3. How many acres are contained in a circle whose diameter is 20 perches ?

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

RULE. Divide the area by .7854, and the square root of the quotient will be the diameter. Or divide the area by .07958, and the square root of the quotient will be the circumference.

EXAMPLES.

1. The area of a circle being 2 acres, 3 roods, and 12 perches, what is the circumference?

2 acres, 3 roods, 12 perches-452 perches. 452.07958 ✓5679.69-75.3657 perches, the circumference.

2. The area of a circle is 5 acres, 3 roods, and 26 perches: what is the diameter ?

3. The area of a circular garden containing 2 acres, what is the length of a brick wall which will enclose it?

PROBLEM XXI. To find the area of an ellipse, the transverse and conjugate diameters being given.

RULE. Multiply the transverse and conjugate diameters together, and the product again by .7854, and the result will be the area.

EXAMPLES.

1. The transverse and conjugate diameters of an ellipse are 60 and 40 perches: how many acres does it contain? (60×40.7854)=188.4.96=11 acres, 3 roods, 4.96 per. 2. The transverse and conjugate diameters of an ellipse are 8 and 6 chains: how many acres does it contain >

3. What is the area of an ellipse whose diameters are 24 and 18 feet?

PROBLEM XXII. The transverse and conjugate diameters being given to find the circumference.

RULE. Multiply the square root of half the sum of the squares of the two diameters by 3.1416, and the product will be the circumference.

EXAMPLES.

1. The transverse and conjugate diameters of an ellipse are 12 and 10 yards; required the circumference.

✓( (144+100)÷2)=√122=11.045, and

11.045×3.1416=34.699.

2. The transverse and conjugate diameters of an ellipse are 30 and 20 feet: what is the circumference?

3. What is the circumference of an ellipse whose transverse and conjugate diameters are 80 and 60 chains?

PROBLEM XXIII. To find the side of a square equal in area to any given superficies.

RULE. The square root of any given superficies will be the side of the square required.

EXAMPLES.

1. If the area of a given triangle be 961 perches, what is the side of a square equal in area thereto ?

✓961-31 perches, the side required.

2. If the area of a given circle be 9876 perches, what is the side of a square equal in area?

3. If an elliptical fish-pond contain one acre (=4840 square yards), required the side of a square fish-pond of equal dimensions.

PROBLEM XXIV. To find the area of the surface of a cube. RULE. Multiply the square of the length of one side by 6, and the product will be the area.

EXAMPLES.

1. The side of a cube is 15 inches: what is the area of its surface?

15×15×6=1350 inches=93 feet.

2. The side of a cube is 21 inches: what is the area of its surface?

3. The side of a cube is 6 feet: what is the area of its surface?

PROBLEM XXV. The area of the surface of a cube being given, to find the length of the side.

RULE. Divide the area by 6, and extract the square root of the quotient.

EXAMPLES.

1. The area of a cube is 2400 square inches: what is the length of the side?

✓(2400÷6)=400-20 inches.

2. The area of a cube is 216 square feet: what is the length of its side?

3. What is the length of a side of a cube whose area conains 96 square feet?

PROBLEM XXVI. To find the solidity of a cube, the height of one of its sides being given.

RULE. Cube the given side.

EXAMPLES.

1. What is the solidity of a cube whose side is 4 feet? 4x4x4-64 solid feet.

2. How many solid feet in a cubic box, whose depth is 23 inches?

3. How many cubic inches are contained in a cubical piece of timber, whose length is 76 inches?

PROBLEM XXVII. To find the side of a cube, the solidity being given.

RULE. Extract the cube root of the solidity.

EXAMPLES.

1. What is the length of the side of a cubic box containing 36 solid feet?

36 3.302 the side required.

2. The height of a cubic box containing 60 bushels, each bushel containing 2150.4 solid inches?

3. The height of a cubic box containing 16 bushels ?

PROBLEM XXVIII. To find the solidity of a parallelopipedon.

RULE. Multiply the length by the breadth, and that product by the depth or altitude, and it will give the solidity required.

EXAMPLES.

