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29. The inner diameter of a circular building is 68 feet 10 inches, and the thickness of the wall is 22 inches: find how many square feet of ground the base of the wall occupies.

30. In a circular riding-school of 100 feet in diameter a circular ride, within the outer edge, is to be made of a uniform width of 10 feet: find the cost of doing this at 4d. per square foot.

31. A circular grass-plot whose diameter is 40 yards contains a gravel walk, one yard wide, running round it one yard from the edge: find what it will cost to turf the grass-plot at 4d. per square yard.

32. A road runs round a circular shrubbery; the outer circumference is 500 feet and the inner 420 feet: find the area of the road.

33. Find the side of a square which is equivalent in area to a circle of 80 feet radius.

34. Find the radius of a circle which is equivalent in area to a square the side of which is 80 feet.

35. The side of a square is 16 feet; a circle is inscribed in the square so as to touch all its sides: find the area between the circle and the square.

36. The side of a square is 18 feet; a circle is described round the square: find the area between the circle and the square.

37.

The sides of a right-angled triangle are 27 and 43 feet respectively: find the area of the circle described on the hypotenuse as diameter.

38. The area of a semicircle is 645 square feet: find the length of the whole perimeter of the semicircle.

39. The radius of a circle is 1 foot; an equilateral triangle is inscribed in the circle: find the area between the circle and the triangle. (See Art. 99.)

40. The sides of a right-angled triangle are 370 feet and 168 feet respectively: find the area of the circle which has the hypotenuse of this triangle for diameter.

41. A rectangle is 8 feet long and 7 feet broad: find the area of the circle which has the same perimeter.

42. The sides of a triangle are 13, 14, and 15 feet: find the area of the circle which has the same perimeter.

If a circle has the same perimeter as a rectangle the circle has the greater area; verify this statement in the following cases:

43. Rectangle 18 feet by 10.

44. Rectangle 27 feet by 13.

If a circle has the same perimeter as a triangle the circle has the greater area; verify this statement in the following cases:

45. Sides of a triangle 9, 10, 17 feet.

46. Sides of a triangle 11, 16, 19 feet.

If a circle has the same area as a rectangle the circle has the less perimeter; verify this statement in the following cases:

47. Rectangle 15 feet by 12.

48. Rectangle 24 feet by 21.

If a circle has the same area as a triangle the circle has the less perimeter; verify this statement in the following cases:

49. Sides of a triangle 5, 6, 7 feet.

50. Sides of a triangle 12, 15, 17 feet.

51. A circle is 4 feet in circumference: find the arca of a square inscribed in it.

52. A circle is 7 feet in circumference: find the area of a square inscribed in it.

XVII. SECTOR OF A CIRCLE AND SEGMENT OF A CIRCLE.

179. To find the area of a sector of a circle.

RULE. As 360 is to the number of degrees in the angle of the sector so is the area of the circle to the area of the sector.

180. Examples:

(1) The radius of a circle is 25 feet, and the angle of the sector is 80 degrees.

The area of the circle = 25 x 25 × 3·1416=1963*5.

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Thus the area of the sector is about 436'3 square feet.

(2) The radius of a circle is 12 feet, and the angle of

the sector is 75 degrees.

The area of the circle = 12 x 12 × 3·1416,

360 75: 12 × 12×31416: the required area,

75 x 12 x 12 × 3.1416 75 x 12 x 3.1416

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= 30 x 31416-94.248.

Thus the area of the sector is 94.248 square feet.

181. The following is another rule for finding the area of a sector of a circle: multiply the arc by the radius and take half the product.

The truth of this Rule will be obvious from the remarks made in Art. 177.

182. Examples:

(1) The radius of a circle is 4 feet, and the arc of the sector is equal to the radius.

1

× 4×4=8. Thus the area of the sector is 8 square feet.

(2) The radius of a circle is 2 feet 6 inches, and the arc of the sector is 1 foot 5 inches.

1

× 30 × 17 = 255. Thus the area of the sector is 255 square inches.

183. Suppose we require the area of a figure which is the difference of two sectors having a common angle. Let OAB be one sector, and OCD the other; so that ABDC is the figure of which the area is required. We may proceed thus:

We may calculate separately the area of each sector and subtract the less from the greater.

A

D

B

Or, we may calculate the area of the entire ring between the two circles, of which AB and CD are arcs; and then use the proportion, as 360 is to the number of degrees in the angle at O so is the area of the ring to the required area.

Or we may use this Rule: multiply the sum of the arcs by the difference of the radii and take half the product.

184. Examples:

(1) The radii are 15 feet and 10 feet, and the arcs 6 feet and 4 feet respectively.

=

1

By Art. 181, the area of the larger sector in square feet

× 15×6=45, and the area of the smaller sector in

1

square feet =

square feet

=

× 10 × 4 20: thus the required area in 45—20= 25.

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Or, using the third Rule of the preceding Article, we have the sum of the arcs 10 feet, and the difference of the radii =5 feet; thus the required area in square feet

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(2) The radii are 7 feet and 5 feet respectively, and the angle at O is 45 degrees.

Here we use the second Rule of the preceding Article. By Art. 173 the area of the entire ring in square feet is 12 × 2 × 31416, that is, 75-3984. Then

360 45: 753984: the required area.

Thus the required area in square feet

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185. Any chord AB of a circle, which is not a diameter, divides the circle into two segments, one greater than a semicircle, and the other less.

When we know the area of the lesser segment, we can, by subtracting this from the area of the circle, determine the area of the greater segment; so that it is sufficient to

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give a Rule for finding the area of a segment less than a semicircle.

Let O be the centre of the circle; then it is obvious that the segment ABC is equal to the difference of the sector OACB and the triangle OAB. Hence we have the Rule which we shall now give.

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