Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

the As BH M, BMH, C MR, and CRM are equal, $ 209

(being measured by halves of equal arcs) ;

..the A BHM and C M R are equal, $ 107 (having a side and two adjacent of the one equal respectively to a side and

two adjacent of the other).

1. ZB=LC,

(being homologous & of equal a ).
In like manner we may prove 2C=_ D, etc.

.. the polygon A B C D, etc., is equiangular.
Since the A BHM, C M R, etc. are isosceles, § 241
(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 8). is 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, y the apothem, a one side, A one angle, and C the angle at the centre; show that

1. In a regular inscribed triangle a = R V3, r = 1 R, A = 60°, C = 120°.

2. In an inscribed square a = RV2, p= } RV2, A = 90°, C = 90°.

3. In a regular inscribed hexagon a = R, r = 1 R V3, A = 120°, C = 60°.

_ R (V5 – 1) 4. In a regular inscribed decagon a = " 2 = { R V10 + 2 V5, A = 144°, C = 36°,

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.

D
AOB

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 D H through the centre 0,

and draw 0 A. HD is I to the chord A B at its middle point C, $ 60 (two points, 0 and D, equally distant from the extremities, A and B, determine the position of a I to the middle point of A B). The 2 HA D is a rt. L,

§ 204 (being inscribed in a semicircle), . AD= D H XDC,

§ 289 (the square on one side of a rt. A is equal to the product of the hypotenuse by

the adjacent segment made by the I let fall from the vertex of the rt. Z).

Now

and

DH= 2 R,
DC=DO CO=R-CO;

· AD= 2 R (R CO).

[blocks in formation]

then 1D*=2 R (R – V+ 2? - 1B),

= 2 R R (V4 R – AB"). :. AD=V2 F2 (V4 Pe – AB).

= 2 R2 –

Q. E. F.

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

omes

the equation 4 D=V2P R (V4 ? – AB”) becomes

AD=V2 – V4-AB.

PROPOSITION XXII. PROBLEM. 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

circle.
Since

§ 376 2 R'

No.

when R= 1, 1 = 2:
It is required to find the numerical value of 7.

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

AD=V2 – V4 - A B?,
when A B is a side of a regular hexagon :
In a polygon of
Sides.
Form of Computation.

Length of Side. Perimeter. 12 AD=V2 - V1 – 12

.51763809 6.21165708 24 A D=V2 - V4 (.51763809)2 .26105238 6.26525722 48 AD= V2 – V4– (.26105238)2.13080626 6.27870041 96 AD=V2 - V4 – (.13080626)2.06543817 6.28206396 192 AD=V2 — V4—(.06543817)2 .03272346 6.28290510 384 AD=V2 - V4 – 0.03272346)2.01636228 6.28311544 768 AD=V2 – V4 – 0.01636228)2.00818121 6.28316941

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

Donal. C 6.28317

2' = 2 ..7 = 3.14159 nearly.

Q. E. F

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 maximum.

В

Let the triangles A B C and EBC have the sides A B

and BC equal respectively to E B and BC; and
let the angle A B C be a right angle.
We are to prove A ABC > A EBC.
From E, let fall the I ED.

The A ABC and E B C, having the same base B C, are to each other as their altitudes A B and ED,

$ 326 (S having the same base are to each other as their altitudes). Now E D is < EB,

$ 52 (a I is the shortest distance from a point to a straight line). But EB= AB,

Hyp .. E D is < A B. .:. A ABC > A EBC.

Q. E. D.

« ΠροηγούμενηΣυνέχεια »