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Then, since the sector is the same part of the o that the arc is of the Oce, ..s:S=c:C.

(294) CXS CX RX

(438) C

C

s={R X C. Therefore, the area of a sector is equal to half the product of its radius and arc.

or

Proposition 10. Theorem. 441. The areas of two similar segments are to each other as the squares of their radii.

Hyp. Let S and S' be the areas of the two similar sectors AOB, A'O'B', T and T the areas of the two similar AS AOB, A'O'B', R and R’ the radii.

с

C'
S - T R?

S - T
Proof. Since
20= 20',

(432) ..S:S' = R': R',

(435) and T:T = R' : R''.

(377) S T R S' T

R'?

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To prove

R'I.

2.

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S - T
S TV

(by division).

Q.E.D.

R'?

EXERCISES. 1. What is the area of a circle whose radius is 40 feet?

2. What is the diameter of a circle whose circumference is 57 yards?

PROBLEMS OF CONSTRUCTION.

rt Zs,

Proposition 11. Problem.
442. To inscribe a square in a given circle.
Given, the O ABCD, with centre 0.
Required, to inscribe a square in it.

Cons. Draw the diameters AC, BD
I to each other.

B
Join AB, BC, CD, DA.
Then ABCD is the square required.
Proof. Since the Zs at O are all

(Cons.) ... the sides AB, BC, etc., are equal, in the same o equal _s at the centre intercept equal chords (196), and the Zs BAD, ADC, etc., are rt. ZS, every Z inscribed in a fo is a rt. 2 (240). ... the figure ABCD is a square.

(126)

Q.E.F. 443. CoR. 1. If tangents be drawn to the circle at the points A, B, C, D, the figure so formed will be a circumscribed square.

444. Cor. 2. To inscribe, and circumscribe, regular polygons of 8 sides, bisect the arcs AB, BC, CD, DA, and proceed as before.

By repeating this process, regular inscribed and circumscribed polygons of 16, 32, . ., and, in general, of an sides, may be drawn.

EXERCISES.

1. What is the area of a square inscribed in a circle whose area is 48 feet ?

2. What is the area of a regular hexagon inscribed in a circle whose area is 560 square feet ?

Proposition 12. Problem. 44.5. To inscribe a regular hexagon in a given circle. Given, the O ABC, with centre 0. Required, to inscribe a regular hexa

gon in it.

D

A

Cons. With any pt. A on the Oce as a centre, and AO as a radius, describe an arc cutting the Oce in B. Join AB, BO, OA.

E Then AB is a side of the hexagon required. Proof. Since the A OAB is equilateral, (Cons.) :: it is equiangular.

(113) .. AOB is 1 of 2 rt. Zs, or f of 4 rt. Zs. (103) .. the arc AB is t of the Oce.

.. the chord AB is a side of the regular inscribed hexagon.

(400) .: the figure ABCDEF, completed by drawing the chords BC, CD, DE, EF, FA, each equal to the radius OA, is the regular inscribed hexagon required.

Q.E.F. 446. COR. 1. If the alternate vertices of the regular hexagon be joined, we obtain the inscribed equilateral triangle ACE.

447. CoR. 2. If tangents be drawn to the circle at the points A, B, C, D, E, F, the figure so formed will be a regular circumscribed hexagon.

448. COR. 3. To inscribe, and circumscribe, regular polygons of 12 sides, bisect the arcs AB, BC, CD, etc., and proceed as before.

By repeating this process, regular inscribed and circumscribed polygons of 24, 48, etc., sides may be drawn.

E

Proposition 13. Problem. 449. To inscribe a regular decagon in a given circle. Given, the O ABCE.

Required, to inscribe in it a regular decagon.

Cons. Draw any radius 0A, and divide it in extreme and mean ratio at D, so that OA: OD = OD: AD. (352)

B With centre A and OD as a radius, describe an arc cutting the ce at B, and join AB.

Then AB is a side of the required inscribed decagon.
Proof. Join BO, BD.
Since OA: OD = OD: AD,

(Cons.) and OD = BA,

(Cons.) .:. OA: BA = BA: AD.

.. the As OAB and BAD are similar,
having _A common and the including sides proportional (314).
.. ZABD = ZAOB. .

(307) Because OA = OB, and BA = OD,

(Radii) and (Cons.) .:. OB: BA

OD: AD, .'. LABD = _DBO.

(303) LABO = 2 LABD = 2 ZAOB. ... BAO also _BAO = LABO = 2 LAOB,

being opp. the equal sides of an isosceles A (111). ... LABO + BAO + ZAOB = 5 ZAOB = 2 rt. Zs. ... ZAOB

} of 2 rt. Zs, or to of 4 rt. Zs. the arc AB is ty of the Oce. ... the chord AB is a side of the regular inscribed decagon.

(400) ... the figure ABCE . . . , formed by applying AB ten times to the Oce, is the regular inscribed decagon required.

Q.E.F

450. Cor. 1. If the alternate vertices of the required decagon be joined, a regular pentagon is inscribed.

451. Cor. 2. If tangents be drawn at the points at which the circumference is divided, the figure so formed will be a regular circumscribed decagon.

452. COR. 3. To inscribe, and circumscribe, regular polygons of 20 sides, bisect the arcs AB, BC, etc., and proceed as before.

By repeating this process, regular inscribed and circumscribed polygons of 40, 80, etc., sides may be drawn.

EXERCISES.

1. The side of an inscribed square is equal to the radius of the circle multiplied by V2.

2. The side of an inscribed equilateral triangle is equal to the radius multiplied by V3.

3. The apothem of an inscribed square is equal to half the radius multiplied by V2.

4. The apothem of an inscribed equilateral triangle is equal to half the radius.

5. The apothem of a regular inscribed hexagon is equal to half the radius multiplied by V3.

6. The area of a circumscribed square is double the area of the inscribed

square. 7. Required the area of an equilateral triangle inscribed in a circle whose radius is 4.

8. Required the area of a square inscribed in a circle whose radius is 5.

9. Required the area of a regular hexagon inscribed in a circle whose radius is 8.

10. Required the area of a circle whose circumference is 100.

11. Required the area of a circle inscribed in a square whose area is 36.

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