An introduction to linear drawing/Chapter 12
2. The circumference of a column is 10,5 what is its radius ? what its diameter ? Ans. R. 1,67 D.3,34.
3. The circumference of a ship's mast is 136**', 15 what is its diameter ? Ans. 43,28 in.
4. The circumference of a wheel is 48,75, what is its radius and diameter ? Ans. R. 7,75 D. 15,5.
SECTION II.
OF SURFACES.
Problem I. To find the surface of a parallelogram#
Rule. The surface of a parallelogram (fig. 16,) or rectangle (fig. 14,) is found by multiplying the base by the height. That of a square is found by multiplying one of the sides by itself.
Example 1. A rectangle has 2in-,24 for its base, 4ln-,31 for its height, what is its surface ?
2,24
4,31
224 67 2 * 896
9,6544 surface.
That is, 9 square inches, and 65 hundredths of a square inch.
2. A room is 154,6 inches long, and 75,3 wide, what is its surface or area ?
8 78
154,6
75,3
4638 7730
10822
11641,38
That is, 11641 square inches, and 38 hundredths of a square inch.
3. A yard of a rectangular form, is 2023 inches long, and 1145 wide, what is its aréa or surface ì
4. A house is 388 inches long, and 146 wide, how many square inches of ground does it cover?
Problem II. The èutface of an upright prism with- out including the two bases, is found by multiplying the iieight by the circumference of the base.
All the làtéràl faces or sides of the body are reCtàn- giés, and còme under the preceding rule.
Example 1. A man wishes to plaster the walls of the room mentioned in No. '1 of the last problem. These walls are 64,6 inches high, how many superficial inches do they contain? The room forms a parallelo- piped, 2023 inches long, 1145 wide, and 84,6 high. Double the width and length, and add them together, and you have the whole length of the walls, 6336 inches. Multiply this by the height, and you have the answer in square inches.
1145 1145 20£3 2023
6336 84,6
Ä>
3S016 25344 bÓÒ88
536025,6 Ans. 2. A man wishes to cover the walls of a room with cloth. The length of all the sides added together is 675 inches, 7 tenths, the height is 98,4 the cloth is 32 inches wide ; how long a piece will cover the walls ?
Multiply the length by the height, to find the surface or superficies to be covered. The cloth then must have a length which multiplied by its breadth will give the same superficies, and this is found by dividing the su- perficial contents of the walls by the width of the cloth. Paper hangings may be measured in the sanje way.
Ans. 2077 inches or 57 yards.
If the prism be oblique, its surface is found by taking that of ail the parallelograms which form it.
Problem III. To find the surface of a triangle.
Rule. Multiply the base by the height, and take half of the result.
If you please, you may take half the base or half the height, before you multiply, and then there will be no need of halving the result. A triangle is always the half of a parallelogram of the same base and height.
The pupil, it is to be hoped, need not be told that 12 inches make a foot, and 3 feet or 36 inches an English yard. We advert to this, because we have hitherto only measured by inches, and it may be well to say that when feet or yards, inches and decimals are named together, the yards or feet must be brought into inches. To do this, multiply the feet by 12, and the yards by 36. This however is not necessary, when only feet ox yards are named, and the decimals are parts of them, and not parts of inches.
Thus, 8 feet, 4,8 inches are the same as 100,8 inches.
6 yds. 4,5 inches are equal to . . 220,5 inches.
Example 1. Required the extent of a field of a trian- gular shape, of which one side taken for the base, is 154 yards long, and the height (a perpendicular drawn from this base to the point or summit of the opposite angle) 83 yards. Multiply 77 (that is, half the base) by 83, and I have-the answer, 6391 square yards of surface.
The superficies, or surface of a polygon, or a pyra- mid, is found by taking separately the surfaces of the triangles of which they are composed.
2. An irregular court has a quadrilateral (four sided) form. To find its surface, I measure one of the diago- nals which I find to be 129,7 yards. I draw perpen- diculars from the angles opposite this diagonal, one of which I find to be 52,5 yds. and the other, 41,8 yds- i consider the court as forming two triangles, and find their superficies separately, thus :
First triangle . . 129,7 Second triangle . . 129,7 Height.....52,5 Height ...... 41,8
6485 10376
2594 1297
6485 5188
' 6809,25> 5421,46
6809,25
2) 12230,71 Square yards, .... 6115,355
But as this requires two multiplications, it is a shorter way to add the two heights, 52,5 and 41,8 which gives 94,3 of which the half, 47,15 multiplied by 129,7 the base, gives 6115,355 square yards as above.
3. A four sided polygon has a diagonal of 66 feet, 3,8 inches ; the height of one triangle is 22 feet, 6,6 inches ; and of the other, 18 feet, 8,2 inches ; requir- ed the superficial contents of the polygon.
Ans. 195880,9 inches, or 1360 square feet. If the polygon be regular, draw lines from the centre to two of the neighbouring angles, find the contents of the triangle thus formed, and multiply by the number of sides, which being all of a size, wifl make triangles of the same size.
4. A hexagonal basin has equal sides of 3,34 inches each. Its width from the centre of one side to the centre of the opposite side, is 4,88 inches. As half of this line drawn from side to side is the height of one of the triangles, the height is, 2,44 inches. Multiply the base, which in this case is the side, by half the height, and you have the answer. Ans. 24,42 square inches.
Problem IV. To find the surface of a trapezoid. (fig. 3.)
Rule. Take half the sum of the two parallel sides, and multiply by the height.
Example 1. A roof in the form of a trapezoid, has one of its parallel sides 44,7 feet, and the other 38,5 feet in length, and the height is 9,4 feet, what is the superficies ? Ans. 367,54 feet.
2. How many slates 15 inches long, and 12 wide, will cover the above roof ? Ans. 294.
Note. No allowance is here made for one slate's projecting over another, &c. This would only increase the size of the roof, but not alter the mode of calcula- tion. Change the feet of the roof into inches ; find the square inches in each slate, and divide the number of inches in the roof by the number in a slate.
Problem V. To find the surface, or superficial eon- tents of a circle.
Rule. Multiply the radius by itself, and then the product by 3^ (or 3,143.)
8* Example I. A circular basin has a radius of 8,3 inches.
8,3 68,89 8,3 249 20667 664 984
68,89 216,51 square in.
2. The radius of a cistern is 3,45 feet, what is its surface ? Ans. 37,40 sq. ft.
3. I have measured round a basin, and find the dis- tance 28,5 inches, and conclude by Problem III, Sect. 1, that the radius is 4,53. The rest of the work is done like the above example. Ans. 64,49 sq. in.
4. The circumference of a circle is found to be 4,85 feet, what is its superficial contents ? Ans, 1,76 sq. ft.
Problem VI. To find the surface of an upright cylinder. (4th Class, fig. 12.)
Rule. Multiply the circumference of the base by the height. As the base is a circle, knowing the radius, it is easy to find the circumference. (Prob. 2, Sect. I.)
Example 1. A painter has painted a circular hall, the walls are 3,4 yards high, and the diameter of the hall is 54,2 yards, how many square yards has he painted ? Ans. 579,156 sq. yds.
If there be doors and windows, they are calculated separately, and subtracted from the amount. To find the contents of mouldings, measure them with a piece of string or parchment, which will yield to their various curvatures, and if you please, add their width and sur- face to the first amount.