Question

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.​

Answer 1

Step-by-step explanation:

this can also be solved by using Brahmagupta theorem. Some part got cropped, inconvenience regretted

Answer 2

$\huge\star{\red{\underline{\underline{\mathbb{EXPLANATION:}}}}}$

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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.