⭐A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segment BD and DC into which BC is divided by the point of contact D are if lengths 8 cm and 6cm respectively. Find the sides AB and AC. ⭐
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Answers
Step-by-step explanation:
Let O be the centre of Δ such that,
CF = CD = 6 cm.
BE = BD = 8 cm.
∴ AE = AF = x.
Now, AB = AE + BE
= x + 8.
BC = BD + CD
= 8 + 6
= 14 cm.
CA = AF + CF
= x + 6.
We know that Semi perimeter s = a + b + c/2 (or) 2s = a + b + c.
⇒ 2s = AB + BC + CA
⇒ 2s = x + 8 + 14 + 6 + x
⇒ 2s = 2(14 + x)
⇒ s = 14 + x.
Now,
Area of ΔAB = √s(s - a)(s - b)(s - c)
⇒ √(14+x){(14+x)-14}{(14+x)-(6+x)}{(14+x)-(8+x)}
⇒ √(14+x)(x)(8)(6)
⇒ √48(14x+x²)
⇒ 4√3(14x+x²)
(i) Area of ΔOBC:
= 1/2 * OD * BC
= 1/2 * 4 * 14
= 28.
(ii) Area of ΔOCA:
= 1/2 * OF * AC
= 1/2 * 4 * (6 + x)
= 12 + 2x.
(iii) Area of ΔOAB:
= 1/2 * OE * AB
= 1/2 * 4 * (8 + x)
= 16 + 2x.
∴ Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB:
⇒ 4√3(14x+x²) = 28 + 12 + 2x + 16 + 2x
⇒ 4√3(14x+x²) = 56 + 4x
⇒ √3(14x+x²) = 14+x
On squaring both sides, we get
⇒ 3(14x+x²) = (14+x)²
⇒ 42x + 3x² = 196 + x² + 28x
⇒ 2x² + 14x - 196 = 0
⇒ x² + 7x - 98 = 0
⇒ x² + 14x - 7x - 98 = 0
⇒ x(x + 14) - 7(x + 14) = 0
⇒ (x-7)(x+14) = 0
⇒ x = 7, -14{Neglect -ve values}
⇒ x = 7
(iv)
AB = x + 8
= 7 + 8
= 15 cm.
(v)
AC = 6 + x
= 6 + 7
= 13 cm.
Therefore, AB= 15 cm and AC = 13 cm.
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