Question

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 ${cm}^2$, what is the value of AB + AC?

• A. 48 cm
• B. 36 cm
• C. 24 cm
• D. 60 cm

Answer 1

The correct option is B

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
$=\sqrt{s(s-a)(s-b)(s-c)}$

Here ,
$s=\frac{a+b+c}{2}$ = $\frac{(4+x)+(20+x)+24}{2}=24+x$

Area =
$\sqrt{x(24+x)\left(20\right)\left(4\right)}=120$

Squaring on both sides, we get,

$x^2+24x-180=0$

$(x-6)(x+30)=0$

$x=6$
(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