Question

ABC is a right angled triangle, right angled at B such that $BC=6cm$ and $AB=8cm$. A circle with centre O is inscribed in $\Delta ABC$. The radius of the circle is:

• A. $1cm$
• B. $2cm$
• C. $3cm$
• D. $4cm$

Answer 1

Answer: (B)

$2cm$

In $\Delta ABC$
$(AC{)}^2=(AB{)}^2+(BC{)}^2$
$(AC{)}^2=(8{)}^2+(6{)}^2$
$(PR{)}^2=100$
$PR=10$
Now, FOEB is a square
So So, $FB=x$
Then, $CF=6-x$
Similarly $BE=x$
Then $EA=8-x$
Also, $CD=CF$
[$\because$ Length of tangents drawn from external point are equal]
$=6-x$
And, $AD=AE=8-x$
[$\because$ Length of tangents drawn from external point are equal]
$AC=AD+DC$
$10=6-x+8-x$
$2x=4\Rightarrow x=2$ cm.