Question

A triangle abc is drawn to subscribe a circle of radius 4cm such that the segments bd and dc in to which bc is divide by the points of contact d are of lenght 8cm and 6cm find sides ab and ac

Answer 1

Let there is a circle having centre O touches the sides AB and AC of the triangle at point E and F respectively.

Let the length of the line segment AE is x.

Now in ΔABC,

CF = CD = 6 (tangents on the circle from point C)

BE = BD = 6 (tangents on the circle from point B)

AE = AF = x (tangents on the circle from point A)

Now AB = AE + EB

=> AB = x + 8

BC = BD + DC

=> BC = 8+6 = 14

CA = CF + FA

=> CA = 6 + x

Now

s = (AB + BC + CA )/2

=> s = (x + 8 + 14 + 6 +x)/2

=> s = (2x + 28)/2

=> s = x + 14

Area of the ΔABC = √{s*(s-a)*(s-b)*(s-c)}

                           = √[(14+x)*{(14+x)-14}*{(14+x)-(6+x)}*{(14+x)-(8+x)}]

                           = √[(14+x)*x*8*6]

                           = √[(14+x)*x*2*4*2*3]

=> Area of the ΔABC = 4√[3x(14+x)]  .............................1

Now area of ΔOBC = (1/2)*OD*BC

                             = (1/2)*4*14

                             = 56/2

                             = 28

Area of ΔOBC = (1/2)*OF*AC

                             = (1/2)*4*(6+x)

                             = 2(6+x)

                             = 12 + 2x

Area of ΔOAB = (1/2)*OE*AB

                             = (1/2)*4*(8+x)

                             = 2(8+x)

                             = 16 + 2x

Now Area of the ΔABC = Area of ΔOBC + Area of ΔOBC + Area of ΔOAB

=> 4√[3x(14+x)] = 28 + 12 + 2x + 16 + 2x

=> 4√[3x(14+x)] = 56 + 4x

=> 4√[3x(14+x)] = 4(14 + x)

=> √[3x(14+x)] = 14 + x

On squaring bothe side, we get

3x(14 + x) = (14 + x)2

=> 3x = 14 + x                   (14 + x = 0 => x = -14 is not possible)

=> 3x - x = 14

=> 2x = 14

=> x = 14/2

=> x = 7

Hense 

AB = x + 8

=> AB = 7+8

=> AB = 15

AC = 6 + x

=> AC = 6 + 7

=> AC = 13

So value of AB is 15 cm and value of AC is 13 cm

Answer 2

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:

2s = AB + BC + CA

= x + 8 + 14 + 6 + x

= 28 + 2x

⇒ 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!!