Problem III. To Find The Area Of A Triangle Whose Three Sides Are Given

Rule. - From the half-sum of the three sides subtract each side severally. Multiply the half-sum and the three remainders together, and the square-root of the product will be the area required.

Problem IV. Any Two Sides Of A Right-Angled Triangle Being Given, To Find A Third Side

Case I. - When two sides are given, to find the hy-pothenuse.

Rule. - Add the squares of the two legs together, and the square-root of the sum will be the hypothenuse.

Case II. - The hypothenuse and one of the legs being given, to find the other leg.

Rule. - From the square of the hypothenuse take the square of the given leg, and the square-root of the remainder will be equal to the other leg.

Problem V. To Find The Area Of Any Regular Polygon

Rule. - Multiply half the perimeter of the figure by the perpendicular falling from its centre upon one of the sides, and the product will be the area of the polygon.

Problem VI. To Find The Area Of A Regular Polygon, When The Side Only Is Given

Rule. - Multiply the square of the given side of the polygon by that number which stands opposite to its name in the following table, and the product will be the area: -

As the foregoing table extends to five places of decimals, it is exact enough for all practical purposes.

Problem VII. The Diameter Of A Circle Being Given, To Find The Circumference

Rule. - Multiply the diameter by 22, and divide the product by 7: the quotient will be the circumference. Or multiply the diameter by 3, and add a seventh part of the diameter to the product: the sum will be the circumference as obtained before. Either of these methods is sufficiently correct for common purposes.

Problem VIII. The Circumference Of A Circle Being Given, To Find The Diameter

Rule. - Multiply the circumference by 7, and divide the product by 22: the quotient will be the diameter.

Problem IX. The Chord And Height Of A Segment Being Given, To Find The Radius Of The Circle

Rule. - To the square of the half-chord add the square of the height, and divide the sum by twice the height of the segment: the quotient will be the radius of the circle when it is less than a semicircle.

Problem X. To Find The Area Of A Circle, The Diameter Being Given

Rule. - Multiply half the circumference by half the diameter, and the product will be the area.

Problem XI. To Find The Area Of A Sector Of A Circle

Rule. - Multiply the radius, or half the diameter, by half the length of the arc of the sector; and the product will be the area.

Problem XII. To Find The Area Of The Segment Of A Circle, The Chord And Height Of The Arc Being Given

Rule. - To two-thirds of the product of the base, multiplied by the height, add the cube of the height divided by twice the length of the segment; and the sum will be nearly the area.

Problem XIII. To Find The Area Of An Ellipsis, The Transverse And Conjugate, Or Long And Short, Diameters Being Given

Rule. - Multiply the transverse axis by the conju-gate, and the product multiplied by .7854 will be the area.

Problem XIV. To Find The Area Of A Prism

Rule. - Multiply the area of the base, or end, by the perpendicular height; and the product will be the solidity.

Problem XV. To Find The Solidity Of A Pyramid

Rule. - Multiply the area of the base, or end, by the perpendicular height; and a third of the product will be the solidity.

Problem XVI. To Find The Solidity Of The Frustum Of A Square Pyramid

Rule. - To the rectangle of the sides of the two ends add the sum of their squares. That sum being multiplied by the height, a third of the product will be the solidity.

Problem XVII. To Find The Solidity Of A Sphere, Or Globe

Rule. - Multiply the cube of the diameter by .5236, and the product will be the solidity.

Problem XVIII. To Find The Solidity Of The Segment Of A Globe

Rule. - To three times the square of half the diameter of the base of the segment add the square of the height of the same. Multiply that sum by the height named, and the product multiplied by .5236 will give the, solidity.

Problem XIX. To Find The Convex Superfice Of A Right Cylinder, The Circumference And Length Being Given

Rule. - Multiply the circumference by the length, and the product will be the area.

Problem XX. To Find The Convex Superfice Of A Right Cone, The Circumference And Slant Side Being Given

Rule. - Multiply the circumference by the slant side, and half the product will be the area.

Problem XXI. To Find The Convex Superfice Of The Frustum Of A Cone, The Circumferences Of Both Ends And The Slant Side Being Given

Rule. - Multiply the sum of the circumferences by the slant side, and half the product will be the area.

Problem XXII. To Find The Superfice Of A Sphere, Or Globe, The Circumference Being Given

Rule. - Multiply the square of the circumference by .3183, and the product will be the superfice.

Problem XXIII. To Find The Convex Superfice Of The Segment Of A Globe, The Diameter Of The Base Of The Segment And Its Height Being Given

Rule. - To the square of the diameter of the base add the square of twice the height, and the sum multiplied by .7854 will give the superfice.

Problem XXIV. To Find The Convex Superfice Of An Annulus, Or Ring, Whose Thickness And Inner Diameter Are Given

Rule. - To the thickness of the ring add the inner diameter. Multiply the sum by the thickness, and the product multiplied by 9.869 will be the superfice.