Q9 pl solve cu the of the is S and the AC length 8 is e the taduS me inner - Maths - Circles

Question

Q9 pl solve

MD Kabeer · Accepted Answer

Dear Student!

Let O be the incentre of ΔABC OD = OE = OF = 4 cm Let BD = 6 cm and CD = 8 cm We know that, the lengths of two tangents drawn from an external point to a circle are equal ∴ BF = BD = 6cm CE = CD = 8cm AE = AF = x cm (say) CA = AE + CE = (x + 8) cm AB = AF + BF = (x + 6) cm BC = 6 cm + 8 cm = 14 cm $Semi-perimeter of Δ ABC, s = \frac{(x + 8) + (x + 6) + 14}{2} cm = (x + 14) cm Area of Δ ABC= \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{(x + 14) [(x + 14) - (x + 8)] [(x + 14) - (x + 6)] [(x + 14) - 14]} {cm}^{2} = \sqrt{(x + 14) \times 6 \times 8 \times x} {cm}^{2} = \sqrt{48 x (x + 14)} {cm}^{2}$ Area of ΔABC = Area of ΔOBC + Area of (ΔOCA) + Area of (ΔOAB) $∴ Area of Δ ABC = \frac{1}{2} \times BC \times OD + \frac{1}{2} \times CA× OE + \frac{1}{2} \times AB×OF Area of Δ ABC = \frac{1}{2} \times 14 \times 4 {cm}^{2} + \frac{1}{2} \times (x + 8) \times 4 {cm}^{2} + \frac{1}{2} \times (x + 6) \times 4 {cm}^{2} Area of Δ ABC = 28 {cm}^{2} + 2 (x + 8) {cm}^{2} + 2 (x + 6) {cm}^{2} Area of Δ ABC = (4 x + 56) {cm}^{2} ∴ \sqrt{48 x (x + 14)} = (4 x + 56) \sqrt{48 x (x + 14)} = 4 (x + 14)$ Squaring on both sides, we get 48 x (x + 14) = 16 (x + 14)2 ∴ 3x (x + 14) = (x + 14)2 ⇒ (x + 14)2 – 3x (x + 14) = 0 ⇒ (x + 14) (x + 14 – 3x) = 0 ⇒ (x + 14) (14 – 2x) = 0 ⇒ x + 14 = 0 or 14 – 2x =0 ⇒ x = –14 or x =7 ∴ x = 7 (x cannot be negative) BC = (x + 8) cm = (7 + 8) cm = 15 cm AB = (x + 6) cm = (7 + 6) cm = 13 cm Thus, the lengths of other two sides of the triangle are 15 cm and 13 cm Regards