1. The length of a parallelopipedon is 4 feet, its breadth 3 feet, and thickness 2 feet: what is the solidity?

4×3×2=24 solid feet.

2. How many bushels are contained in a bin 5 feet 6 inches long, 4 feet 9 inches wide, and 3 feet 9 inches deep? 3. What is the solidity of a block of marble whose length is 12 feet, breadth 5 feet, and thickness 2 feet?

PROBLEM XXIX. To find the solidity of a prism.

RULE. Find the area of the base, and multiply it by the height, the product will be the solidity.

EXAMPLES.

1. What is the solidity of a triangular prism, whose length is 12 feet, and each of its equal sides 2 feet?

The area of the base found by problem XI.=2.7063.

And 2.7063x12=32.4756 solid feet.

2. What is the value of a triangular prism, whose height is 48 feet, and each side of the base 4 feet, at 16 cents per solid foot?

3. What is the solidity of a square prism, each of its sides being 7 feet, and height 24 ?

PROBLEM XXX. To find the surface of a cylinder.

RULE. Multiply the circumference of the end by the length, and the product will be the convex surface, to which add the area of each end, and the sum will be the whole surface of the cylinder.

EXAMPLES.

1. What is the convex surface of a cylinder, whose length is 12 feet, and the diameter of its end 3 feet?

3.1416x3=9.4248 circum. and 9.4248×12=113.0976 convex surface.

2. What is the whole surface of a cylinder, whose length is 16 feet, and the circumference of its base 8 feet? 8x16 128 convex surface.

8x8x.07958x2=10.18624 area of both ends.

Then 128.+10.18624=138.18624 the whole surface. 3. How many yards are there in the convex surface of a cylinder 13 feet long, and diameter 4 feet?

PROBLEM XXXI. To find the solidity of a cylinder.

RULE. Multiply the square of the diameter by .7854, or the square of the circumference by .07958, and that product by the length will give the solidity.

EXAMPLES.

1. How many solid inches are there in a grindstone, whose diameter is 20 inches, and thickness 4 inches? .7854×400×41=1413.72 solid inches.

2. How many gallons of water, each containing 282 solid inches, can be put into a cylindrical vessel 18 inches in diameter, and 2 feet in height?

3. The circumference is 12 feet 4 inches, and length 18 feet what is its solidity?

PROBLEM XXXII. To find the whole surface of a cone. RULE. Multiply the circumference of the base by the slant height, or the length of the side of the cone, and half the product will be the area of the convex surface, to which add the area of the base.

EXAMPLES.

1. The slant height of a cone is 20 feet, and the diameter is 4 feet: what is the whole surface of the cone? (4×3.1416×20)÷2=125.664 convex surface. .7854×16=12.5664 area of the base. 125.664+12.5664=138.2304 whole surface.

2. The slant height of a cone is 18 feet, and the diameter 5 feet required the convex surface of the cone?

3. The circumference of a cone is 12 feet, and the slant height 10: what is the whole surface of the cone?

PROBLEM XXXIII. To find the solidity of a cone, the diameter of the base and perpendicular altitude given.

RULE. Multiply the square of the diameter of the base by .7854, or the square of the circumference of the base by .07958, and that product by of the perpendicular altitude, the product will be the solidity.

EXAMPLES.

1. The diameter of the base of a cone is 4 feet, 6 inches, and the perpendicular altitude is 10 feet: what is the solidity? 4.5x4.5X.7854×(10÷3)=53.0145 solidity.

2. The circumference of the base is 12 feet, and the perpendicular altitude is 16 feet: what is the solidity?

3. The altitude of a cone is 8 feet, and the diameter of the base 6 feet what is its solidity?

PROBLEM XXXIV. To find the solidity of a pyramid. RULE. Multiply the area of the base by of the altitude, and the product will be the solidity.

EXAMPLES.

1. What is the solidity of a pyramid, the area of whose base is 418 feet, and the altitude is 18 feet?

418x(18-3)=2508 feet solidity.

2. What is the solidity of a triangular pyramid, the area of whose base is 270.6 yards, and the altitude is 13 yards?