Please suggest me the best way to solve the problem of question from circles chapter of exercise 10.2 of ncert math of question no.12 how i solve this problem in easy and simple way 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.

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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 Fig. 10.14). Find the sides AB and AC. Q.12

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The figure given is as follows:

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 =

xNow, 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+2

xs = 14+

xBy solving this we get,

= √(14+

x)48x……… (i)Again, the area of △ABC = 2 × area of (△AOF + △COD + △DOB)

= 2×[(½×OF×AF)+(½×CD×OD)+(½×DB×OD)]

= 2×½(4

x+24+32) = 56+4x…………..(ii)Now from (i) and (ii) we get,

√(14+

x)48x= 56+4xNow, square both the sides,

48

x(14+x) = (56+4x)^{2}48

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

x =16(14+x)48

x =224+16x32

x =224x =7 cmSo, AB = 8+x

i.e. AB = 15 cm

And, CA = x+6 =13 cm.