MENSURATION OF SUPERFICIES. Q. What is meant by Superficial Measure? A. The measurement of the surface or outside of any body, as of a piece of land, the outside of a building, of painting, paving, &c. Q. How are superficies or surfaces measured? A. By the superficial foot, yard, rod, or acre, according to the measure of different artificers. Q. How are the superficial contents of every surface found? A. By the proper rule of its figure, whether square, triangular, or circular. ARTICLE I. TO MEASURE A SQUARE AND PARALLELOGRAM. Q. What is the RULE for finding the superficial contents of a square of equal sides? A. Multiply the length of one side into itself, and the product will be the superficial contents or area, in the same denomination of the given sides, which must be brought to the denomination required. Q. How do you find the superficial area of a parallelogram, or long square? A. Multiply the length by the breadth, and the product will be the superficial area as before. EXAMPLES. 1. There is a garden, laid out in an exact square, measuring, on each side, 160 feet; how many square yards does it contain? Operation. 160x160 25600-9-28444 yds. Ans. 2. There is a farm, lying in the form of a parallelogram, measuring 240,5 rods in length, and 87,5 rods in width; how many acres does it contain? Ans. 131 acres, 2 roods, 3 rods. 3. How many feet of boards will lay the floor of a room, measuring 16 by 18 feet? Ans. 288 feet. 4. A man bought a tract of land, measuring 320 rods in length, and 175 rods in breadth, at $12 an acre; how many acres did it contain, and what was the amount of his purchase? Ans. 350 acres. Amount, $4375. ARTICLE II. TO MEASURE A RHOMBUS AND RHOMBOID. Q. What is a Rhombus ? A. It is a figure in the form of a diamond, having four equal sides, its opposite angles be. ing equal; two of which are greater, and two less than the angles of a square. U* Q. What is a Rhomboid? A. It is a four sided figure, whose opposite sides and ends are equal, but meeting each other obliquely, making two of its angles greater, and two less than those of a parallelogram. Q. What is the RULE for finding the superficial area of a Rhombus, or of a Rhomboid? A. Multiply the length of one side by the perpendicular width of the figure; as the dotted lines across the figures, in the following examples. EXAMPLES. 1. Let the annexed figure, A B C D represent a field of four equal sides, measuring 45 rods, each; the dotted perpendicular, E measures 36 rods; how many square rods are contained in the field? B 45 D E Ans. 45×36-1620 square rods, A 2. Let the annexed rhomboidal figure represent a field, whose sides, A B and C D, measure 75 rods, and whose perpendicular width, E, is 45 rods; how many c rods, and how many acres does it contain? 3. There is a lot, ly. ing between two streets, which run in such direc tion as to bring it into the form of a rhomboid; the lot is 160 feet deep, and measuring from corner A D 75 B 75 4:5 D E Ans. 3375 rods 21 acres, 15 rods. E sh STREET 160 с to B, and also, from corner D to E, the distance is 75 feet; but the perpendicular distance, A C, measures but 60 feet; how many square feet less are there in the lot, than there would be, if it lay in an exact parallelogram of 75 feet in width? Ans. It contains 2400 feet less. ARTICLE III. TO MEASURE A TRAPEZOID. Q. What is a Trapezoid? A. It is a four sided figure, having its sides parallel, but not equal; and its ends equal, but not parallel, and two of its angles greater and two less than 90 degrees. 1 Q. What is the RULE for measuring a Trapezoid? A. Add the length of the two longest sides together, and multiply the sum by the perpendicular breadth, and half the product will be the answer. EXAMPLES. C A 1.20 B 180 09 1. In a lot of land described by the figure A B C D, the line A B, is 120 rods; the line CD, is 180 rods; the two lines AC, and B D, are 70 rods; but the dotted perpendicular A, is but 60 rods; how many acres are contained in the lot, and how many acres would there be, if the lines A C, and B D, were raised perpendicular to C D, and the line A B, etxended to ⚫ them ? It now contains 56 It would contain 784 acres. Ans.{ 2. A garden, lying in the form of the annexed figure, measures as fol. lows the line A B, 60 feet; the line C D, 90 feet; and the end B D, 50 feet; how many square feet, and square yards does it contain ? Ans. 3750 ft. and 416 square yds. A 60 90 acres. 