Question

A triangle PQR is drawn to circumscribe a circle of radius 6cm and QR is divided by the point of contact T. QT and TR are 12cm and 9cm respectively. If the area of Triangle PQR = 189 sq. cm, then find the lengths of sides PQ and PR.
Please help me with this. Its very urgent. Thank you :-)

Answer 1

Since the triangle is circumscribing the circle, the circle is the incircle of the triangle. All the sides of the triangle are tangents to the circle.

Given RT= 9cm and QT = 12cm
Since RT and RU are tangents to a circle from a point R, they are equal
Thus RT = RU = 9cm
similarly, QT = QS = 12cm
Let PS = PU = x cm
radius of circle, r = 6cm

sides of triangle are:
PQ = (12+x) cm
QR = 12+9 = 21cm
PR = (9+x) cm 

$s= \frac{PQ+QR+PR}{2}= \frac{12+x+21+9+x}{2}= \frac{42+2x}{2}=21+x$

Given that area = 189cm²

$area=sr\\ \Rightarrow 189=s \times6\\ \Rightarrow s= \frac{189}{6}=31.5cm$

thus 21+x = 31.5
⇒ x = 31.5 - 21
⇒ x = 10.5

Sides of triangle are:
PQ = (12+x) cm = 12+10.5 = 22.5cm
QR = 12+9 = 21cm
PR = (9+x) cm = 9+10.5 = 19.5cm

Answer 2

Answer: let,

a=12+x

b=12+9=21cm

c=9+x

According to heron's formula

S=a+b+c/2

=12+x+12+9+9+x/2

=2x+42/2

=2(x+21)/2

=x+21

Area given=189 cm^2

Area=sr

189=x+21*6

189/6=x+21

31.5=x+21

X=31.5-21

X=10.5

QR=12+9=21cm

PQ=12+x

=12+10.5=22.5 cm

PR=x+9

=10.5+9=19.5 cm