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
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