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# 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).

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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.

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Consider the triangle ABC,

We know that the length of any two tangents which are drawn from the same point to the circle is equal.

(i) CF = CD = 6 cm

(ii) BE = BD = 8 cm

(iii) AE = AF = x

Now, it can be observed that,

(i) AB = EB + AE = 8 + x

(ii) CA = CF + FA = 6 + x

(iii) BC = DC + BD = 6 + 8 = 14

Now the semi perimeter 's' will be calculated as follows

2s = AB + CA + BC

By putting the respective values we get,

2s = 28 + 2x

s = 14+x

Area of ΔABC = $$\sqrt{S(s-a)(s-b)(s-c)}$$

By solving this we get,

= √(14 + x)48x ……… (i)

Again, the 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)

Now from (i) and (ii) we get,

√(14 + x)48x = 56 + 4x

Now, square both the sides,

48 x (14 + x) = (56 + 4x)2

48x = [4(14 + x)]2/(14 + x)

48x = 16(14 + x)

48x = 224 + 16x

32x = 224

x = 7 cm

So, AB = 8 + x

i.e. AB = 15 cm

CA = x + 6 = 13 cm

This post was modified 2 years ago by admin
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