123. The error which arises from the use of the Rule in Art. 121 is, as we have said, much less for a small arc than for a large arc. It may therefore be expedient in some cases to calculate by the Rule the length of half the arc, and to double this result instead of calculating the length of the whole arc immediately. We should proceed thus: from eight times the chord of one fourth of the arc subtract the chord of half the arc; multiply the remainder by two, and divide the product by three.

124. The following Rule for finding the length of an arc is much more accurate than that in Art. 121, and may be used when a very close approximation is required to 256 times the chord of one fourth of the arc add the chord of the arc; subtract 40 times the chord of half the arc, and divide the remainder by 45.

This Rule gives the length of the arc a little larger than it ought to be. If the arc subtend at the centre of the circle an angle of 45 degrees the error is less than

1 of the arc: the error increases rapidly as the

80000000

angle increases, and diminishes rapidly as the angle diminishes.

125. We will now solve some exercises.

(1) The radius of a circle is 1 foot: find the whole perimeter of a sector of 60 degrees.

Since the radius is 1 the circumference of the circle is 2 × 31416, that is, 6·2832; then

360 60 6'2832: the length of the arc.

Thus the length of the arc is 1.0472. Add the length of the two radii, that is 2; thus the whole perimeter is 3·0472 feet.

(2) The whole perimeter of a sector of 60 degrees is 20 feet: find the radius.

Use the result of the preceding exercise. Thus we have the proportion

3-0472 20:1 the required radius.

EXAMPLES. IX.

1. The radius of a circle is 10 inches, and the angle subtended by an arc at the centre is 72o: find the length of the arc.

2. The radius of a circle is 19 feet 7 inches, and the angle subtended by an arc at the centre is 10° 24': find the length of the arc.

3. The radius of a circle is 2 feet, and the length of an arc is 15 inches: find the angle subtended at the centre by the arc.

4. The radius of a circle is 1 foot, and the length of an arc is equal to the radius: find the angle subtended at the centre by the arc.

5. The chord of an arc is 36 inches, and the chord of half the arc is 19 inches: find the arc.

6. The chord of an arc is 56 inches, and the radius of the circle is 197 inches: find the arc.

7. The chord of an arc is 6 inches, and the radius of the circle is 9 inches: find the arc.

8. The radius of a circle is 5 inches: find the whole perimeter of a sector, the angle of which is 90°.

9. The radius of a circle is 16 inches: find the whole perimeter of a segment, the arc of which subtends an angle of 90° at the centre of the circle.

10. The radius of a circle is 1 foot: find the whole perimeter of a semicircle.

11. The whole perimeter of a semicircle is 100 feet: find the radius.

12. The radius of a circle is 25 inches, and the angle subtended by an arc at the centre is 32° 31′ 12′′4: find the length of the arc.

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THIRD SECTION. AREAS.

X. TABLE OF SQUARE MEASURE.

126. It will be convenient to place here the Table of Measures of Area, which is usually called the Table of Square Measure.

144 square inches make 1 square foot.
9 square feet make 1 square yard.
36 square feet make 1 square fathom.

272 square feet or 30 square yards make 1 square
rod or pole.

1600 square poles make 1 square furlong. 64 square furlongs make 1 square mile.

Hence we obtain the following results:

127. The following terms are also used in expressing areas: a square link, a square chain, a rood, and an acre.

A square chain contains 22 × 22, that is, 434 square yards. A rood is 40 poles, that is, 1210 square yards. An acre is 4 roods, that is, 4840 square yards: thus an acre is equal to 10 square chains.

A square chain contains 100 x 100, that is, 10000 square links; so that an acre is equal to 100000 square links.

XI. RECTANGLE.

128. Suppose we have a rectangle which is 4 inches long and 3 inches broad. Draw

straight lines, an inch apart, parallel to the sides. The rectangle is thus divided into 12 equal figures, each of which is a square being an inch long and an inch broad: such a square is called a square inch. The rectangle then contains 12 square inches; this fact is also expressed thus: the area of the rectangle is 12 square inches.

The number 12 is the product of the numbers 4 and 3, which denote respectively the length and the breadth of the rectangle.

129. If a rectangle be 8 inches long and 5 inches broad, we can shew in the same manner that its area is 8 times 5 square inches, that is 40 square inches. Similarly, if a rectangle be 9 inches long and 7 inches broad, its area is 9 times 7 square inches, that is, 63 square inches. And

so on.

130. In the same manner, if a rectangle be 4 feet long and 3 feet broad, its area is 12 square feet; that is, the rectangle might be divided into 12 equal figures, each being a foot long and a foot broad. If a rectangle be 4 yards long and 3 yards broad, its area is 12 square yards. And so on.

131. The beginner should observe very carefully the way in which areas are measured; it is a case of the general principle which applies to all measurable things. For example, when we measure lengths we fix on some length for a standard, as an inch or a foot, and we compare other lengths with the standard; thus when we say that a

certain line is 17 inches long, we mean that the line is 17 times as long as our standard, which is one inch. In like manner when we measure areas we fix on some area for a standard, and we compare other areas with the standard. The most convenient standard is found to be the area of a square; it may be a square inch, or a square foot, or any other square.

132. In order then to find the area of a rectangle we must express the length and the breadth in terms of the same denomination; and then the product of the numbers which denote the length and the breadth will denote the area. If the length and the breadth are both expressed in inches, the area will be expressed in square inches; if the length and the breadth are both expressed in feet, the area will be expressed in square feet; and so on.

133. The student will now be able to understand the way in which we estimate the areas of figures, and to use correctly the rules which will be given; the rules will be stated with brevity, but this will present no difficulty to those who have read the foregoing explanations.

134. To find the area of a rectangle.

RULE. Multiply the length by the breadth, and the product will be the area.

Sometimes the words base and height are used respectively for the length and breadth of a rectangle.

135. Examples.

(1) The length of a rectangle is 3 feet 4 inches, and its breadth is 2 feet 6 inches.

3 feet 4 inches=40 inches, 2 feet 6 inches = 30 inches ; 40 × 30=1200.

Thus the area is 1200 square inches.

Or thus: 3 feet 4 inches=33 feet, 2 feet 6 inches = 21 ft. ; 10 5 25

3x2= X =

3 2

3

Thus the area is 8 square feet.