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
[tex]s= \frac{PQ+QR+PR}{2}= \frac{12+x+21+9+x}{2}= \frac{42+2x}{2}=21+x [/tex]
Given that area = 189cm²
[tex]area=sr\\ \Rightarrow 189=s \times6\\ \Rightarrow s= \frac{189}{6}=31.5cm [/tex]
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
