EXAMPLES. XVIII.

Apply Simpson's Rule to find in square feet the areas of figures having the following dimensions:

1. Ordinates 3, 8, 15, 24, 35, 48, 63 feet; common distance 1 foot.

2. Ordinates 4, 14, 36, 76, 140 feet; common distance 1 foot.

3. Ordinates 0, 20, 32, 36, 32, 20, 0 feet; common distance 2 feet.

4. Ordinates 0, 1.25, 4, 675, 8, 625, 0 feet; common distance 1 foot.

5. Ordinates 6'082, 6·164, 6'245, 6325, 6'403, 6'481, 6.557 feet; common distance 1 foot 6 inches.

6. Ordinates 2.714, 2.759, 2802, 2.844, 2.884 feet; common distance 9 inches.

7. Ordinates 14°2, 14′9, 15′3, 15°1, 14′5, 14·1, 13.7 feet; common distance 3 feet.

8. Ordinates 0, 111, 248, 417, 6·24, 875, 11.76, 15:33, 1952 feet; common distance 1 foot.

9. Ordinates 10-204, 9804, 9'434, 9'090, 8771, 8475, 8197, 7937, 7-692 feet; common distance 1 foot.

10. Ordinates 2.4849, 25649, 26391, 27081, 2.7726, 2.8332, 28904, 2.9444, 2.9957 feet; common distance 1 foot.

11. Ordinates 0, 4359, 6, 7141, 8, 8660, 9165, 9539, 9798, 9950, 1 foot; common distance 1 of a foot.

12. Ordinates

10 10 10 10 10 10 10 10 10' 11' 12' 13' 14' 15' 16' 17'

feet; common distance 1 of a foot.

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XIX. SIMILAR FIGURES.

204. In Chapter VI. we have drawn attention to the nature of Similar Figures; and we have now to point out the relation which holds between the areas of Similar Figures. We shall state a most important proposition, and then proceed to apply it to various problems.

205. The areas of similar figures are as the squares of corresponding lengths.

For example, suppose we have two similar triangles, and that the side of one triangle is three times the corresponding side of the other; then the area of the larger triangle is nine times the area of the smaller, the number 9 being the square of the number 3. And it is easy to see the reason for this fact: the larger triangle has its base three times the base of the smaller, and, because the triangles are similar, the height of the larger triangle is also three times the height of the smaller; but the area is half the product of the base into the height; and therefore the area of the larger triangle is 9 times the area of the smaller.

In like manner, if two triangles are similar, and the side of one triangle is five times the corresponding side of the other, the area of the larger triangle is twenty-five times the area of the smaller.

206. We have found in Art. 154, that the area of an equilateral triangle, of which the side is 1 foot, is 4330127 square feet: suppose we require the area of an equilateral triangle, of which the side is 7 feet.

The square of 1 is 1, and the square of 7 is 49; therefore we have the proportion

Thus the required area in square feet

=49 x 4330127=21°2176223.

In the same way we may proceed with other examples, and it is obvious that we shall have the following RULE for finding the area of any equilateral triangle: Multiply the square of the length of a side by 4330127.

207. Circles are similar figures; and the areas of two circles are in the same proportion as the squares of their radii, see Art. 178. So also sectors of circles having the same angle are similar figures; and the corresponding segments are similar figures; the areas of two similar sectors are in the same proportion as the squares of the radii, and so also are the areas of two similar segments.

208. Suppose we require the radius of a circle, such that the area of the segment corresponding to an angle of 60° shall be 20 square inches.

In Art. 187 we have found that if the radius is 10 inches, the area of the segment corresponding to this angle is 9'06 square inches; thus we have the proportion

9'06 20:100: the square of the required radius. Therefore the square of the required radius

Thus the required radius is 14.857 inches.

Again; suppose we require the radius of a circle such that the area of the segment corresponding to an angle of 50° shall be 10 square feet.

By Art. 187 we have the proportion

4:5664 10: 16 the square of the required radius. Therefore the square of the required radius

the square root of this number = 5'92.

Thus the required radius is 5·92 feet.

of ADE will be one-fourth of the area of ABC.

Thus we have the proportion

49A

1: the square of AB: the square of AD.

Therefore the square of AD=of the square of AB = 100=25; the square root of this number=5. Thus AD=5 feet.

In like manner, if FG be the next straight line

1 the square of AB: the square of AF.

Therefore the square of AF=2 of the square of AB =2×100=50; thus the number of feet in AF- the square root of 50=7·0710678.

In like manner, if HK be the next straight line, we find that the number of feet in AH = the square root of 75 8'6602540.

210. ABCD is a trapezoid; the perpendicular distance of the parallel sides AB and DC is 3 feet; AB=10 feet, DC-6 feet: it is required to divide the trapezoid into two equal parts by a straight line parallel to AB. Produce AD and BC to meet at O. Let EF denote the required straight

line. Draw OM perpendicular to AB, meeting EF at L,

and DC at K.

Since EF divides the trapezoid into two equal parts, the triangle OEF will be equal to half the sum of the triangles OAB and ODC. The three triangles ODC, OEF, and OAB are similar; and their areas are therefore as the squares of the corresponding lengths OK, OL, and OM.

Hence the square of OL must be equal to half the sum of the squares of OK and OM.

Now by Art. 77, we have OK=4•5; therefore OM=7.5. The square of 4.5=2025; the square of 7.5=56°25. Thus the square of OL of 765-38 25; and therefore the

number of feet in OL

=

the square root of 38°25= 6·1846.

The number of feet in KL=6·1846−4·5=1'6846. Thus the position of EF is determined.

211. We will now solve some exercises:

(1) A plan of an estate is drawn on the scale of 1 inch to 20 feet: find what space on the plan will correspond to 8000 square yards of the estate.

The scale is that of 1 inch to 240 inches. The required space will be obtained by dividing 8000 square yards by the square of 240.

(2) If a square inch on a plan corresponds to 4 squaro yards of the original, find the scale.

4 square yards = 4 × 9 × 144 square inches. The square root of 4 x 9 x 144-72. Thus the scale is that of 1 inch to 72 inches.

(3) The sides of a rectangle are in the proportion of 4 to 5, and the area is 180 square feet; find the sides.