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406. PROBLEM. To inscribe a square in a circle.

Draw two diameters AE, CG (Fig. 40) perpendicular to each other, and join the extremities of these diameters. We have the square ACEG (§ 400), for its sides are equal, since they are chords subtending equal arcs, and its diagonals are equal, since they are diameters of the same circle.

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407. By bisecting the four arcs, and joining the points of division, we shall have a regular octagon inscribed.

All the sides of the regular octagon are equally distant from the centre O (§ 367), and therefore the perpendiculars OI and OK let fall from the centre upon the sides have the same length, and each of them is called the apothem of the polygon.

By bisecting these eight arcs, and joining the points of bisection, we shall have a regular polygon of 16 sides, and by continuing this process we may inscribe regular polygons of 32 sides, 64 sides, and so on.

408. PROBLEM. To inscribe in a given circle a regular hexagon.

The side of a regular hexagon is equal to the radius. A regular hexagon is obtained then by applying six times

F

A

upon the circumference an opening of the compasses equal to the radius. By joining the alternate vertices B, D, F of a regular hexagon, we have an equilateral triangle inscribed. By bisecting the arcs AB, BC,a regular polygon of 12 sides may E be inscribed, and by continuing the process regular polygons of 24 sides, 48 sides, 96 sides, may be inscribed.

D

LENGTH OF THE CIRCUMFERENCE.

FIG. 41.

B

409. In inscribing in the same circle polygons of a greater and greater number of sides, for instance, polygons of 12, 24, 48, 96, etc., sides, we see that the lengths of their perimeters differ less and less from the length of the circumference.

If then we wish to measure the length of a circumference of a circle, we inscribe in it a regular polygon of a great number of sides, and measure its perimeter. We shall then have a value a little less than, but approximating very closely to, the length of the circumference. The error in this approximation will be diminished as the number of sides of the regular inscribed polygon is increased.

410. Below are given results of very exact computations which are employed instead of direct measuring.

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411. It is found that the length of the circumference whose diameter is 1 yd. is 3.1416 yds. The Greek letter is used to designate the number 3.1416; that is, T=3.1416. Hence the rule.

412. To find the length of a circumference, multiply its diameter by the number 3.1416.

413. If C denotes the circumference, R the radius, 2 R will denote the diameter, and we shall have the formula,

C=2πR.

414. The letter is called the ratio of the circumference to the diameter.

APPLICATION. Find the circumference of a circle whose radius is 20 ft.

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Find the circumference of a circle when the radius is

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Find the circumference when the diameter is equal to :

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NOTE. Since the reciprocal of 3.1416 is 0.31831, the diameter is

found by multiplying the circumference by 0.31831.

19. If the driving wheels of a locomotive have a diameter of 6 ft., how many revolutions will they make in going from Portland to Boston, 106 miles?

20. What is the diameter of a circular reservoir, if a man in walking round it takes 100 steps, each 2 ft.? 21. The diameters of two circles having the same centre are 8 ft. and 20 ft.; find the length of the circumference lying midway between them.

22. If the diameter of a circle is 3 yds., find the length of an arc of 160°.

23. The length of an arc of 40° is 10 ft.; find the diameter. 24. Find the distance traversed in an hour by a point in the circumference of a water-wheel 7 ft. in diameter, if the wheel makes 1400 revolutions in one hour. 25. If the radius of a circle is 2 ft., find the number of degrees in an arc of the circle 2 ft. in length.

AREAS.

415. The area of a surface is the number of square (square inches, feet, yards, or rods) which it contains.

units

416. Two figures are equal when they can be applied to each other so as to coincide; two figures are equivalent

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into two equal parts by the straight line EF, which joins the middle points of the sides AC and BD, and place the two rectangles as represented in the right-hand figure, we obtain a rectangle equivalent to the square, but which cannot be made to coincide with it.

417. THEOREM. The area of a rectangle is equal to the product of its base by its altitude (§ 160).

418. THEOREM. The area of a square, therefore, is equal to the square of the number which measures its side.

419. THEOREM. The area of a parallelogram is equal to F C the product of its base by its altitude.

E D

A

FIG. 43.

B

Upon the base AB of the parallelogram ABCD construct the rectangle ABEF

The parallelogram ABCD=

trapezoid ABFD + triangle BCF.

The rectangle ABEF= trapezoid ABFD + triangle AED.

Triangles BCF and AED can be applied to each other, and are equal.

Then ABCD is equivalent to ABEF.

Therefore, the area of the parallelogram is equal to AB multiplied by BF; that is, it is equal to the product of its base by its altitude.

420. THEOREM. The area of a triangle is equal to onehalf of the product of its base by its altitude.

A

D

For a triangle is half of a parallelogram of the same base and altitude. Thus, if through the point C of the triangle ABC (Fig. 44) we draw CD parallel to AB, and through the point A, AD parallel to BC, we shall have a parallelogram ABCD, of which AC will be the diagonal.

B

FIG. 44.

We can make the two triangles ABC and ACD coincide by turning ABC 180° about the point O, the middle of AC. Hence, ABC ACD, and ABC = { ABCD.

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