12. 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 Fig. 10.14). Find the sides AB
and AC.
Answers
Step-by-step explanation:
this can also be solved by using Brahmagupta theorem. Some part got cropped, inconvenience regretted
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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.
So,
- (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,
s = 28 + 2x
⇒ s = 14 + x
Area of △ABC = √s (s – a)(s – b)(s – c)
By solving this we get,
= √(14 + x) 48 x ....… (i)
Again, the area of △ABC = 2 × area of (△ AOF + △ COD + △ DOB)
= 2 × [(½ × OF × AF) + (½ × CD × OD) + (½ × DB × OD)]
= 2 × ½ (4x + 24 + 32) = 56 + 4x ….... (ii)
Now from (i) and (ii) we get,
√(14 + x) 48 x = 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
And, CA = x + 6 = 13 cm.