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

Step by Step Explanation:
  1. In ΔABC, we have B=90,AB=7 cm  and BC=24 cm
  2. 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, OMAB,ONBC,OPCA. A M B N C O P r r r
  3. 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=CA2CA=25 cm.
  4. 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=56rr=16856=3 cm. Hence, the radius of the circle is 3 cm.

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