Page images
PDF
EPUB

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).

[blocks in formation]

§ 241

.. 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 O are equal),
the sides BH, BM, CM, CR, etc. are equal,
(being homologous sides of equal isosceles

).

Ax. 6

.. the sides A B, BC, C D, etc. are equal, 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 apothem, a one side, A one angle, and C the angle at the centre; show that

1. In a regular inscribed triangle a

[blocks in formation]

=

R √3,

2. In an inscribed square a = R√√2, r =

C = 90°.

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

=

4. In a regular inscribed decagon

1'=

=

r

=

R,

R √2, A = 90°,

R, r

[blocks in formation]

R V10+2 √5, A = 144°, C=36°,

[blocks in formation]

R (√5
2

[ocr errors]

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 AB 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 D H 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 to the middle point of A B).

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

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

Let C be the circumference and R the radius of a

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

384 AD

=

=

=

Length of Side.
.51763809

Perimeter.

6.21165708

2 √4-(.51763809)2 .26105238 6.26525722

2-V4-(.26105238)2 .13080626

6.27870041

(.13080626)2 .06543817 6.28206396

2 √4-(.06543817)2 .03272346 6.28290510

V2

[ocr errors]

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.

[blocks in formation]

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 ABC 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 ED,

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

[blocks in formation]

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

[blocks in formation]

$326

§ 52

Hyp.

Q. E. D.

« PreviousContinue »