Problem 7. To find the Area of a Trapezium. Def. A Trapezium consists of four unequal Sides, and four unequal Angles. Rule. * Add the two Perpendiculars together, and multiply that Sum by Half the Diagonal; (or multiply the Diagonal by Half the Sum of the Perpendiculars) and the Product will be the Area. Example. Suppose ABCD be a Trapezium, whose Diagonal A C is 108 Inches; the Perpendicular BP 38 Inches; and the Perpendicular DE 34.5 Inches; what is the Area? Operation. 72.5 × 54 = 3915.0 Inches, the Area. * Every Trapezium being a double Triangle, is confequently equal to Half its circumfcribing Parallelogram. Problem 8. To find the Area of a Regular Polygon. Def. All Figures that have more than four Sides, and those all equal, are called Regular Polygons. Of fuch Figures, those of five Sides are called Pentagons; those of fix Sides, Hexagons; of seven Sides, Heptagons; of eight Sides, Octagons; of nine Sides, Nonagons; of ten Sides, Decagons, &c. whence it is evident, that they take their Names from the Number of Angles within them. Rule. * Multiply Half the Sum of the Sides by the nearest Diftance of any Side from the Center. Example. Let ABCDE be a Pentagon, each of whose Sides is 25 Inches, and the nearest Distance from the Center to the Middle of any Side is 17.2 Inches; what is its Area ? Operation. 62.5 × 17.5 = 1075.00 Inches, the Area fought. * Every Regular Polygon is equal to a Triangle, whose Base is equal to the Sum of the Sides, and whose Height is the Distance of any Side from the Center. To find the Center, only bisect 2 of its Angles, as D and C, and the Point of Meeting of these Lines will be the Center required. Problem 9. To find the Area of a Circle. Def. A Circle is a plain Figure, whose Area is bounded by one continued Line, called the Circumference, or Periphery; and it is every where equally diftant from a Point within, called its Center. Note. The Line going through the Center of the Circle, dividing it into two equal Parts, is called the Diameter; Half of which is called the Semi-diameter, or Radius. Rule. * Square the Diameter, and multiply it by .7854 (a Decimal) and that Product will be the Content. Example. Suppose the Diameter of the Circle A B be 41.2 Inches, what is its Area? Operation. 41.2×41.2=1697.44.7854=1333.169376 Inches, the Area. * The Area of a Circle whose Diameter is 1, is .7854; and the Areas of all Circles are to each other, as the Square of their Diameters; whence the above Rule.- The Square of a Number is the Product fing from multiplying that Number by itself. Problem 10. To find the Area of a Circle by another Method. Rule. * Multiply half the Circumference by half the Diameter, and the Product is the Area. Note. If the Diameter of a Circle be given, the Cir cumference of it may be easily found by the following Proportions: As 7'is to 22, so is the Diameter to the Circumference. Or, as 113 is to 355, so is the Diameter to the Circumference. Or, as 1 is to 3.14159, so is the Diameter to the Circum ference. The Squaring of a Circle, as it is usually called, or finding a Square exactly equal to a Circle given, is what many have endeavoured to perform; but none as yet have absolutely compleated it; because none have found out the exact Proportion of the Diameter to the Circumference. Archimedes, a very famous Mathematician, first discovered that the Diameter was in Proportion to the Circumference, as 7 is to 22 nearly; which serves very well for common Ufe. After him, one Metius found that the Diameter was (more nearly) to the Circumference, as 113 to 355. * Every Circle, as a Polygon of an infinite Number of Sides, is equal to a Triangle, whose Base is the Circumference and Height to half the Diameter. + The Emperor Charles V. offered a Reward of One Hundred Thousand Crowns to the Perfon who should solve this celebrated Pro. blem; and the States of Holland have proposed a Reward for the fame Purpose. But the Moderns have computed the Proportion of the Diameter to the Circumference to greater Exactness. For, fuppofing the Diameter to be 100000, 000, the Periphery they find will be more than 31415, and less than 31416; but Ludolphus Van Geulen has exceeded the Labours of all who went before him; for by immenfe Application he found, that suppofing the Diameter to be 1.00000.000.000.000.000.000.00০.০০০.০০০.০০০ the Periphery or Circumference will be less than 3.14159.265.358.979.323.846.264.338.327.951 but greater than 3.14159.265.358.979.323.846.264.338.327.950 This able Mathematician has carried on his Calculation to 36 Places of Decimals; and to eternalize the vast Work, they are engraven upon his Tomb-ftone in St. Peter's Church at Leyden, in Holland, of which large Number the first fix Places 3.14159, answering to the Diameter of 1.00000, are fufficient in any Calculation. But fince Van Ceulen's Time, our Countryman Mr.. Machin has carried the Proportion on to 100 Places. He finds that if the Diameter be 1, the Circumference will be 3.14159.26535.89793.23846.26433.83279.50288.47971. 69399.37510.58209.74944.59230.78164.05286.20899. 86280.34825.34211.70679 of the fame Parts. But the Ratios generally used are those of Archimedes, Metius, or four or five leading Figures of Van Geulen. |