Question

In a right angled triangle ABC, right-angled at B, \[BC=12\,cm\] and \[AB=5\,cm\]. The radius of the circle inscribed in the triangle (in cm) is
(a) 4
(b) 3
(c) 2
(d) 1

Answer 1

Hint: First we will draw the figure from the given details in the question and then using Pythagoras theorem we will find the length of side AC. Then we will calculate the area of triangle ABC and then we will equate it with the summation of areas of small triangles inside the figure.

Complete step-by-step solution -
Drawing the figure from the given details in the question,

As the triangle is right angled triangle, so from Pythagoras theorem, \[\text{AC=}\sqrt{\text{A}{{\text{B}}^{\text{2}}}\text{+B}{{\text{C}}^{\text{2}}}}........(1)\]. Now substituting the values of AB and BC in equation (1) we get,
\[\text{AC=}\sqrt{{{5}^{\text{2}}}\text{+1}{{\text{2}}^{\text{2}}}}=\sqrt{25+144}=\sqrt{169}=13\,cm\]
Now let the radius of the circle inscribed in the triangle be x.
So from the figure we can see that the area of triangle ABC(A) \[=\dfrac{1}{2}\times \text{AB }\!\!\times\!\!\text{ BC}......(2)\] (
Putting the values of AB and BC in equation (2) we get,
\[\text{A=}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ 5}\times 12.=30\,c{{m}^{2}}.....(3)\]
Now the area of ADB \[=\dfrac{1}{2}\times ED\times AB......(4)\]
Now putting the value of ED and AB in equation (4) we get,
\[\Rightarrow \dfrac{1}{2}\times x\times 5=\dfrac{5x}{2}......(5)\]
Now the area of BDC \[=\dfrac{1}{2}\times DF\times BC......(6)\]
Now putting the value of DF and BC in equation (6) we get,
\[\Rightarrow \dfrac{1}{2}\times x\times 12=6x......(7)\]
Now the area of ADC \[=\dfrac{1}{2}\times DG\times AC......(8)\]
Now putting the value of DG and AC in equation (8) we get,
\[\Rightarrow \dfrac{1}{2}\times x\times 13=\dfrac{13x}{2}......(9)\]
From the figure we know that the summation of area of ADB, area of BDC and area of ADC is equal to area of ABC. So using this information we equate equation (3) with the summation of equation (5), equation (7) and equation (9) we get.
\[\Rightarrow 30\,=\dfrac{5x}{2}+6x+\dfrac{13x}{2}.....(10)\]
Now taking the LCM and rearranging in equation (10) we get,
\[\begin{align}
  & \Rightarrow 30\,=\dfrac{5x+12x+13x}{2} \\
 & \Rightarrow \dfrac{30x}{2}=30..........(11) \\
\end{align}\]
Now solving for x in equation (11) we get,
\[\Rightarrow x=2\,cm\]
Hence the correct answer is option (c).

Note: Here we have to remember the formula of Pythagoras theorem and formula of area of triangle. Area of triangle \[=\dfrac{1}{2}\times base\times height\]. We in a hurry can make a mistake in solving equation (10) so we need to be careful while doing this step.