In this case we must consider that two sides of a triangle are given to find the third: The ladder, 120 feet, is the hypothenuse, the distance the foot stands from the steeple, 20 feet, is the short leg of the triangle, and that part of the steeple from the foundation to the top of the ladder is the leg sought; subtract this leg, when found, from the height of the steeple, the difference will be the answer. Diff. OPERATION. The short leg 20 20 Square 400 ✓ 1,40,00(118.32 feet from the foundation to the 1 21)040 21 228) 1900 [top of the ladder. Height of steeple, 120. ft. 118.32 Difference, 1.68 If I have 300 apple trees, and wish to set them two rods or 50 links apart, on a piece of land in form of a parallelogram, whose length and breadth shall be as 4 to 3, how many trees must stand on the length, and how many on the breadth? And, how many acres of ground will they require? Answer, 20 trees in length, and 15 in breadth. 6 acres, 2 roods, 24 perches. RULES. 1. As the ratio in breadth, 2. As the ratio in length, is to the ratio in breadth; So is the number of trees, to a fourth number. 3. Extract the square root of each fourth number, and you will have the number of trees sought. Multiply the two fourth numbers together, the product will be the quantity of land. Trees in breadth 15-1-14. Distance apart 50 links. Trees in length 20-1-19. Distance apart 50 links. 4th number for the breadth,700 4th no. for length, 950 950 length. 700 breadth. 6165000 links = 6 A. .65 4 Roods 2.60 40 Perches 24.00 LESSON 12. Let 6400 men be formed in solid column, with 4 times as many in file as in rank; how will they stand? Ans. 40 in rank and 160 in file. RULE. The Square Root of one fourth of the men will give the number in rank; and the number in rank multiplied by the ratio 4, will give the number in file. Let 17298 men be so formed that the number in rank will be double the number in file. Answer, 186 in rank and 93 in file. MODE OF OPERATION. 1729828649. 864993 in file. 93 in file x 2 = 186 in rank. LESSON 14. The area of a circle contains 176.4 acres; how long is the side of a square farm which will contain the same quantity of land? Ans. 42 chains. In this case reduce the acres into links by annexing four cyphers to the decimal .4, which will be the same as annexing five cyphers to whole numbers of acres. OPERATION. ✓ 17,64,00,00(4200 links = 42 chains for one side 16 82).164 164 [of a square field. 8400)...0000 LESSON 15. The area of a circle given to find the diameter.* RULE. Multiply the Square root of the area by 1.12837: or, Multiply the area by 1.273241, the product will be the square of the diameter; consequently the Square Root of that product will be the diameter. EXAMPLE. The area of a circle contains 113.09 superficial inches; how long is the diameter? Ans. 11.999 inches, or 12 inches nearest. * Diameter is the longest line across a circle. Semidiameter is half the diameter. Periphery, or Circumference, is the line round the circle. The diameter of a circle given to make another circle 2, 3 or 4 times greater. RULE 1. Multiply the square of the diameter by the given pro portion, and extract the square root of that product for an answer. EXAMPLE 1. If the diameter of a circle be 12 inches, what will be the length of the diameter of a circle 4 times as large? Answer, 24 inches. 12 x 12 = 144 the square of the diameter. 4 the proportion. ✔576(24 inches, answer. 4 44)176 176 RULE 2. If the required circle be less than the one given, then divide the square of the diameter of the given circle by the proportion and extract the square root of that quotient for an answer. |