendicularly on a horizontal plane, was broken by the wind, so that, as it fell, while the part broken off remained in contact with the upright portion, the top reached the ground 40 feet from the foot of the tree; what is the length of each part? Ans. The part broken off, 58 ft.; the upright, 42 ft. PROBLEM XII. 626. The area and the base of a TRIANGLE being given, to find the altitude; or the area and altitude being given, to find the base. Divide double the area by the base, and the quotient will be the altitude; or divide double the area by the altitude, and the quotient will be the base. 627. Scholium. This problem is the converse of Prob. VIII. EXAMPLES. 1. The area of a triangle is 1300 square feet, and the base 65 feet; what is the altitude? 1300 X 22600; 2600 ÷ 65 = 40 ft., altitude required. 2. The area of a right-angled triangle is 17,272 yards, of which one of the sides about the right angle is 136 yards; required the other perpendicular side. 3. The area of a triangle is 46.25 chains, and the altitude 5.2 chains; what is the base ? 4. A triangular field contains 30 A. 3 R. 27 P.; one of its sides is 97 rods; required the perpendicular distance from the opposite angle to that side. Ans. 102 rods. PROBLEM XIII. 628. To find the area of a TRAPEZOID. Multiply half the sum of its parallel sides by its altitude (Prop. VII. Bk. IV.). EXAMPLES. 1. What is the area of the trapezoid ABCD, whose parallel sides, A B, D C, are 32 and 24 feet, and the altitude, EF, 20 feet? 322456; 56228; 28 × 20 560 sq. ft., = [the area required. 2. How many square feet in a board in the form of a trapezoid, whose width at one end is 2 feet 3 inches, and at the other 1 foot 6 inches, the length being 16 feet? 3. Required the area of a garden in the form of a trapezoid, whose parallel sides are 786 and 473 links, and the perpendicular distance between them 986 links. Ans. 6 A. 33 P. 3 yd. 4. How many acres in a quadrilateral field, having two parallel sides 83 and 101 rods in length, and which are distant from each other 60 rods? PROBLEM XIV. 629. To find the area of a REGULAR POLYGON, the perimeter and apothegm being given. Multiply the perimeter by half the apothegm, and the product will be the area (Prop. VIII. Bk. VI.). 630. Scholium. This is in effect resolving the polygon into as many equal triangles as it has sides, by drawing lines from the center to all the angles, then finding their areas, and taking their sum. 2. What is the area of a regular pentagon, whose sides. are each 25 feet, and the perpendicular from the center to a side 17.205 feet? 3. A park is laid out in the form of a regular heptagon, whose sides are each 19.263 chains; and the perpendicular |