Page images
PDF
EPUB

PROPOSITION XV. THEOREM.

412. The radius and the value of the chord of an arc being given, we can find the value of the chord of half that arc.

E

A

B

Given: In circle ACBE, a radius OC perpendicular at D to AB, the chord of arc ACB;

To Prove: The value of the chord of one half arc ACB can be found.

Since OC is to AB at D,

AD = AB, and arc AC = arc ACB.
Then since AC < a quadrant,*

Join 04, AC.
OA,

[merged small][ocr errors][merged small][merged small][merged small][merged small]

(Hyp.) (172)

(350)

(Hyp.)

2

(347)

OD2 OA2 — AD2 ≈ R2 — 1 AB2;

4

[blocks in formation]

(OD, R, AB, here denoting numerical measures;)

... AC2 = 2 R2 - 2 R. √4 R2-AB2;

[ocr errors]

ΑΒ

[ocr errors][subsumed]

If, as is usually done, we take R = 1, then

AC=√2-V4-AB2.
ᎪᏴ .

*At most AC of the circumference, since ACB =, at most, of the circumference,

PROPOSITION XVI. PROBLEM.

413. To find an approximate value of T, the ratio of circumference to diameter.

FORMULA: 82n = √2-√4 — Sn

where s denotes the value of a side of a regular inscribed polygon of n sides, R being taken = 1.

[blocks in formation]

If we take n = 6,

[blocks in formation]

Taking this last value of the perimeter as an approximation towards the value of the circumference whose radius is

C

[ocr errors]

1, since π= (396), we obtain = (6.28317) = 3.14159

2 R

nearly, a result correct to the last decimal.

By the useless expenditure of much time and labor, the value of has been calculated as far as 700 places of decimals. The first fifteen, more than sufficient for any useful purpose, are π = 3.141592653589793. Ten decimals are sufficient to give the circumference of the earth to the fraction of an inch, and thirty decimals would give the circumference of the whole visible universe to a quantity imperceptible with the most powerful microscope.

PROPOSITION XVII. PROBLEM.

414. The diameter being given, to find a line approximately equal to the circumference.

[blocks in formation]

Required: To find a straight line approximately equal to the circumference of that circle.

Draw indefinite lines AM, AN, making any angle.

Upon AM lay off AB=113, and AC=355, units of length,* and upon AN lay off AD = D. Join BD and draw CX to BD.

Since AB AC AD AX = 113: 355,

[blocks in formation]

(274)

Q.E.F.

The line thus obtained is greater than the circumference by about one four millionth of the diameter, as the student may easily prove for himself.

415. In the inscribed equilateral triangle ABC, draw AD 1 to BC, and join CD. Then AD bisects arc BDC (172), and DCR, the radius (408);

... AC2 = DA — DC4 R2-R2 = 3 R2;

[ocr errors]

.. AC = R√3.

B

A

* Lengths respectively equal to 1.13 and 3.55 units can be taken from a diagonal scale; or see Exercise 836.

416. In the inscribed square ADBC, joining AB, CD,

AC2 = 402+002 = 2 R2 (347);

.. AC=R√2.

417. In the inscribed pentagon, the relation of the side to the radius is most readily obtained from the expres- E sion for the side of the inscribed deca

[blocks in formation]

E

[merged small][ocr errors][merged small]
[blocks in formation]

418. In the inscribed hexagon, the side DC = R (408), a result that can be obtained also by means of the formula referred to in the next article.

419. In the inscribed octagon (see diagram for Art. 416) the value of a side is most readily found by means of the formula obtained in Prop. XV.; that is

[ocr errors]

In this, putting AE for AC, and R√2, the value of a side of the inscribed square, for AB, we obtain

[merged small][merged small][merged small][merged small][merged small][ocr errors]

420. In the inscribed decagon AF... E (see diagram for Art. 417),

[merged small][merged small][merged small][ocr errors]

.. BF = R(√5-1), solving as a quadratic.

(409)

(322)

-2

In the formula 4C = √2 R2 — R √4 R2 — AB3, putting R(√5-1), the value of a side of the inscribed decagon, for AC, and squaring, we obtain

1 R2 (√5 — 1)2 = 2 R2 - RV4 R2 - AB";

whence 2 V4 R2
2 √4 R2 — AB2 = R(√5 +1);
whence ABR (10 −2 √5).

EXERCISES.

QUESTIONS.

584. Into how many equal parts, up to 15, are we able to divide a circumference by means of the straight line and the circle?

585. What is the ratio of the radius of the circle circumscribed about an equilateral triangle to that of the inscribed circle?

586. By which of the constructions of this book can a right angle be divided into five equal parts?

587. How many different stellate polygons can be formed if we have a circumference divided into 7 equal parts? 9 equal parts? 11 equal parts? 13 equal parts ?

588. If two nonadjacent sides of a regular pentagon be produced to meet, what angle will they contain ?

589. What is the value of the interior angles of a regular octagon ? Of a regular decagon? Of a regular pentadecagon ? What is the limit towards which each of the interior angles of a regular polygon tends as the number of sides increases indefinitely? What is the limit for each exterior angle?

590. What is the circumference and area of a circle whose diameter is 10 inches, supposing = 3.1416?

591. What is the circumference and area of a circle whose radius is 2 ft. 6 in. ?

592. What is the circumference of a circle whose area is 100 sq. in. ?

« PreviousContinue »