3. How many feet in a board 24 feet, 6 inches long; 2 feet, 4 inches wide, at one end, and 16 inches wide at the other? Ans. 44 feet, 11 inches. ARTICLE IV. TO MEASURE TRIANGLES. Q. What are the three sides, that enclose a right angled triangle, called? A. They are called the Base, the Perpendicular, and the Hypothenuse. The base and perpendicular are also called legs. Q. Which is the base of a triangle? In A. The longest side, in all oblique angles, is called the base. right angled triangle, the longest side is called the hypothenuse, and the longest of the two legs is called the base, and the shortest leg is called the perpendicular. Q. What is the perpendicular in all oblique angles? A. It is a line drawn perpendicularly to the base, from the angle opposite to the base. Q. What known proportion do the sides of right angled triangles bear to each other? A. The square of the hypothenuse is always equal to the sum of the squares of the other two legs; therefore, by having any two sides of an angle given, the other is easily found. Q. When the two legs of a right angled triangle are given to find the hypothenuse, what is the RULE? A. Square each given side, and add them together, then extract the square root of this sum and you will have the length of the hypothenuse, in the same denomination of the given sides. Q. When the hypothenuse, and one of the legs are given to find the other leg, what is the RULE ? A. Subtract the square of the given leg, from the square of the hypothenuse, and the remainder is the square of the unknown side; extract the square root of this remainder, and you will have the length of the required side. Q. How do you find the superficial contents or area of a right angled triangle? A. Multiply the base and perpendicular together, and half the product will be the area or contents. Q. How can you measure an oblique angled triangle, as an isosceles, or scalene triangle? A. Multiply the base, by a perpendicular, let fall upon the base, from the angle opposite to it, and half the product will be the answer. Or in measuring any angle, multiply the base by half the perpendicular, or the perpendicular by half the base, and the product is the area. Q. What reason can you give for this RULE of measuring triangles ? A. Every triangle is just one half of a square, or a parallelogram, of equal base and perpendicular height; and as in measuring those figures we multiply the length by the breadth, it is evident, that to measure an angle, which contains just one half of a square, or of a parallelogram, we must multiply one half of the length by the breadth; or, if the whole length and breadth are multiplied together, it must also be evident, that half the product will be the area of the angle. EXAMPLES. 1. In the annexed right angled triangle, the base A B, is 45 rods; the perpendicular B C, is 30 rods; what is the length of the hypothenuse, A C, and how many square rods does the angle contain? Hypothenuse 54 rods+. A. 2. In the annexed angle, the base A B, is 40 rods; the perpendicular CD, is 20 rods; what is the length of the sides, A D and B D, and how many square rods does it contain? Ans. The lines A D, and B D, 28,28 A. rods. Superficial area 400 rods. 46 B 39 3. In the annexed angle, the line A B, is 46 rods; the line B C, is 38 rods, and the line DC, is 30 rods; what is the perpendicular height from D to B, and A also, what is the distance from A to D, and how many acres are contained in the whole triangular field? Ans. Perpendicular B D, 23,32 rods. The distance A D, 39,64 rods. Superficial area, 5 acres, 12 rods. D 30 4. An eagle, sitting on the top of a tree, measuring 144 feet in height, was shot by a boy, from behind a wall, 48 yards distant from the root of the tree; how many yards distance did the boy shoot? Ans. 67 yds. 3 qr. 2 na. 5. Two ships part at sea, one sails due north, 64 leagues, the other due west, 81 leagues; how far are the two ships from each other? Ans. 103 leag. 0 m. 5 fur. 22 rds. 4 yds. nearly. 6. If a house be 36 feet wide, how long must the rafters be, to make the perpendicular height of the roof 16 feet; and how many square feet, are contained in each gable end of the house? Ans. Length of rafters, 24,08 feet. Area of each end, 288 ft. 7. There is a well 6 feet in diameter, and the post which supports the sweep, is 20 feet high, and stands 16 feet from the curb of the well, the top of which is 3 feet from the ground; what length of well-sweep, will it require, if hung in the centre, for the end of the sweep, when drawn down, to strike the centre of the curb ?. Ans. 50,98 feet. SUPPLEMENT TO ARTICLE FOURTH. Q. In what proportion does the width of a triangle decrease, as you recede or measure from the base? A. The width of a triangle decreases in exact proportion to the perpendicular distance from the base. Q. If the length of the base and the width of the angle, a certain distance from the base, be given, how can you find the length of the whole perpendicular, from the base to the opposite angle, or the distance from the base to the point, where the other two sides meet? |