As rad. 90 = is to ÁD = 55 = 1,74036 is to AF = 36 So is fine 40° 30' 9,81254 So is fine 48° 20′ - 9,87334 10,00000| As rad. 90° 10.00000 = 1.55630 To Dd 35,72 1,55290 To Ff 26.89 = 1,42964 Now, to find the area by bafes and perpendiculars. Ex. 3. Required the area of the following polygon, whereof the fides are as follow, viz. AF 31.5, ,FE 33.5 ED 25.5, DC 38.5, CB 43.5, BA 34.5, AE 60.5, AD 81.7, BD 74.3 English chains. Anf. 277 acres 3 ro. 12 perches. PRO PROBLEM XII. To find the angles of any regular polygon. By cor. 1ft, I. 32. Euclid. All the anterior angles of any. rectilineal figure, together with four right angles, are equal to` twice as many right angles as the figure has fides. Hence the following rule. RULE. From double the number of fides fubtract 4, and the remainder is the number of right angles contained by all the fides of the polygon. Multiply the remainder by 90, and divide the product by the number of fides, the quot gives the degrees in any of the angles. 108 degrees in each angle. Ex. 2. Required the angle of a heptagon. Anf. 128° 34' PROBLEM XIII. To find the area of a regular polygon. RULE. Find the area of a triangle, conftructed on one of the fides of the polygon, and whofe vertex is in the centre; then multiply this area by the number of fides, and the product will be the area of the polygon. Or, Multiply the perimeter by the radius of the infcribed circle, and half the product is the area of the polygon. EXAMPLE I. Required the area of a pentagon, whofe fide is 10. Ex, 2. Required the area of a hexagon, whofe fide is 30. Anf. 2338.2 Ex. 3. Required the area of a heptagon, whofe fide is 1. Anf. 3.633912. Ex. 4. Required the radius of the infcribed circle, area, internal angles, and angles at the centre, for the following polygons: viz. The trigon, tetragon, pentagon, hexagon, heptagon, octagon, enneagón, decagon, undecagon, and dodecagon, the fide of each being 1, Regular polygons of the like number of fides are fimilar, and fimilar furfaces are to one another in the duplicate ratio of their homologous fides; but the fides of the polygons in the foregoing table are each of them I; therefore, as the fquare of i is to the tabular area, fo is the fquare of the fide of any given polygon to the area required: Hence the following RULE. Multiply the fquate of the fide of any given polygon into the tabular area of the like polygon, and the product will be the area of the polygon. Ex. 5. Required the area of a pentagon whofe fide is 20 feet. Ex. 6. What is the area of a hexagon, whose fide is 50 yards? Anf. 6495.18 Square yards. Ex. 7. Required the area of a hexagon, whose fide is 20 Feet. Anf. 1039.2288 Ex. 8. Required the area of a pentagon, whofe fide is 4. Anf. 27.5276. Ex. 9. How many square yards are in a decagon, whose fide is 12 feet? Anf. 123 yards 11 inches 6 pts, PROBLEM XIV. The area of a polygon being given to find the fide, RULE. Divide the area of the given polygon by the tabular area of the like polygon, and the square root of the quotient will be the fide of the given polygon. EXAMPLE I. Required the fide of a pentagon, whofe area is 61.9371. 1.720473)61.937100(36 51 61425 10322850 36-6 the fide of poly. 10322850 Anf. 20 yards. Required the fide of a decagon, whofe area is 3077.682. (quare yards. Ex. 3 What is the fide of a trigon, whofe area is 173.2025 Anf. 20. fquare yards? Ex. 4. Required the fide of a pentagon, whofe area is 27.5276 square yards. Anf. 4 yards. OF |