Note.-As the standard thickness is 1 brick thick, then 16 feet long, 1 foot thick and 1 feet high == 24.75 feet, or 1 perch, hence we multiply the length, breadth and thickness of the wall together, and divide by 24.75 for the number of perches required.

DIGGING.

Cellars, vaults, clay for brick, canals, &c., are measured by the solid yard of 27 feet.

RULE.-Multiply the length, width and depth, together and divide the product by 27, for the number of cubic yards.

How many yards of digging in a cellar 25 feet long, 20 feet wide and 10 feet 6 inches deep? Ans. 1943 cub. yds.

Note.-A solid yard of clay will make 7 or 800 brick, and 3 bushels of lime and half a load of sand will be sufficient to lay 1000 brick.

To find how many thousand brick will be required for building a house of any given dimensions.

Suppose a house of the following dimensions, viz: 84 feet long, 40 feet wide, 20 feet high and the walls to be 1 foot.thick? Ans. 105,408 bricks. RULE.-Deduct the thickness of the wall, from the length of each side, because the inner sides are 1 foot less in height than the outer sides. This rule is unquestionably correct.

wide, and 2 inches thick, there would be 80 cubic inches

in a brick, hence 843264(0

=

105,408 brick, Ans.

8(0

2. How many thousand brick 8 inches long, 4 inches wide, and 24 inches thick, will build a wall in front of a church which is to be in compass 240 feet long, 6 feet high and 1 foot 6 inches wide? Ans. 51.840 bricks.

3. How many shingles will it take to cover the roof of a barn 40 feet long, allowing the length of the rafters to be 16 feet 6 inches, and 6 shingles to cover square foot; what will they cost at $1.25 per 1000?

Ans. 7,920 shingles; cost $9.90.

MENSURATION OF SUPERFICES AND SOLIDS.

PROBLEM 1.

To find the area of a square.

RULE.-Square the side: and the pro

duct or rectangle will be the superficial

content.

A

D

1. A lot of ground is 10 perches square, what is the area? Ans. 100 ps. = 2 r. 20 p.

PROBLEM 2.

2. What is the content of a board 15 feet long and 2 feet wide? Ans. 30 feet. 3. What is the difference between a floor 40 feet square Ans. 800 feet.

and 2 others, each 20 feet square?

4. There is a square of 3600 yards area; what is the side of a square, and the breadth of a walk along each side of the square at each end, which may take up just one half the square?

Ans. 42.42 yds. side of the sq. 8.78 + yds. breadth of walk.

PROBLEM 3.

To find the area of a rhombus. RULE.-Multiply the length of the base by the perpendicular height. 5. The base of a rhombus is 14 C feet C and its height 6 feet, required the area?

Ans. 84 feet.

PROBLEM 4.

To find the area of a triangle. RULE.-Multiply half the base

by the perpendicular height, or if

the perpendicular is not given, A

add the three sides together, take

D

B

half that sum, subtract each side severally; from the half sum multiply the half sum and the three differences together, and the square root of the product will be the area. 1. Required the area of a right angled triangle whose base is 40 and perpendicular 30 perches? Ans. 600.

2. Required the area of a triangle whose sides are 10, 12 and 18 perches respectively? Ans. 56.57 perches.

PROBLEM 5.

CASE I.

By having the diameter of a circle to

find the area.

RULE.-Square the diameter, and multiply the product by .7854 for the

area.

CASE II.

Diameter.

By having the circumference of a circle to find the area. RULE.-Square the circumference and multiply that square by .07958.

1. The diameter of a circle is 24, required the area? Ans. 452.4904.

2. The circumference of a circle is 80, required the area? Ans. 509.312.

CASE III.

By having the diameter to find the circumference. RULE-Multiply the diameter by 3.1416, and you get the circumference.

If the diameter of a circle be 24, what is the circumference? Ans. 75.3984.

CASE IV.

By having the circumference of a cube to find the diameter. RULE.-Multiply the circumference by .31831 and the product is the diameter.

PROBLEM 6.

I have a circular field 50 rods in diameter, what is the side of a square field, that shall contain the same area? Ans. 44.31.

SOLIDS

Are figures having length breadth and thickness.

PROBLEM 7.

To find the content of a cube or paral- C lelopipedon, whose side is 18 inches. RULE.-Multiply the length, height,

and breadth continually together, and the product is the content.

A

Ans. 3 feet.

Ꭰ

B

1. How many cubic feet in a cube whose side is 18 inches? 2. What is the content of a parallelopipedon whose length is 6 feet, height 2 feet, breadth 13 feet? Ans 18ft.

3. A cellar is 50 feet long, 38 feet wide and 12 feet deep, how many cubic yards of earth has been taken out in digging, and what was the expense of digging it at 10 cts. per cubic yard?

Ans. 844.44 + cubic yds.; expense $84.45 nearly.

PROBLEM 8.

To find the solidity of a Prism. RULE.-Multiply the area of a base or end by the height.

4. Required the solidity of a triangular prism whose length is 10 feet, and the three sides of its triangular base are 5, 4 and 3 ft. Ans. 60ft.

PROBLEM 9.

To find the solidity of a Cylinder.
RULE.—Multiply the area of the base by the

length.

5. The diameter of the base of a cylinder is 10 inches and its length 24 feet, required the solidity? Ans. 13.09 feet.

PROBLEM 10.

D

To find the solidity of a cone or pyramid. RULE.-Multiply the area of the base by of its height. 6. What is the solid content of a cone whose heighth is 12 feet, and the diameter of the base 24 feet?

To find the superfices of a Cone. RULE.-Multiply the circumference of the

base by half its slant height.

7. What is the convex surface of a cone, whose slant height is is 20 feet, and the circumference of its base 9 feet?

To find the solidity of the frustum of a cone or pyramid. RULE.-Multiply the diameters of the two bases together, and to the product add of the square of the difference of the diameters; then multiply this sum by .7854 and