Question

A triangle ABC is drawn to circumscribe a circle of radius 4cm such that the segment BD and DC into which BC is divided by the point of contact D are length 8cm and 6cm respectively. Find the sides AB and Ac

Answer 1

✏ Fig. Of the question

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✴ A triangle ABC is drawn to circumscribe a circle of radius 4cm such that the segment BD and DC into which BC is divided by the point of contact D are length 8cm and 6cm respectively. Find the sides AB and Ac.

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✒ The length of AB = 15cm & AC = 13cm

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★ In the given ΔABC,

☛ Length of two tangents drawn from the same point to the circle are equal.So,

• CF = CD = 6cm

• BE = BD = 8cm

• AE = AF = xcm

❈ We observed that ,

• AB = AE + EB = x + 8.........[1]

• BC = BD + DC = 8 + 6 = 14

• CA = CF + FA = 6 + x........[2]

Now, semi perimeter of circle (s),

➠ 2s = AB + BC + CA

➠ x + 8 + 14 + 6 + x

➠ 28 + 2x

➠ s = 14 + x

Area of ΔABC = √s (s - a)(s - b)(s - c)

➠ √(14 + x) (14 + x - 14)(14 + x - x - 6)(14 + x - x - 8)

➠ √(14 + x) (x)(8)(6)

➠√(14 + x) 48 x ... (i)

Also, Area of ΔABC = 2×area of (ΔAOF + ΔCOD + ΔDOB)

➠ 2×[(1/2×OF×AF) + (1/2×CD×OD) + (1/2×DB×OD)]

➠ 2×1/2 (4x + 24 + 32) = 56 + 4x ... (ii)

✰ Equating equation (i) and (ii) we get,

√(14 + x) 48 x = 56 + 4x

Squaring both sides,

✯ 48x (14 + x) = (56 + 4x)2

➠ 48x = [4(14 + x)]2/(14 + x)

➠ 48x = 16 (14 + x)

➠ 48x = 224 + 16x

➠ 32x = 224

➠ x = 224/32

➠ x = 7 cm

Hence, from 1 & 2

➡AB = x + 8 = 7 + 8 = 15 cm

➡CA = 6 + x = 6 + 7 = 13 cm

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✴ Refers to the attachment ✔✨

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