Given a right-angled ΔABC. The lengths of the sides containing the right angle are 7 cm and 24 cm. A circle is inscribed in ΔABC. Find the radius of the circle.
Answer:
3 cm
- In ΔABC, we have ∠B=90∘,AB=7 cm and BC=24 cm
- A circle is inscribed in ΔABC. Let O be its centre and M, N and P be the points where it touches the sides AB, BC and CA respectively. Then, OM⊥AB,ON⊥BC,OP⊥CA.
- Let r cm be the radius of the circle.
Then, OM=ON=OP=r cm.
Now, AB2+BC2=CA2 [ By pythagoras' theorem ]
⟹(7)2cm2+(24)2cm2=CA2⟹CA=25 cm. - Now, ar(ΔABC)=ar(ΔAOB)+ar(ΔBOC)+ar(ΔCOA)⟹12×AB×BC=(12×AB×OM)+(12×BC×ON)+(12×CA×OP)⟹12×7×24=(12×7×r)+(12×24×r)+(12×25×r)⟹168=7r+24r+25r=56r⟹r=16856=3 cm. Hence, the radius of the circle is 3 cm.