EXAMPLES. 1. To find the area of a square, whose side is 6 inches 2. To find the area of a rectangle, whose length is and breadth 4 inches, or feet, &c. RULE 2. If any two sides of a parallelogram be multiphed together, and the product again by the natural sine of their included angle, the last product will give the area of the par allelogram. DEMONSTRATION. For having drawn the perpendicular AP, the area, by the first rule, is APX BC; but as rad. 1 (s. P) 3. To find the area of a rhombus, whose length is 6:20 chains, and perpendicular height 5'45. 4. To find the area of the rhomboid, whose length is 12 feet 3 inches, and breadth 5 feet 4 inches. 5. To find the area of a rectangular board, whose length is 12.5 feet, and breadth 9 inches. Ans. 9 ft. 6. To find the square yards of painting in a rhomboid, whose length is 37 feet, and breadth 54fect. Ans. 21 square yards. PROBLEM 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 * A triangle is half a parallelogram of the same base and altitude, (Euc. I. 41) and therefore the truth of the rule is evident. DEMONSTRATION. Take any triangle ABC, and let amp be its inscribed circle, whose centre is o; join Ao and Co, and let fall the perpendiculars on, om, and op; in CA produced take AI Bn, and erect the perpendicular Now, as CpCm, Ap-An, and BBn, it is evident, that CH, or CL, will half the perimeter of the triangle, and that HA, Ap, and ¿C will be the differences between the half petimeter and each side respectively. And since CH=CL, CG VOL. II. E common, three remainders continually together; and the square root of the last product will be the area of the triangle. EXAMPLES. common, and the angle HCG angle LCG, therefore GL= GH, and the angle GLC angle GHC a right angle. Also, as GH=GL, HI=LB, and the angles H and L are right angles, therefore GI=GB. In like manner, as GI GB, AI AB, and GA common, therefore the angle GAI= angle GAB. = But the points A, n, o, p, fall in the circumference of a circle, therefore the angle IAB angle pon; (Euc. III. 22) and consequently their halves HAG and Acp are also equal to each other, and the triangle AHG similar to the triangle Apo. And, as the triangles Cpo and CHG are also similar, we shall have HG po HC: Cp and pA: HG :: po: AH; whence HGXpA: pox HG :: HCxpo: CpxAH or pA : po :: HCxpo: CpxAH, or pA×CH: poxCH :: HCxpo: CpXAH, which is the same as the rule; for if this be expressed algebraically, it will be CHXpAX Cpx AH, which is the rule =√CH1×po2=CHxpo = the area. Q. E. D. COR. I. If the triangle be right-angled, the rectangle of the half perimeter, and the difference between the half perimeter and the hypotenuse will be the area; because when CAB is a right angle, BL will be equal to po. COR. 2. If the triangle be equilateral, 43, multiplied by the square of the side, will be the area; because, in that case, the perimeter is three times the side, and the three differences are all equal to each other. RULE 3. Any two sides of a triangle being multiplied together, and the product again by half the natural sine of their included angle, will give the area of the triangle. DEMONSTRATION. This follows from Rule 2, mentioned in the note under Prob. I. for a triangle is half of a parallelogram of the same base and height. |