N.B.-An imperial gallon contains 277,274 cubical inches.

3. Show that less tin will be used in making a vessel of the dimensions given in the last example, than

in making one of any other dimensions but of a cylindrical form and the same capacity.

4. Investigate Thomas Simpson's rule for determining the area of a plane surface bounded by an irregular line.

5. Investigate a formula for determining the quantity of earth to be taken out of a cutting for a road, the width of the road and the slope of the banks being given.

SECTION IV.

1. Describe Gunter's chain.

2. Construct a field-book for a three-sided field of which one side has an irregular form, and the other two are straight lines; assuming any dimensions whatever.

3. Describe and explain the vernier.

4. Describe the spirit level and its adjustments.

SECTION I.

1. "Prove the rule of cross multiplication." Let us find the area of the floor in ques. 1. sec. ii. to exemplify the rule.

Here the product of the 6 by 15 is divided by 12, and the remainder put in the inches' place; the quotient is carried to the square feet. In multiplying by the 3 in., the first remainder from the division of 12 is put one place further to the right; the process is then continued, dividing each product by 12 and placing the remainder in order towards the left hand. Now the result which is generally termed 266 feet, 10 inches, 6 parts, is really 266 feet, 10 parts, 6 inches; the parts being rectangles of 1 by 12 in., which, of course, are each 12 square inches in area. The result expressed in feet and fractions of a foot would stand thus,

Let us now prove that this is the true result, by working with the 17 ft. 6 in. and the 15 ft. 3 in. in the form of their equivalents, 17 and 15 ft.; thus,

The first product is obtained by multiplying by the 15 whole numbers, and the second by the fraction, care being taken to retain the duodecimal scale, or the decrease of the numbers from the left in a twelvefold ratio.

66

2. Prove the rule for determining the number of standard rods of brickwork in a wall."

Rule:-Multiply the number of feet in the surface of the wall by the number of half bricks in the thickness of the wall, and divide the result by 3 × 272.

The standard rod is 272 feet of surface of brickwork,

of a brick and a half thick. This rod is in surface nearly the same as that in land measure for the square of 5 yards, or 16.5 feet equals 272.25.

Let a ft. = the length of the wall

b ft. = the height

And с

is in thickness

the number of half bricks the wall

the standard thickness.

ахъхс

3 x 272

the number of feet in the wall of

the number of standard rods.

3. "Prove a rule for determining the area of a trapezoid."

"The area of a trapezoid is equal to half the area of a rectangle, having the same altitude, and whose base equals the sum of the parallel sides of the trapezoid."— Tate's Geometry. Art. 43.

1.

"How many

SECTION II.

cubical feet of timber are there in the flooring of a room in. thick and 17 ft. 6 in. in length by 15 ft. 3 in. in breadth?"

2. "In a wall 10 ft. high, 15 ft. long, and 2

bricks thick, there is an arched doorway 4 ft. wide and

6 ft. high to the springing of the arch, which is semicircular; how many standard rods of brick-work are there in the wall?

15 X 10 X 5

3 x 272

doorway.

6 x 4 x 5

3 x 272

number of rods including the

number of rods in the door-way to

3. "What is the weight of a circular iron ring whose inner diameter is 18 in., and whose section is a circle 2 in. in diameter, the weight of a cubic foot of iron being 450 lbs. ?"

in.

in.

18+ 2 = 20 diameter of ring to the centre of circular rod.

20 × 3·1416 = 62·832 = the length of circular rod in inches.

22 x 7854 = 3·1416 = area of section of circular rod in inches.

62.832 X 3.1416 =

197.393

1728

197.393, solid content in in.

× 450 = 51·4044 lbs. weight.

SECTION III.

1. "After measuring a piece of cloth, to contain 90 yards, I find that the yard measure that I have used is

too short by 1-30th part; what is the true measure of of the cloth?"

Every length measured contained 1 yard minus. 1-30th yard.

.. 90 lengths 90 yards minus 90 X =

1

30

yards.

= 90 387 yards, the true measure.

2. "How many square inches of tin plate are required to make an open cylindrical vessel to contain a gallon whose height is equal to one half its diameter ?" "N.B.-An imperial gallon contains 277 274 cubical inches."

Let x the diameter of the vessel

But the vessel is to contain 277.274 cubical inches.

Having found the diameter of the vessel we can now find the surface.

8.9 8.92 x 7854 + 8.9 x 3.1416 X = surface in 2 186 6346, the number of square inches re

inches quired

3. "Show that less tin will be used in making a vessel of the dimensions given in the last question, than in making one of any other dimensions but of a cylindrical form and the same capacity."