Question

In a right triangle $ABC$, right-angled at $B,BC=12cm$ and $AB=5cm$. The radius of the circle inscribed in the triangle (in cm) is

• A. $3$
• B. $4$
• C. $1$
• D. $2$

Answer 1

Answer: (D)

Given:

AB = 5 cm, BC = 12 cm

Using Pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2$

= $5^2+{12}^2$

= $25+144$

= $169$

$AC=13$.

We know that two tangents drawn to a circle from the same point that is exterior to the circle are of equal lengths.

So, $AM=AQ=a$

Similarly $MB=BP=b$ and $PC=CQ=c$

We know

$AB=a+b=5$

$BC=b+c=12$ and

$AC=a+c=13$

Solving simultaneously we get $a=3,b=2$ and $c=10$

We also know that the tangent is perpendicular to the radius

Thus $OMBP$ is a square with side $b$.

Hence the length of the radius of the circle inscribed in the right angled triangle is $2cm$.