# An introduction to linear drawing/Chapter 11

​﻿When you have obtained two or three decimal fig- ures in the quotient, it is useless to carry the calcula- tion any further, as they will be too small.
We shall now endeavour to apply these principles.
SECTION I.
OF LINES.
Problem I. To find a side of a rectangular trian- gle, the two others being known.
Rule. Multiply by itself each of the known sides, then add them together if you wish to find the greater side ; and subtract the lesser number from the greater if you wish to find one of the lesser sides. Then you will have the same result as if you had multiplied the unknown side by itself. Of course, you have only to find what number, multiplied by itself, will give this result.
Example 1. The smaller sides of a rectangular tri- angle, (figs. 12 and 13, Classl.) are one 3, and the oth- er 4 inches, find the larger side.
3 times 3 are 9
4 times 4 are 16
These added make 25
5 multiplied by itself makes 25, and the greater side or side required, must be 5 inches.
Example 2. \n a rectangle, (1st Class, fig. 14,) it is known that the base is 8 inches, 54 hundredths, the ​﻿height is unknown, but the diagonal (a right line drawn from corner to corner) is found by measurement to be 15 inches, 32 hundredths. What is the height ?
Note. A diagonal cuts the oblong square or rect- angle into two rectangular triangles, of which the height above required is one of the smaller sides.
8,54 15,32 As a smaller side is re-
8,54 15,32 quired, subtract the known
3416 3064
4270 4596 234,7024
6832 7660 72,9316
72,9316 ----161,7708
234,7024
It remains to find a number, which multiplied by it- self will give 161,770. A few trials will show that this is (as near as possible) 12,719 as may be found by multiplying this number by itself. The height then, is 12 inches, and 719 thousandths of an inch.
Example 3. Find the height of an isoceles triangle, (1st Class, Prob. 27.) whose base is 52 and the equal sides, 87. The perpendicular, drawn from the summit, cuts the base in halves, and is one side of a rectan- gular triangle, of which
the base i§ half the larger one, or . . 26 The great side of the new angle, which 26
was one of the equal sides of the -
isoceles, is........ 87 156
87 52
From 7569 - -
Take 676 609 676
--696
6893 • -
7569 ​﻿A few trials will show, that the height required is, 83 nearly, for 83 times 83 are 6889.
4. The smaller sides of a rectangular scalene trian- gle, (1st Class, fig. 13,) are 5 and 7 inches, required the larger side. Ans. 8,6 in. nearly.
5. One small side of a rectangular scalene triangle is, 8in-,3 and the other 4in-,8 what is the size of the third side ? Ans. 9,59 in. nearly.
6. The two equal sides of a rectangular isoceles tri- angle, are 4,8 in length, what is the length of the base ? Ans. 6,785 in. nearly.
7. The base of a rectangle (Ex.2) is 7,15; the diagonal 13,25 ; required the height (that is, the third side of the triangle.) Ans. 14,14 in. nearly.
8. The base of a rectangle is 4,75 and the height 7,25 required the diagonal, or longest side of the trian- gle. Ans. 8,67 in. nearly.
9. What is the height of an isoceles triangle, of which the base is 6 inches, and the equal sides 3in-,8 each i Ans. 2,33 in. nearly.
The instructor may increase these examples at pleas- ure.
Problem II. To find the circumference of a circle . when lengthened out into a right line.
Rule. Multiply its diameter hy 3, and add a sev- enth of a diameter.
Example 1. The diameter of a circle is 4lu-,523.
4,523 3
13,569
A seventh of 4,523 is ,646
The circumference is 14,215 ​﻿2. The width of a basin is how long must a string be to reach round it ?
5,5 3
16,5
A seventh of 5,5 . . 0,786
Length of thestring, . . 17,286 the circumference.
3. The diameter of a circle is 4,45, what is the cir- cumference? Ans. 13,98 inches.
4. The diameter of a ring is 1,5, what is the circum- ference ? Ans. 4,7 inches.
5. The diameter is 4,17, what is the circumference ? Ans. 13,1 in.
Problem III. The circumference being known, to find the radius.
Rule. Multiply the circumference by 0,159 and you will have the radius.
Example 1. To find the thickness of a column, its circumference has been measured with a string and found to be 12,542—required the radius.
12,542 0,159
0 112878 62710 12542
1,994178 radius.
If 1,994 thousandths be the radius or half diameter, 3,988 will be the whole diameter of the column. Six figures are separated, because there are three in the multiplier, and 3 in the multiplicand. The three right hand decimals are unimportant.