\[\Delta PQR\] is a right angles triangle with \[PQ=3cm\] and \[QR=4cm\] . A circle which touches all sides of the triangle is inscribed in the triangle. Calculate the radius of the circle.

Hint: We solve this problem by using a simple formula of properties of triangle. If \['r'\] is the radius of circle inscribed in a triangle that touches all the sides and \['\Delta '\] is the area of the triangle then \[r=\dfrac{\Delta }{s}\] , where \['s'\] is the semi perimeter of the triangle given as \[s=\dfrac{\text{sum of sides}}{2}\] . By using this property we can calculate the radius of Incircle.

Complete step by step answer:
We are given that \[PQ=3cm\] and \[QR=4cm\] .
Let us find the third side of the triangle \[\Delta PQR\] .
We know that the Pythagoras theorem states that the square of hypotenuse is equal to sum of squares of other two sides that is for the triangle shown below

The Pythagoras theorem is given as \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}\] .
By using this theorem to \[\Delta PQR\] , we can write
 \[\Rightarrow P{{R}^{2}}=P{{Q}^{2}}+Q{{R}^{2}}\]
By substituting the required values we get
 \[\begin{align}
  & \Rightarrow P{{R}^{2}}={{3}^{2}}+{{4}^{2}} \\
 & \Rightarrow P{{R}^{2}}=9+16 \\
 & \Rightarrow PR=\sqrt{25}=5cm \\
\end{align}\]
Now, we know that if \['r'\] is the radius of circle inscribed in a triangle that touches all the sides and \['\Delta '\] is the area of the triangle then \[r=\dfrac{\Delta }{s}\] , where \['s'\] is the semi perimeter of the triangle given as \[s=\dfrac{\text{sum of sides}}{2}\] .

Now, let us calculate the semi perimeter of the triangle \[\Delta PQR\] as
 \[\begin{align}
  & \Rightarrow s=\dfrac{PQ+QR+RP}{2} \\
 & \Rightarrow s=\dfrac{3+4+5}{2} \\
 & \Rightarrow s=6cm \\
\end{align}\]
Now, let us calculate the area of \[\Delta PQR\] as
 \[\begin{align}
  & \Rightarrow \Delta =\dfrac{1}{2}\left( base \right)\left( height \right) \\
 & \Rightarrow \Delta =\dfrac{1}{2}\left( QR \right)\left( PQ \right) \\
 & \Rightarrow \Delta =\dfrac{1}{2}\times 3\times 4 \\
 & \Rightarrow \Delta =6c{{m}^{2}} \\
\end{align}\]
Now, by using the properties of triangles let us calculate the radius of incircle as follows
 \[\Rightarrow r=\dfrac{\Delta }{s}\]
By substituting the required values we get
 \[\begin{align}
  & \Rightarrow r=\dfrac{6}{6} \\
 & \Rightarrow r=1cm \\
\end{align}\]

Therefore, the radius of the circle inscribed in a triangle \[\Delta PQR\] which touches all the sides is 1cm.

Note: Students may make mistakes while taking the formula of semi perimeter. The formula of semi perimeter is given as \[s=\dfrac{\text{sum of sides}}{2}\] , but due to some confusion they take the formula as \[s=\dfrac{\text{sum of sides}}{3}\] because of three sides. This will result in a wrong answer. This is the only point where one can make mistakes. The remaining calculations part has to be done with care.