Question

PLS SOLVE IT.....
A triangle ABC is drawn to circumscribe a circle of radius 4cm such that the segments BD and DC into which BC is divided by the point of contact D are of the lengths 8cm and 6cm respectively .find the sides AB and AC.

Answer 1

Area of triangle = ½ x base x height 

Therefore, Area of ΔOBC = ½ x OD x BC = ½ x 4 x 14 = 28 

Area of ΔOCA = ½ x OF x AC = ½ x 4 x (6 + x) = 12 + 2x 

Area of ΔOAB = ½ x OE x AB = ½ x 4 x (8+x) = 16 + 2x 

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

4 underoot 3(14x+x square)  = 28 + 12 + 2x + 16 + 2x 

4 underoot 3(14x+ x square)  = 56 + 4x 

= underoot 3 (14x + x square )  = 14 + x  

On squaring both sides, we get  

Either x +14 = 0 or x –7  = 0 

Therefore, x = -14 and 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 = 15cm 

CA = 6 + x = 6 + 7 = 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 Δ 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 have all the sides of a triangle, so its area can be found 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
Area of Δ ABC = under root s ( s - a ) (s - b) (s - c )
= under root { 14 + x} { (14 + x) - 14} { ( 14 + x ) - ( 6 + x) } { (14 + x) - (8 + x) }  
= under root (14 + x) (x) (8) (6)
= 4 under root 3 (14 x + x2)
Again, area of triangle is also equal to

1/2 * base * height

Therefore,
Area of ΔOBC = 

1/2 * OD * BC 

= 1/2 * 4 * 14 

=28
Area of ΔOCA =
1/2 * OF * AC 

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

= 12 + 2x


Area of ΔOAB = 

1/2 * OE * AB

1/2 * 4 * (8 + x )

= 16 + 2x

Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB
4 under root 3 (14x + x2) = 28 + 12 + 2x  + 16 +2x
 4 (under root 3 (14x + x2)) = 56 + 4x
 under root 3(14x+x2) = 14+ x On squaring both sides, we get3(14x + x2) = (14 + x )2
42x + 3x2 = 196 + x2 + 28x
2x2 = 14x - 196 = 0
x2 + 7 - 98 = 0
x2 + 14x - 7x - 98=0
x(x+14)-7(x+14)=0
x+14(x-7)=0
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