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157. What is the area of a triangle, of which the base is 30 rods, and the perpendicular 10 rods? Ans. 150 rods. 158. If the area be 150 rods, and the base 30 rods, what is the perpendicular? Ans. 10 rods. 159. If the perpendicular be 10 rods, and the area 150 rods, what is the base? Ans. 30 rods. When the legs (the base and perpendicular) of a rightangled triangle are given, how do you find its area?
When the area and one of the legs are given, how do you find the other leg?
Note. Any triangle may be divided into two right-angled triangles, by drawing a perpendicular from one corner to the opposite side, as may be seen by the annexed figure.
Here A B C is a triangle, divided into two right-angled triangles, Ad C, and d B C; therefore the whole base, A B, multiplied by one half the perpendicular d C, will give the area of the whole. If A B = 60 feet, and Ans. 480 feet.
d C = 16 feet, what is the area? 160. There is a triangle, each side of which is 10 feet; what is the length of a perpendicular from one angle to its opposite side? and what is the area of the triangle ?
Note. It is plain, the perpendicular will divide the opposite side into two equal parts. See ¶ 109.
Ans. Perpendicular, 8'66 + feet; area, 43'3 + feet. 161. What is the solid contents of a cube measuring 6 feet on each side? Ans. 216 feet. When one side of a cube is given, how do you find its olid contents?
When the solid contents of a cube are given, how do find one side of it?
162. How many cubic inches in a brick which is 8 inches long, 4 inches wide, and 2 inches thick? in 2 bricks? in 10 bricks? Ans. to last, 640 cubic inches. 163. How many bricks in a cubic foot? in 40 cubie feet? Ans. to last, 27000. 164. How many bricks will it take to build a wall 40 feet in length, 12 feet high, and 2 feet thick? Ans. 25920.
in 1000 cubic feet?
165. If a wall be 150 bricks, = 100 feet, in length, and 4 bricks, = 16 inches, in thickness, how many bricks will lay one course?
10 courses? If the
wall be 48 courses, = 8 feet, high, how many bricks will build it? 150 X 4 = 600, and 600 X 48 28800, Ans. 166. The river Po is 1000 feet broad, and 10 feet deep, and it runs at the rate of 4 miles an hour; in what time will it discharge a cubic mile of water (reckoning 5000 feet to the mile) into the sea? Ans. 26 days, 1 hour.
167. If the country, which supplies the river Po with water, be 380 miles long, and 120 broad, and the whole land upon the surface of the earth be 62,700,000 square miles, and if the quantity of water discharged by the rivers into the sea be every where proportional to the extent of land by which the rivers are supplied; how many times greater than the Po will the whole amount of the rivers be? Ans. 1375 times.
168. Upon the same supposition, what quantity of water, altogether, will be discharged by all the rivers into the sea in a year, or 365 days?
Ans. 19272 cubic miles.
169. If the proportion of the sea on the surface of the earth to that of land be as 104 to 5, and the mean depth of the sea be a quarter of a mile; how many years would it take, if the ocean were empty, to fill it by the rivers running at the present rate? Ans. 1708 years, 17 days, 12 hours.
170. If a cubic foot of water weigh 1000 oz. avoirdupois, and the weight of mercury be 13 times greater than of water, and the height of the mercury in the barometer (the weight of which is equal to the weight of a column of afr on the same base, extending to the top of the atmosphere) be 30 inches; what will be the weight of the air upon a square foot? a square mile? and what will be the whole weight of the atmosphere, supposing the size of the earth as in questions 166 and 168?
Ans. 2109'375 lbs. weight on a square foot.
of the whole atmosphere.
171. If a circle be 14 feet in diameter, what is its circumference?
Note. It is found by calculation, that the circumference of a circle measures about 34 times as much as its diameter, or, more accurately, in decimals, 3'14159 times. Ans. 44 feet. 172. If a wheel measure 4 feet across from side to side, how many feet around it? Ans. 124 feet 173. If the diameter of a circular pond bè 147 feet, what is its circumference? Ans. 462 feet.
174 What is the diameter of a circle, whose circumference is 462 feet? Ans. 147 feet. 175. If the distance through the centre of the earth, from side to side, be 7911 miles, how many miles around it?
7911 X 3'1415924853 square miles, nearly, Ans. 176. What is the area or contents of a circle, whose diameter is 7 feet, and its circumference 22 feet?
Note. The urea of a circle may be found by multiplying the diameter into the circumference. Ans. 381 square feet. 177. What is the area of a circle, whose circumference is 176 rods? Ans. 2464 rods. 178. If a circle is drawn within a square, containing 1 square rod, what is the area of that circle?
Note. The diameter of the circle being 1 rod, the circumference will be 3'14159. Ans. "7854 of a square rod, nearly. Hence, if we square the diaseter of any circle, and multiply the square by 7854, the product will be the area of the circle.
