by the base A C = 680 links, the product will be the quantity required.

Bb 160+ Da 274 434 ÷ 2 = 217 × Ac 680 = 1.476, or la. 1R. 36p.

Note.-This calculation may in all cases be shortened by using the scale, Fig. 1, Plate 39. Observe it is divided one side by 40, that being the scale the plan is plotted, and in computing quantities is applied to measure the base of all triangles; on the other edge of the scale it is divided by 20, being one half, and is used to measure the perpendiculars of all triangles, as shown by the following example; it saves adding the whole perpendiculars, and taking half.

Bb 80+ Da 137 =217 × 680 = 1.476, as before

The same by natural sines, Fig. 65, Plate 4.

Rule 7. Multiply the two diagonals together, and multiply half the product by the natural sine of the included angle.

To find the area of a rhombus or rhomboid, Fig. 66.

Rule 8. Multiply the base by the perpendicular height. Given the base A B = 400 links; the perpendicular Ca 342 links.

A B 400 x C a 342 = 1.368, or la. 1R. 19P.

Required the area of a rhomboid, Fig. 67.

Given the base Ą B = 600 links; the perpendicular Ca 176

Problem 83.

To find the area of a trapezoid, Fig. 68.

Rule 9. Add together the two parallel sides, and half that sum, multiplied by the height, will give the required area. Given A B 440 links, CD 325 links, ba 280 links.

AB 440+ CD 325 762 ÷ 2 = 382 x 280 1.0696, or la. OR. 11p.

Also draw the line ab through the middle of AC parallel to BD, measure the length a B, multiply it by the height ab equal to BD, the result will be the same.

Note.-For practice calculate the last four figures by Rule 6, page 58.

Problem 84.

To find the area of a pentagon, Fig. 69.

Rule 1. Multiply the perimeter or sum of its sides by a perpendicular from its centre to one of its sides, and take half the product for the area.

Rule 2. Multiply the square of the side of a polygon by the number in the third column of areas (see Table, No. 14) opposite the number of the sides required, the product will be the area required.

Required the area of a regular pentagon, each side being 210, and the perpendicular F G 144.

A B 210 × 5 = 1050 × FG 144 = 1.51200 ÷ 2 = .75600

The same by Rule 2.

2102 = 44100 x tab. num. 1.720.75852, or OA. 3R. 1p.

· Problem 85.

Required the area of an octagon, each side being 150 and the perpendicular 195, Fig. 70.

A B 150 x 8 = 1200 x a A 195 =234000

The same by Rule 2.

= 1.170

CIRCLES.

Problem 86.

To find the area of a circle.

Rule 1. Multiply the square of the diameter by .7854, or the diameter by the circumference, and divide by 4.

To find the area of a sphere.

Rule 2. Multiply the square of the diameter by 3.1416.

To find the solidity of a sphere.

Rule 3. Multiply the cube of the diameter by .5236.

To find the circumference of a circle.

Rule 4. Multiply the diameter by 3.1416.

To find the surface of a spherical segment or zone. Rule 5. Multiply the diameter by the height, and then by 3.1416.

To find the solidity of a spherical segment.

Rule 6. To three times the square of the radius (or half the diameter) add the square of its height and the product by .5236.

The area being given to find the diameter.

Rule 7. Divide the area by .7854, extract the square root of the quotient for the diameter required.

The diameter given to find the area.

Rule 8. Multiply the square of the diameter by .7854.

Problem 87.

The diameter of a circle given to find the circumference, or the circumference given to find the diameter.

Rule 1. As 7 is to 22 so is the diameter to the circumference. Or: As 22 is to 7 so is the circumference to the diameter. Rule 2. As 1 is to 3.1416 so is the diameter to the circumference.

Or: As 3.1416 is to 1 so is the circumference to the diameter. Example 1. Required the circumference of a circle whose diameter is 30.55.

30.55 × 22 = 67210 ÷ 7 = 96.014 the circumference

Example 2.

96.01 × 767210 22 = 30.55 the diameter

Example 3. Required the circumference of a circle whose diameter is 30.55.

3.1416 x 30.55 95.975880 the circumference

Example 4. Required the diameter of a circle whose circumference is 95.975880.

95.975880 3.1416 30.55 the diameter

AREAS AND PROPERTIES OF CIRCLES.

Problem 88.

Examples to the Table, No. 19, of the relative proportions of the circle; its equal and inscribed squares.

Example 1. The diameter of a circle is 33.25; required the side of a square equal in area to the given circle.

33.25 × 8862 = 29.466150 the side of the square

Example 2. The circumference of a circle being 104.5, required the side of a square equal in area.

104.5 x .2821 = 29.4794 the side

Example 3. The diameter of a circle being 33.25, required the side of the greatest square that can be inscribed therein.

33.25 × 7071 = 23.511075 the side of inscribed square

Example 4. The circumference of a circle being 104.5, required the inscribed square.

104.5 × 2251 = 23.522 the inscribed square

Example 5. The area of a circle being 868.309, required the area of the greatest square that can be inserted therein.

868.309 × 6366 = 552.7655 the area required

Example 6. The side of a square being 23.511, required the diameter of its circumscribing circle.

23.511 × 1.4142 = 33.2492562

Example 7. Required the circumference of a circle to circumscribe a square, each side being 23.511.

23.511 × 4.443 104.456373

Example 8. The side of a square being 29.4794, required the diameter of a circle equal in area to the given square.

29.4794 × 1.128 33.25276

Example 9. The side of a square being 29.4794, required the circumference of a circle equal in area to the given square.

29.4794 × 3.545 104.5044

SOME OF THE PROPERTIES OF A CIRCLE.

1. It is the most capacious of all plane figures, or contains the greatest area within the same perimeter or outline.

2. The areas of circles are to each other as the squares of the diameters, or of their radii.

3. Any circle whose diameter is double that of another, contains four times the area of the other.

4. The area of a circle is equal to the area of a triangle, whose base is equal to the circumference and perpendicular equal to the radius.

5. The area of a circle is equal to the rectangle of its radius, and a right line equal to half its circumference.

6. The area of a circle is to the square of the diameter as .7854 to 1; or multiply half the circumference by half the diameter, the product will be the area.

Problem 89.

To find the area of the sector of a circle when the degrees of the arc are given, Fig. 73, Plate 5.

Rule 1. The area of a sector depends on the proportion its arc bears to the whole circle, therefore divide 360° by the number of degrees in the arc of the sector for the first quotient; divide the area of the whole circle by the quotient thus found, and the last quotient will be the area of the sector.

Example 1. What is the area of a sector whose radius. BO= 2.50, and the arc A B C 60°.

360° 60° = 6 the first quotient

Then for the area of the whole circle. (See Prob. 86.)