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Set 1 upon B to 12 upon A, and against 9 on A is .75, the answer, on B.
What is the decimal of ?
Set 1 on B to 28 on A, and against 18 on A stands .643 on B, the decimal required.
To find a multiplier to a divisor that shall perform the same by multiplication as the divisor would do by division; the proportion is, as the divisor upon A is to unity or 1 upon B, so is unity on A to the multiplier required on B.
Suppose 25 to be the divisor, what will be the multiplier to that number?
Set 1 upon B to 25 upon A, and against 1 on A is .04, the multiplier on B.
What will be the multiplier to 80?
Set 1 on B to 80 on A, and against 1 on A is .0125 on B.
What will be the multiplier to 40?
Set 1 on B to 40 on A, and against 1 upon A is .025, the answer, upon B.
Having a multiplier given to find a divisor.-The proportion is, as the multiplier upon A is to 1 upon B, so is the divisor upon B to 1 upon A.
Let .04 be the multiplier given to find a divisor.
Set 1 upon B to .04 upon A, and against 1 upon A is 25 upon B.
What will be the divisor for .0125?
B to .0125 upon A, and against 1 on A is 80
What will be the divisor for .7854?
Set 1 upon B to .7854 upon A, and against 1 on A is 1.273, the answer, on B.
5. EXTRACTION OF THE SQUARE ROOT.
The square root of any number is that number which, multiplied by itself, or squared, will produce the given number. Thus
The square root of 1 4 9
16 25 36 49
To find the square root of any number :
Divide the given number into periods of two figures each, by putting a dot over the unit figure, and every second figure from the place of units, and the number of dots will be equal to the number of figures in the required root.
Then seek the greatest square number in the left-hand period, and place it in the quotient, for the first figure of the root, and place its square under the left-hand period, and subtract it from that period, and to the remainder annex the figures in the following period.
Double the root, and place it on the left hand for a divisor, and seek how often the divisor is contained in the dividend, (always excepting the right-hand figure of the dividend,) and place the figure in the quotient, and also annex it to the divisor; then multiply the divisor with the annexed figure of the quotient by that figure, and place the product under the dividend, which subtract from the dividend, and to the remainder annex the following period, and proceed as before, and so on till all the periods are exhausted.
1. What is the square root of 119025?
2. What is the square root of 436.5?
3. What is the square root of 80 ?
4. What is the square root of 22071204?
5. What is the square root of 2.7109?
fraction, extract the
To extract the square root of a vulgar root of the numerator and denominator; or reduce it to a deci
mal, and then extract its root.
6. What is the square root of %? 7. What is the square root of f?
To find a mean proportional between two numbers :Extract the square root of the product of the given numbers. 1. What is the geometrical mean proportional between 3 and 12? Ans. 6.
2. What is the mean proportional, or, strictly, the geometrical mean proportional between 20 and 30? Ans. 24.49.
T 6. RIGHT-ANGLED TRIANGLE.
In any right-angled triangle the square of the hypotenuse is equal to the sum of the squares of the base and perpendicular. To demonstrate this useful proposition,-Let ABC be a rightangled triangle. On the base, BC, draw the square BesC; and h
on the perpendicular, AC, draw the square ACnh; and on the hypotenuse, AB, draw the square AgfB; then from A, draw Ad parallel to CB, and equal to the same, and join dB; then from n, draw nm parallel to Cs, and equal to Cs, and join ms.
Now since the four right-angled triangles, ABd, Bef, fmg, and ghA are equal, their sides and angles being equal; and since the rectangle AdBC is equal to two of these triangles;
and the rectangle Csmn is equal to the rectangle ABC, its sides being of the same length, it is manifest that the sum of these two rectangles is equal to the sum of the four triangles, AdB, Bef, fgm, and Agh. Now if we subtract the sum of the two rectangles, AdBC and Csmn, from the square hdem, there will remain the squares drawn on the base and perpendicular of the triangle ABC, viz., the squares ACnh and CBes; and if we subtract from the same square the four triangles, ABd, Agh, gfm, and Bfe, there will remain the square ABfg on the hypotenuse. But these four triangles are equal to the sum of the two rectangles, AdBC and Csmn: since, therefore, we have subtracted equal quantities, in both cases, from the square hdem, it follows that the remainders will be equal; and consequently, the square on the hypotenuse is equal to the sum of the squares on the base and perpendicular.
The base and perpendicular of a right-angled triangle being given, to find the hypotenuse:
Extract the square root of the sum of the squares of the base and perpendicular.
1. The wall of a town is 25 feet in height, and being surrounded by a moat 30 feet in breadth, I desire to know the length of a ladder which will reach from the outside of the moat to the top of the wall. Ans. 39.05+ feet.
2. One side of a rectangle being 45 rods, and the other side 60 rods, I require the length of the diagonal, or the distance between the opposite corners. Ans. 75 rods.
3. If a room is 9 feet in height, and 25 feet long, and 15 broad, what will be the length of a diagonal between the opposite corners ? Ans. 30.512+ feet.
To find the diagonal in the above example, add together the squares of the length, breadth, and height of the room, and extract the square root of the sum.