CHAPTER XV. PROPERTIES OF A TRIANGLE. 198. Area of a given triangle. Let ABC be any triangle and AD the perpen dicular drawn from A upon the opposite side. Through A draw EAF parallel to BC and draw BE and CF perpendicular to it. By Euc. I. 41, the area of the triangle ABC E B = rectangle BF = {BC.CF = a.ĄD. But ADAB sin B = c sin B. The area of the triangle ABC therefore ca sin B. This area is denoted by A. Hence Aca sin B = lab sin C = bc sin A.....(1). 2 By Art. 169, we have sin A = bc √s (s − a) (s — b) (s — c), so that Abc sin A = √√s (s-a) (s-b) (s—c)...(2). This latter quantity is often called S. EXAMPLES. XXXV. Find the area of the triangle ABC when 1. a=13, b=14, 3. a=25, b=52, 5. a=15, b=36, 7. a=35, b=84, c=30. and c=15.2. a=18, b=24, and c= 9. If B=45°, C=60°, and a= = 2 (√3+1) inches, prove that the area of the triangle is 6+2/3 sq. inches. 10. The sides of a triangle are 119, 111, and 92 yards; prove that its area is 10 sq. yards less than an acre. 11. The sides of a triangular field re 242, 1212 and 1450 yards; prove that the area of the field is 6 acres. 12. A workman is told to make a triangular enclosure of sides 51, 41, and 21 yards respectively; having made the first side one yard too long, what length must he make the other two sides in order to enclose the prescribed area with the prescribed length of fencing? 13. Find, correct to 0001 of an inch, the length of one of the equal sides of an isosceles triangle on a base of 14 inches having the same area as a triangle whose sides are 13·6, 15, and 15·4 inches. 14. Prove that the area of a triangle is a2. sin B sin C sin A If one angle of a triangle be 60°, the area 10/3 square feet, and the perimeter 20 feet, find the lengths of the sides. 15. The sides of a triangle are in A.P. and its area is equal triangle of the same perimeter; prove that its sides are in the ratio 3:5:7, and find the greatest angle of the triangle. 16. In a triangle the least angle is 45° and the tangents of the angles are in A.P. If its area be 3 square yards, prove that the lengths of the sides are 3/5, 6/2, and 9 feet, and that the tangents of the other angles are respectively 2 and 3. 17. The lengths of two sides of a triangle are one foot and √2 feet respectively and the angle opposite the shorter side is 30°; prove that there are two triangles satisfying these conditions, find their angles, and shew that their areas are in the ratio /3+1:√3-1. 18. Find by the aid of the tables the area of the larger of the two triangles given by the data A=31° 15', a=5ins. and b=7ins. 199. On the circles connected with a given triangle. The circle which passes through the angular points of a triangle ABC is called its circumscribing circle or, more briefly, its circumcircle. The centre of this circle is found by the construction of Euc. IV. 5. Its radius is always called R. The circle which can be inscribed within the triangle so as to touch each of the sides is called its inscribed circle or, more briefly, its incircle. The centre of this circle is found by the construction of Euc. IV. 4. Its radius will be denoted by r. The circle which touches the side BC and the two sides AB and AC produced is called the escribed circle opposite the angle A. Its radius will be denoted by r1 Similarly r2 denotes the radius of the circle which touches the side CA and the two sides BC and BA produced. Also r, denotes the radius of the circle touching AB and the two sides CA and CB produced. 3 200. To find the magnitude of R, the radius of the circumcircle of any triangle ABC. Bisect the two sides BC and CA in D and E respectively, and draw DO and EO perpendicular to BC and CA. By Euc. IV. 5, 0 is the centre of the circumcircle. The point O may either lie within the triangle as in Fig. I., or without it as in Fig. II., or upon one of the sides as in Fig. III. Taking the first figure, the two triangles BOD and COD are equal in all respects, so that If A be obtuse, as in Fig. II., we have < BOD=< BOC = 2 BLC = 180°-A (Euc. III. 22), so that, as before, sin BOD = sin A, If A be a right angle, as in Fig. III., we have Substituting this value of sin A in (1), we have giving the radius of the circumcircle in terms of the sides. 202. To find the value of r, the radius of the incircle of the triangle ABC. Bisect the two angles B and C by the two lines BI and CI meeting in I. A F area of ▲ IBC = 1⁄2ID. BC = 1⁄2r. a, 1r.a, area of ▲ ICA = 1⁄2IE. CA = 1⁄2r. b, area of ▲ IAB = ¿IF. AB = {r.c. D பப |