Last updated at Feb. 25, 2017 by Teachoo

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Ex 10.2,12 A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC. Given: A circle with centre O with OD = radius = 4 cm Let Δ ABC circumscribe the circle Also, BD = 8 cm & CD = 6 cm To find: AB & AC Construction: Join OA, OC& OB Let AC, AB intersect circle at E & F respectively Solution: From theorem 10.2, lengths of tangents drawn from external point are equal Hence, CE = CD = 6 cm BF = BD = 8 cm AE = AF = x Now, our three sides are CD = CD + DB = 6 + 8 = 14 cm AC = AE + EC = (x + 6) cm AB = AF + FC = (x + 8) cm Now, we find Area of Δ ABC using Herons formula Area of triangle = √(𝑠(𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐)) Here, AC = a = (x + 6) cm AB = b = (x + 8) cm BC = c = 6 + 8 = 14 cm. s = (𝑎 + 𝑏 + 𝑐)/2 = ((𝑥 + 6) + (𝑥 + 8) + 14)/2 = (2𝑥 + 28)/2 = (2(𝑥 + 14))/2 = x + 14 Area Δ ABC= √(𝑠(𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐)) Putting values = √((𝑥+14)(𝑥+14−(𝑥+6))(𝑥+14−(𝑥+8))(𝑥+14−14)) = √((𝑥+14)(𝑥+14−𝑥−6)(𝑥+14−𝑥−8)(𝑥+14−14)) = √((𝑥+14)(8)(6)(𝑥)) = √((𝑥+14)(48)(𝑥)) = √((𝑥+14)(48𝑥)) = √(𝑥 (48𝑥)+14(48𝑥)) = √(48𝑥2+672𝑥) ∴ Area Δ ABC = √(48𝑥2+672𝑥) Join OE & OF Here, OD = OF = OE = radius = 4 cm Also, we know that tangent is perpendicular to the radius, So, OD ⊥ BC , OF ⊥ AB & OE ⊥ AC Also, Area of Δ ABC = Area Δ AOC + Area Δ AOB + Area Δ BOC We find Area Δ AOC , Area Δ AOB & Area Δ BOC Now, Area of Δ ABC = Area Δ AOC + Area Δ AOB + Area Δ BOC Putting values √(48𝑥2+672𝑥) = (2x + 12) + (2x + 16) + 28 √(48𝑥2+672𝑥) = 4x + 56 Squaring both sides (√(48𝑥2+672𝑥))2 = (4x + 56)2 48x2 + 672x = (4x)2 + 562 + 2 × 4x × 56 48x2 + 672x = 16x2 + 3136 + 448x 48x2 – 16x2 + 672x – 448x – 3136 = 0 32x2 + 224x – 3136 = 0 32(x2 + 7x – 98)= 0 x2 + 7x – 98 = 0 x2 + 14x – 7x – 98 = 0 x(x + 14) – 7(x + 14) = 0 (x – 7) (x + 14) = 0 So, x = 7 & x = –14 Since x cannot be negative, So, x = 7 Now, AC = x + 6 = 7 + 6 = 13 cm & AB = x + 8 = 7 + 8 = 15 cm

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.