MEASURING BY THE SQUARE OF 100 FEET, As Flooring, Partitioning, Roofing, Tyling, &c.

EXAMPLES.

31. In 173 feet 10 inches in length, and 10 feet 7 inches in height of partitioning, how many squares?

Ans. 18 squares, 39 feet, 8 inches, 10 p. 32. If a house of three stories, besides the ground floor, was to be floored at £6...10...0 per square, and the house measured 20 feet 8 inches, by 16 feet 9 inches: there are 7 fire places, whose measures are two of 6 feet, by 4 feet 6 inches each, two of 6 feet, by 5 feet 4 inches each, and two of 5 feet 8 inches, by 4 feet 8 inches, and the seventh of 5 feet 2 inches, by 4 feet, and the well-hole for the stairs is 10 feet 6 inches, by 8 feet 9 inches, what will the whole come to? Ans. £53...13...3.

33. It a house measures within the walls 52 feet 8 inches in length, and 30 feet 6 inches in breadth, and the roof be of a true pitch, what will it come to roofing at 10s. 6d. per square? Ans. £12...12...11.

Note. In tyling, roofing, and slating, it is customary to reckon the flat, and half of any building within the wall, to be the measure of the roof of that building, when the said roof is of a true pitch, i. e. when the rafters are of the breadth of the building; but if the roof is more or less than the true pitch, they measure from one side to the other, with a rod or string.

34. What will the tyling of a barn cost, at 25s. 6d. per square; the length being 43 feet 10 inches, and breadth 27 feet 5 inches on the flat, the eve boards projecting 16 inches on each side? Ans. £24...9...5.

MEASURING BY THE ROD.

Note. Bricklayers always value their work at the rate of a brick and a half thick; and if the thickness of the wall is more or less, it must be reduced to that thickness by this.

RULE. Multiply the area of the wall by the number of halt bricks the thickness of the wall is of the product, divided by 3, gives the area.

EXAMPLES.

35. If the area of a wall be 4085 feet, and the thickness two bricks and a half, how many rods doth it contain ?

Ans. 25 rods. 36. If a garden wall be 254 feet round, and 12 feet 7 inches high, and 3 bricks thick, how many rods doth it contain? Ans. 23 rods, 137 feet, 7 in. 37. How many square rods are there in a wall 621 feet long, 14 feet 8 inches high, and 24 bricks thick ?

P

Ans. 5 rods, 166 feet, 6 in.

38. If the side walls of an house be 28 feet 10 inches in length, and the height of the roof from the ground 55 feet 8 inches, and the gable (or triangular part at top) to rise 42 course of bricks, reckoning 4 course to a foot. Now, 20 feet high is 24hricks thick, 20 feet more, at 2 bricks thick, 15 feet 6 inches more, at 14 brick thick, and the gable at 1 brick thick, vhat will the whole work come to at £5...16...0 per rod ? Ans. £48...12...74. Multiplying several figures by several, and the product to be produced on the line only.

RULE. Multiply the units of the multiplicand by the units of the multiplier, setting down the units of the product, and carry the tens; next multiply the tens in the multiplicand by the units of the multiplier, to which add the product of the units of the multiplicand multiplied by the tens in the multiplier and the tens carried; then multiply the hundreds in the multiplicand by the units of the multiplier, adding the product of the tens in the multiplicand multiplied by the tens in the multiplier, and the units of the multiplicand by the hundreds in the multiplier; and so proceed till you have multiplied the multiplicand all through, by every figure in the multiplier

First. 4x416, that is 6 and

Third

carry i. Secondly 3x4+4 X2 and 1 that is carried is 21, set down 1 and carry 2. y, 2×4+3×2+4×4+2 carried-32; that is 2 and carry 3. Fourthly, 5x4+2×2+3×4+4×2+3 carried=47; set down 7 and carry 4. Fifthly, 3x4+5×2+2×4+3×2+4×5+4 earried 60; set down 0 and carry 6. Sixthly, 3x2+5×4 +2x2+3x5+6 carried 51; set down 1 and carry 5. Seventhly, 3×4+5x2+2×5+5 carried=37, that is 7 and carry Eighthly, 3x2+5×5+3_carried = 34; set down and arry 3. Lastly, 3x5+3 carried=18; which being multiplied by the last figure in the multiplier set the whole down, and the work is finished.

3.

THE

TUTOR'S ASSISTANT.

PART V.

THE MENSURATION OF CIRCLES, &c.

A CIRCLE is a plain figure, contained under one line, which is called a circumference, unto which all lines drawn from a point in the middle of the figure, called the centre, and falling upon the circumference, are equal the one to the other. The circle contains more space than any plain figure of equal compass.

