Inscribed and circumscribed Polygons.

214. To find the perimeter and area of a regular polygon of n sides inscribed in a circle.

Let be the radius of the circle, and

AB a side of the polygon.

Join OA, OB, and draw OD bisecting LAOB; then AB is bisected at right angles in D.

Perimeter of polygon=nAB=2nAD=2nOA sin AOD

215. To find the perimeter and area of a regular polygon of n sides circumscribed about a given circle.

Let be the radius of the circle, and AB a side of the polygon. Let AB touch the circle at D. Join OA, OB, OD; then OD bisects AB at right angles, and also bisects LAOB.

Area of polygon=n (area of triangle AOB)

216. There is no need to burden the memory with the formulæ of the last two articles, as in any particular instance they are very readily obtained.

Example 1. The side of a regular dodecagon is 2 ft., find the radius of the circumscribed circle.

Let r be the required radius. In the adjoining figure we have

Example 2. A regular pentagon and a regular decagon have the same perimeter, prove that their areas are as 2 to √√5.

Let AB be one of the n sides of a regular

polygon, O the centre of the circumscribed circle, OD perpendicular to AB.

By increasing the number of sides without limit, the area and the perimeter of the polygon may be made to differ as little as we please from the area and the circumference of the circle. Hence

Let

218. To find the area of the sector of a circle.

be the circular measure of the angle of the sector;

then by Euc. VI. 33,

1. Find the area of a regular decagon inscribed in a circle

whose radius is 3 feet; given sin 36°='588.

2. Find the perimeter and area of a regular quindecagon described about a circle whose diameter is 3 yards; given

tan 12° = 213.

3. Shew that the areas of the inscribed and circumscribed circles of a regular hexagon are in the ratio of 3 to 4.

4. Find the area of a circle inscribed in a regular pentagon whose area is 250 sq. ft.; given cot 36° = 1.376.

5. Find the perimeter of a regular octagon inscribed in a circle whose area is 1386 sq. inches; given sin 22° 30′ =·382.

6. Find the perimeter of a regular pentagon described about a circle whose area is 616 sq. ft.; given tan 36° = '727.

7. Find the diameter of the circle circumscribing a regular quindecagon, whose inscribed circle has an area of 2464 sq. ft.; given sec 12°=1′022.

8. Find the area of a regular dodecagon in a circle about a regular pentagon 50 sq. ft. in area; given cosec 72° = 1·0515.

9. A regular pentagon and a regular decagon have the same area, prove that the ratio of their perimeters is 5: √2.

10. Two regular polygons of n sides and 2n sides have the same perimeter; shew that the ratio of their areas is

11. If 2a be the side of a regular polygon of n sides, R and r the radii of the circumscribed and inscribed circles, prove that

12. Prove that the square of the side of a regular pentagon inscribed in a circle is equal to the sum of the squares of the sides of a regular hexagon and decagon inscribed in the same circle.

13. With reference to a given circle, A1 and B1 are the areas of the inscribed and circumscribed regular polygons of n sides, A and B are corresponding quantities for regular polygons of 2n sides prove that

2

(1) A, is a geometric mean between A1 and B1;
(2) B2 is a harmonic mean between A, and B1.

The Ex-central Triangle.

*219. Let ABC be a triangle, I, I, I, its ex-centres; the III, is called the Ex-central triangle of ABC.

Let I be the in-centre; then from the construction for finding the positions of the in-centre and ex-centres, it follows that:

(i) The points I, I, lie on the line bisecting the angle BAC; the points I, I lie on the line bisecting the angle ABC; the points I, I, lie on the line bisecting the angle ACB.

(ii) The points I, I lie on the line bisecting the angle BAC externally; the points I, I lie on the line bisecting the angle ABC externally; the points I1, I lie on the line bisecting the angle ACB externally.

(iii) The line AI is perpendicular to II; the line BI, is perpendicular to 1,11; the line CI, is perpendicular to II. Thus the triangle ABC is the Pedal triangle of its ex-central triangle Ill [See Art. 223.]

(iv) The angles IBI, and ICI, are right angles; hence the points B, I, C, I, are concyclic. Similarly, the points C, I, A, I and the points A, I, B, I, are concyclic.

(v) The lines AI1, BI2, CI, meet at the in-centre I, which is therefore the Orthocentre of the ex-central triangle I12.

(vi) Each of the four points I, I, I, I, is the orthocentre of the triangle formed by joining the other three points.