217. To find the area of a circle. Let be the radius of the circle, and let a regular polygon of n sides be described about it. Then from the adjoining figure, we have area of polygon=n (area of triangle AOB) 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 218. To find the area of the sector of a circle. Let be the circular measure of the angle of the sector; 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 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 inscribed in a circle whose regular inscribed pentagon has an area of 50 sq. ft. 9. A regular pentagon and a regular decagon have the same area, prove that the ratio of their perimeters is √2 : 4/5. 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, A, 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 (1) A is a geometric mean between A1 and B1; 2 The Ex-central Triangle. *219. Let ABC be a triangle, I1, I2, I, its ex-centres; then 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 I, I, lie on the line bisecting the angle ACB externally. (iii) The line AI is perpendicular to II; the line BI, is perpendicular to 1,1; the line CI, is perpendicular to 12. Thus the triangle ABC is the Pedal triangle of its ex-central triangle III. [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, I2 and the points A, I, B, I, are concyclic. (v) The lines AI1, BI, CI, meet at the in-centre I, which is therefore the Orthocentre of the ex-central triangle III. (vi) Each of the four points I, I, I, I, is the orthocentre of the triangle formed by joining the other three points. *220. To find the distances between the in-centre and ex centres. A B 2 2 =4R (sin cos cos-sin sin sin cosec 4 COS COS A B 2 2 ) *221. To find the sides and angles of the ex-central triangle. With the figure of Art. 219, LBI1C= L BI1I+ ≤ CI1I = LBCI+ LCBI [Euc. III. 21] *222. To find the area and circum-radius of the ex-central triangle. (product of two sides) x (sine of included angle) 2 *224. To find the sides and angles of the pedal triangle. In the figure of the last article, the points K, O, G, B are concyclic; .. LOGK=LOBK=90° – A. Also the points H, O, G, C are concyclic; ... LOGH LOCH=90° - A; .. LKGH=180° – 2A. Thus the angles of the pedal triangle are Again, LAKH=180° – ▲ BKH= L BCH, since the points B, K, H, C are concyclic; .. LAKH=C. |