A triangle ABC is drawn to circumscribe a circle such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 4 cm and 20 cm respectively. If the area of the triangle is 120 cm2, what is the value of AB + AC?
36 cm
Let P,Q,R be the points where sides AB,BC and AC meet the circle respectively.
The tangents drawn from an external point to the circle are equal in length.
So, BP = BQ = 4 cm
Also, CR = CQ = 20 cm
Let AP = AR = x cm
Then,
AB = 4+x
AC = 20+x
BC = 24
Area of the traingle
=√s(s−a)(s−b)(s−c)
Here ,
s=a+b+c2 = (4+x)+(20+x)+242=24+x
Area =
√x(24+x)(20)(4)=120
Squaring on both sides, we get,
x2+24x−180=0
(x−6)(x+30)=0
x=6
(As x = - 30 cannot be length od a side)
So, the sides are
BC = 24 cm
AB = 4+x = 10 cm
AC = 20+x = 26 cm
Therefore , AB + AC = 36 cm