The proportion of the diameter of a circle to the circumference was never yet exactly found, notwithstanding many eminent and learned men have laboured very far therein; among whom the excellent Van Ceulen has hitherto outdone all, in his having calculated the said proportion to thirty-six places of decimals, which are engraven upon his tomb-stone in St. Peter's church in Leyden.

Let it be required to find the area of a circle, whose diameter is an unit. By the proportion of Van Ceulen, if the diameter be 1, the circumference will be 3,14159265, &c. of which 3,1416 is sufficient in most cases. Then the rule teaches, to multiply half the circumference by half the diameter, and the product is the area: that is, multiply 1,5708 by,5, (viz. half 3.1416 by half 1) and the product is ,7854, which is the area of the circle, whose diameter is 1.

Again, if the area be required when the circumference is 1, first find what the diameter will be, thus, as 3,1416 to 1 :: I: 318309, which is the diameter when the circumference is 1. Then multiply half ,318309 by half 1, that is ,159154 by,5, and the product is ,079577, which is the area of a circle whose circumference is 1.

If the area be given to find the side of the square equal, you need but extract the square root of the area given, and it is done. So that the square root of ,7854 is ,8862, which is the side of a square equal when the diameter is 1. And if you extract the square root of ,079577 it will be ,2821, which is the side of the square equal to the circle whose circumference is 1.

If the side of a square within a circle be required, if you square the semi-diaineter, and double that square, and out of that sum extract the square root, that shall be the side of the square, which may be inscribed in that circle: so if the diameter of the circle be 1, then the half is ,5, which squared is ,25, and this doubled is,5, whose square root is ,7071, the side of the cuare inscribed.

From what has been here said, the ingenious scholar will easily perceive how all other proportional numbers are found, and may examine them at pleasure. We shall now proceed to the different problems.

Problem 1. Having the diameter and circumference, to find the area.

Every circle is equal to a parallelogram, whose length is equal to half the circumference, and half the breadth equal to half the diameter; therefore multiply half the circumference by half the diameter, and the product is the area of the circle.

Thus, if the diameter of a circle, that is, the line drawn across the circle through the centre, be 22,6; and if the circumference be 71, the half of 71 is 35,5, and the half of 22,6 is 11,3, which multiplied together, the product is 401,15, which is the area of the circle.

Problem 2. Having the diameter of a circle to find the circumference.

As 7 to 22, so is the diameter to the circumference. Or, as 113 to 355, so is the diameter to the circumference. Or, as 1 to 3,141593, so is the diameter to the circumference.

Let the diameter, as in the first problem, be 22,6. This mul tiplied by 22, and the product divided by 7, gives, 71.028 for the circumference; but the other two proportions are more exact, as appear by the following work.

Problem 3. Having the circumference of a circle, to find the diameter.

As 1 is to .318309, so is the circumference to the diameter. Or, as 355 to 113, so is the circumference to the diameter. Or, as 22 to 7, so is the circumference to the diameter.

Let the circumference be 71, and then proceed with either of the three proportions, as follow:

Thus, by the second proportion, the diameter is 22.6; but by the other two it falls something short.

Problem 4. Having the diameter of a circle, to find the area. All circles are in proportion one to another, as are the squares of their diameters, (by Euclid, lib. 12, prop. 2.) Now the area of a circle, whose diameter is 1, will be ,785398, according to Van Ceulen's proportion beforementioned; but for practice,7854 will be sufficient. Therefore, as I (the square of the diameter 1) is to 7854, so is 510,76 (the square of 22.6, the diameter of the given circle) to 41,15 the area of the given circle.

Problem 5. Having the circumference of a circle to find the

area

Because the diameters of circles are proportional to their circumferences; that is, as the diameter of one circle is to its circumference, so is the diameter of another circle to its circum ference therefore the areas of the circles are to one another, as the squares of the circumferences. And if the circumference of a circle be 1, the area of that circle will be',07958; then the square of I is 1, and the square of 71, (the circumference of the former circle) is 5011. Therefore it will be, as 1:07958;;5041 : 401,16278.

Problem 6. Having the diameter, to find the side of a square equal in erea to that circle.

If the diameter of a cirele be 1, the side of a square equal thereto will be ,8862. Therefore as 1: 8862::22.6 (the diameter) : 20,02812, the side of the square.

Problem 7. By having the circumference, to find the side of the square equal thereto.

If the circumference of a circle be 1, the side of the square equal will be ,2821. Therefore as 1: 2821::71 (the circumference) 20,0291, the side of the square

Problem 8. Having the diameter, to find the side of a square, which may be inscribed in that circle.

If the diameter of a circle be 1, the side of the square inscribed will be 7071. Therefore, as 1:7071 :: 22.6: 15.98046, the side inscribed. Or, if you square the semi-diameter, and double that square, the square root of the double square will be the side of the square inscribed.