Question

the length of the hypotenuse of a right triangle is 18cm.its inradius is 3cm.find the perimeter of the triangle. find the area of the triangle

Answer 1

Let ABC be the right angled triangle in which ∠B=90°

Let I be incentre and the incircle touches the sides AB,BC,AC at points D,E,F respectively

As we know that IB is the angle bisector .Hence ∠IBD=∠IBE=90°2=45°Now in ∆IBD ,∠BDE =90° (since ID⊥AB )Now in △IBD ,∠IBD =45°,∠BDE=90°∴∠BID = 180°−(∠IBD +∠BDE)=180°−(45°+90°)=45°Hence ∠BID=∠IBD ∴BD=ID(since sides opposite to equal angles in a triangle are equal)∴BD=1 cmSimilarly BE = 1cmLet AD = x cmAD =AF =x cm (since tangents from exterior point to a circle are equal in length)Similarly CF=CE

Now using pythagoras theorem in ∆ABC we get

AB2+BC2=AC2⇒⎛⎝⎜x+1⎞⎠⎟2+⎛⎝⎜11−x⎞⎠⎟2=102⇒x2+1+2x+121+x2−22x=100⇒2x2−20x+22=0⇒x2−10x+11=0⇒x=−(−10)±(−10)2−4×1×11√2=10±100−44√2=10±56√2=10±7.482=8.74 or 1.26If x=8.74 then AB =x+1=8.74+1=9.74 cm and BC = 11−x=11−8.74 =2.26 cm and perimeter = 9.74 cm+2.26 cm+10 cm=22 cmIf x = 1.26 then AB = 1+1.26 = 2.26 cm and BC = 11−1.26=11−1.26 =9.74 cm perimeter = 2.26 cm+9.74 cm+10 cm=22 cmHence in both cases perimeter = 22 cm

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