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 given figure). Find the sides AB and AC.

explain this some in a little easy maNNER.PLS EXPERTSS

Firstly, consider that the given circle will touch the sides AB and AC of the triangle at point E and F respectively.

Let AF = *x*

Now, in ABC,

CF = CD = 6cm (Tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point C)

BE = BD = 8cm (Tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point B)

AE = AF = *x* (Again, tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point A)

Now, AB = AE + EB

= *x* + 8

Also, BC = BD + DC = 8 + 6 = 14 and CA = CF + FA = 6 +* x*

Now, we get all the sides of a triangle so its area can be find out by using Heron's formula as:

2*s = *AB + BC + CA

= *x* + 8 + 14 + 6 +* x*

*=* 28 + 2*x*

*⇒ ***Semi-perimeter*** = s* = (28 + 2x)/2 = 14 +

*x*

Again, area of triangle is also equal to the . Therefore,

Area of ΔOBC =

Area of ΔOCA =

Area of ΔOAB =

Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB

On squaring both sides, we get

Either *x+*14 = 0 or *x* − 7 =0

Therefore*, x = −*14and 7

However, *x *= −14 is not possible as the length of the sides will be negative.

Therefore, *x* = 7

Hence, AB = *x* + 8 = 7 + 8 = 15 cm

CA = 6 +* x *= 6 + 7 = 13 cm

Hope you get it!!

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