2. 'To find the area of a rectangle, whose length is 9, and breadth 4 inches, or feet, &c. 9 4 Ansr. 36 3. To find the area of a rhombus, whose length is 6 chains, and perpendicular height 5. 5 5 Ansr. 30 Prob. 2. To find the Area of a Triangle. Rule 1. Multiply the base by the perpendicular height, and half the product will be the area. Rule 2. When the three sides only are given: Add the three sides together, and take half the sum; from the half sum subtract each side separately; multiply the half sum and the three remainders continually together; and the square root of the last product will be the area of the triangle. Ex. Required the area of the triangle whose base is 6 feet, and perpendicular height 5 feet. 6 5 2) 30 (15 Ansr. Prob. 3. To find one Side of a right-angled Triangle, having the other two Sides given. The square of the hypotenuse is equal to both the squares of the two legs. Therefore, 1. To find the hypotenuse; add the squares of the two legs together, and extract the square root of the sum. 2. To find one leg; subtract the square of the other leg from the square of the hypotenuse, and extract the root of the difference. Ex. 1. Required the hypotenuse of a right-angled triangle, whose base AB is 40, and perpendicular BC 30. 4 4 3 3 - 16 9 25 25 (5 the square root of the sum of the two squares, being the hypotenuse AC. 2. What is the perpendicular of a right-angled triangle, whose base AB is 56, and hypotenuse, AC 65 ? 9 1089 (33 The perpendicular, which is the root of the remainder of the square of the hypotenuse AC, when the square of AB has been subtracted. 63 189 Prob. 4. To find the Area of a Trapezoid. Multiply the sum of the two parallel sides by the perpendicular distance between them, and half the product will be the area. Ex. In a trapezoid, the parallel sides are AB 7, and CD 12, and the perpendicular distance AP or CN is 9 : required the area. 7 12 19 9 171 85 the area. Prob. 5. To find the Area of a Trapezium, Case for any trapezium.-Divide it into two triangles by a diagonal; then find the areas of these triangles, and add them together. Note. If two perpendiculars be let fall on the diagonal, from the other two opposite angles, the sum of these perpendiculars being multiplied by the diagonal, half the product will be the area of the trapezium. Ex. To find the area of the trapezium ABCD, the diagonal AC being 42, the perpendicular BF 18, and the perpendicular DE 16. 18 16 34 42 68 136 2) 1428 714 the answer. Prob. 6. To find the Area of an Irregular Polygon. Draw diagonals dividing the figure into trapeziums and triangles. Then find the areas of all these separately, and their sum will be the content of the whole irregular figure. Ex. To find the content of the irregular figure ABCDEF, in which are given the following diagonals and perpendiculars namely, c.a = 10 polygon. Prob. 7. To find the Area of a Regular Polygon. Rule. Multiply the perimeter of the figure, or sum of its sides, by the perpendicular falling from its centre upon one of its sides, and half the product will be the area. Prob. 8. In a Circular Arc, having any two of the following lines, viz. the chord AB, the versed sine DP, the chord of half the arc AD, and the diameter, or the radius AC or CD given, to find the others. If any two of these lines be given, two sides of one of the rightangled triangles, APC or APD, will be known, and from them the remaining side, and other lines in the arc, may be found by Prob. 3. Suppose AB and PD be given, then, by Prob. 3., the half of AB, or AP is a mean proportional between DP and PC + CD; for PC + CDPD is the diameter of the circle, half of which is the radius or CA, and by Prob. 3, AC2 - AP2 = CP2, and AP2 + PD2 =AD2. Suppose CD and AB be given, then half of AB = AP, and CP, and CD CD AC; therefore ✓ CD2 ✔ PD2 + AP2 = AD. AP2 CP PD. Prob. 9. To find the Diameter and Circumference of a Circle, the one from the other. Rule 1. As 7 is to 22, so is the diameter to the circumference. Rule 2. As 113 is to 355, so is the diameter to the circumference As 355 is to 113, so is the circumference to the diameter Rule 3. As I is to 3·1416, so is the diameter to the circumference. As 3 1416 is to 1, so is the circumference to the diameter. Er. 1. To find the circumference of a circle, whose diameter AB is 10. 2. To find the diameter when the circumference is 100. By Rule 1. 7 × 25 = 175 22:7:50: = 15 = 15'9090 ans 11. = 11 |