179. What is the area of a circle whose diameter is 10 rods? 102 X 7854 = 78'54. Ans. 78'54 rods. 180. How many square inches of leather will cover a ball 34 inches in diameter ?
Note. The area of a globe or ball is 4 times as much as the area of a circle of the same diameter, and may be found, therefore, by multiplying the whole circumference into the whole diameter. Ans. 38 square inches. 181. What is the number of square miles on the surface of the earth, supposing its diameter 7911 miles?
7911 X 24853:
196,612,083, Ans. 182. How many solid inches in a ball 7 inches in diame ter?
Note. The solid contents of a globe are found by multiplying its area by part of its diameter.
Ans. 1793 solid inches. 183. What is the number of cubic miles in the earth, supposing its diameter as above?
Ans. 259,233,031,435 miles. 184. What is the capacity, in cubic inches, of a hollow globe 20 inches in diameter, and how much wine will it contain, 1 gallon being 231 cubic inches?
Ans. 4188'8+ cubic inches, and 18'13+ gallons. 185. There is a round log, all the way of a bigness; the areas of the circular ends of it are each 3 square feet;
how many solid feet does 1 foot in length of this log contain? 2 feet in length? A solid of this form is called a Cylinder.
How do you find the solid content of a cylinder, when the area of one end, and the length are given?
186. What is the solid content of a round stick, 20 feet long and 7 inches through, that is, the ends being 7 inches in diameter?
Find the area of one end, as before taught, and multiply it by the length. Ans. 5'347 cubic feet.
If you multiply square inches by inches in length, what parts of a foot will the product be? if square inches by feet in length, what part?
187. A bushel measure is 18'5 inches in diameter, and 8 inches deep; how many cubic inches does it contain? Ans. 2150'4+.
It is plain, from the above, that the solid content of all bodies, which are of uniform bigness throughout, whatever may be the form of the ends, is found by multiplying the area of one end into its height or length.
Solids which decrease gradually from the base till they come to a point, are generally called Pyramids. If the base be a square, it is called a square pyramid; if a triangle, a triangular pyramid; if a circle, a circular pyramid, or a cone. The point at the top of a pyramid is called the vertex, and a line, drawn from the vertex perpendicular to the base, is called the perpendicular height of the pyramid.
The solid content of any pyramid may be found by multiplying the area of the base by of the perpendicular height. 188. What is the solid content of a pyramid whose base is 4 feet square, and the perpendicular height 9 feet?
42 X = 48.
Ans. 48 feet. 189. There is a cone, whose height is 27 feet, and whose base is 7 feet in diameter; what is its content?
Ans. 3464 feet. 190. There is a cask, whose head diameter is 25 inches, bung diameter 31 inches, and whose length is 36 inches; how many wine gallons does it contain? how many
Note. The mean diameter of the cask may be found by adding 2 thirds, or, if the staves be but little curving, 6 tenths, of the difference between the head and bung diame
ters, to the head diameter. The cask will then be reduced to a cylinder.
Now, if the square of the mean diameter be multiplied by 7854, (ex. 177,) the product will be the area of one end, and that, multiplied by the length, in inches, will give the solid content, in cubic inches, (ex. 185,) which, divided by 231, (note to table, wine meas.) will give the content in wine gallons, and, divided by 282, (note to table, beer meas.) will give the content in ale or beer gallons.
In this process we see, that the square of the mean diameter will be multiplied by '7854, and divided, for wine gallons, by 231. Hence we may contract the operation by only multiplying by their quotient (3850034;) that is, by '0034, (or by 34, pointing off 4 figures from the product for decimals.) For the same reason we may, for beer gallons, multiply by (7854 = '0028, nearly,) ‘0028, &c.
Hence this concise RULE, for guaging or measuring casks,Multiply the square of the mean diameter by the length; multiply this product by 34 for wine, or by 28 for beer, and, pointing off four decimals, the product will be the content in gallons and decimals of a gallon.
In the above example, the bung diameter, 31 in. - 25 in. the head diameter 6 in. difference, and of 6 = 4 inches; 25 in. + 4 in. 29 in. mean diameter.
Then, 292841, and 841 × 36 in. — 30276.
30276 × 34 = 1029384. Ans. 102'9384 wine gals. 30276 X 28 = 847728. Ans. 847728 beer gals. 191. How many wine gallons in a cask whose bung diameter is 36 inches, head diameter 27 inches, and length 45 inches? Ans. 166 617.
192. There is a lever 10 feet long, and the fulcrum, or prop, on which it turns, is 2 feet from one end; how many pounds weight at the end, 2 feet from the prop, will be balanced by a power of 42 pounds at the other end, 8 feet from the prop?
Note. In turning around the prop, the end of the lever 8 feet from the prop will evidently pass over a space of 8 inches, while the end 2 feet from the prop passes over a space of 2 inches. Now, it is a fundamental principle in mechanics, that the weight and power will exactly balance each other, when they are inversely as the spaces they pass over. Hence, In this example, 2 pounds, 8 feet from the prop, will balance