Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

the BHM, BMH, CMR, and CRM are equal, § 209 (being measured by halves of equal arcs);

.. the ABHM and CMR are equal,

§ 107

(having a side and two adjacent of the one equal respectively to a side and two adjacent of the other).

..LB=LC,

(being homologous ▲ of equal ▲ ).

=

In like manner we may prove Z CZ D, etc.

.. the polygon A B C D, etc., is equiangular.
Since the ABHM, CMR, etc. are isosceles,
(two tangents drawn from the same point to a are equal),
the sides BH, BM, CM, CR, etc. are equal,

§ 241

(being homologous sides of equal isosceles §).

.. the sides A B, BC, C D, etc. are equal,

Ax. 6

and the polygon A B C D, etc. is equilateral.

Therefore the circumscribed polygon is regular and similar

to the given inscribed polygon.

$ 372

Q. E F.

Ex. Let R denote the radius of a regular inscribed polygon, r the apothegm, a one side, A one angle, and C the angle at the centre; show that

1. In a regular inscribed triangle a = R √3, r =

[blocks in formation]

60°, C:

=

120°.

2. In an inscribed square a

C = 90°.

R,

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

3. In a regular inscribed hexagon a = A=120°, C = 60°.

4. In a regular inscribed decagon a

7=

= 1 R √10 + 2 √5, A = 144°, C=36°.

1)

PROPOSITION XXI. PROBLEM.

401. To find the value of the chord of one-half an arc, in terms of the chord of the whole arc and the radius of the circle.

[blocks in formation]

Let A B be the chord of arc A B and AD the chord

of one-half the arc A B.

It is required to find the value of A D in terms of A B and R (radius).

From D draw DH through the centre 0,

and draw O A.

HD is to the chord A B at its middle point C, § 60 (two points, O and D, equally distant from the extremities, A and B, determine the position of a 1 to the middle point of A B).

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

402. COROLLARY. If we take the radius equal to unity,

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

403. To compute the ratio of the circumference of a circle to its diameter, approximately.

Ө

Let C be the circumference and R the radius of a

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

It is required to find the numerical value of π.

We make the following computations by the use of the formula obtained in the last proposition,

[blocks in formation]

(.51763809)2 .26105238

6.26525722

24 AD=√2 −√4 — (.51763809)2

48 AD 2-√4-(.26105238)2 .13080626

=

6.27870041

96 AD=√2-√4-(.13080626)2 .06543817 6.28206396 192 AD=√2-4-(.06543817)2 .03272346 6.28290510 384 AD= 2-4-(.03272346)2 .01636228 6.28311544 768 AD=√2-√4-(.01636228)2 .00818121

6.28316941

Hence we may consider 6.28317 as approximately the circumference of a O whose radius is unity.

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

ON ISOPERIMETRICAL POLYGONS.

-SUPPLEMENTARY.

404. DEF. Isoperimetrical figures are figures which have equal perimeters.

405. DEF. Among magnitudes of the same kind, that which is greatest is a Maximum, and that which is smallest is a Minimum.

Thus the diameter of a circle is the maximum among all inscribed straight lines; and a perpendicular is the minimum among all straight lines drawn from a point to a given straight line.

PROPOSITION XXIII. THEOREM.

406. Of all triangles having two sides respectively equal, that in which these sides include a right angle is the maxi

[blocks in formation]

Let the triangles A B C and EBC have the sides A B and BC equal respectively to EB and BC; and let the angle ABC be a right angle.

[blocks in formation]

The AABC and E B C, having the same base B C, are to

each other as their altitudes A B and E D,

§ 326

(A having the same base are to each other as their altitudes).

Now

ED is < EB,

$ 52

But

(a is the shortest distance from a point to a straight line).

EB A B,

.. ED is < A B.

..A ABC>A EBC.

Hyp.

Q. E. D.

« ΠροηγούμενηΣυνέχεια »