Question

The hypotenuse of a right angled triangle is $10cm$ and the radius of its inscribed circle is $1cm$. Therefore, perimeter of the triangle is

• A. $22cm$
• B. $24cm$
• C. $26cm$
• D. $30cm$

Answer 1

Answer: (A)

$22cm$

Let ABC is aright angled triangle in which $\angle B={90}^{\circ }$
Let I be the in centre and incircle touches the side AB,BC.AC at point D,E,F.
As we know that IB is the angle bisector
Hence $\angle IBD=\angle IBF=\frac{90}{2}={45}^{\circ }$
Now in $\Delta IBD,\angle BED={90}^{\circ }$
now in $\Delta IBD,\angle IBD={45}^{\circ }.\angle BDE={90}^{\circ }$
$\therefore \angle BID={180}^{\circ }-(\angle IBD+\angle BDE)$
$={180}^{\circ }-({45}^{\circ }+{90}^{\circ })={45}^{\circ }$
Hence $\angle BID=\angle IBD$
$\therefore BD=ID$
$\therefore BD=1cm$
similarly $BE=1cm$
Let$AD=xcm$
$AD=AF=1cm$
Since$CF=CE$
then using Pythagoras theorem in $\Delta ABC$
$=(AB{)}^2+(BC{)}^2=(AC{)}^2$
$=(x+1{)}^2+(11-x{)}^2={10}^2$
$=x^2+1+2x+121+x^2-22x=100$
$=2x^2-20x+22=0$
$=x^2-10x+11=0$
$x=\frac{-(-10\pm \sqrt{-{10}^2-4\times 1\times 11})}{2}=\frac{10\pm \sqrt{56}}{2}=\frac{10\pm 7.48}{2}=8.74or1.26$
If x=8.74 cm then $AB=x+1=8.74+1=9.74cm$
and $BC=11-x=11-8.74=2.26cm.$
Then $perimeter=9.74+2.26+10=22cm$