Question

8. In the given figure, if B = 90°, BC = 15 cm and AC = 17 cm, then
find the radius of the circle with centre O.
​

Answer 1

Answer:

Answer:

Let ABC be the triangle with AB =8 cm, BC = 15 cm and AC = 17cm. Consider, 82 + 152 = 64 + 225 = 289 That is 82 + 152 = 172 Hence ABC is a right angled triangle Therefore area of ΔABC = (1/2) × 8 × 15 = 60 sq cm Let r be the radius of the incircle whose centre is I. Now consider, ar(ΔAIB + ΔBIC + ΔCIA) = [(1/2) × AB × r] + [(1/2) × BC × r] + [(1/2) ×AC × r] ( 1/2 ) × r × ( 8 + 15 + 17 ) = ( 1/2 ) × r × (40) = 20r But, ar(ΔAIB + ΔBIC + ΔCIA) = ar(ΔABC )= 60 sq cm Hence 20r = 60 r = 3 cm

Answer 2

Answer:

Answer:

Let ABC be the triangle with AB =8 cm, BC = 15 cm and AC = 17cm. Consider, 82 + 152 = 64 + 225 = 289 That is 82 + 152 = 172 Hence ABC is a right angled triangle Therefore area of ΔABC = (1/2) × 8 × 15 = 60 sq cm Let r be the radius of the incircle whose centre is I. Now consider, ar(ΔAIB + ΔBIC + ΔCIA) = [(1/2) × AB × r] + [(1/2) × BC × r] + [(1/2) ×AC × r] ( 1/2 ) × r × ( 8 + 15 + 17 ) = ( 1/2 ) × r × (40) = 20r But, ar(ΔAIB + ΔBIC + ΔCIA) = ar(ΔABC )= 60 sq cm Hence 20r = 60 r = 3 cm

Answer 3

Answer:

Answer:

Let ABC be the triangle with AB =8 cm, BC = 15 cm and AC = 17cm. Consider, 82 + 152 = 64 + 225 = 289 That is 82 + 152 = 172 Hence ABC is a right angled triangle Therefore area of ΔABC = (1/2) × 8 × 15 = 60 sq cm Let r be the radius of the incircle whose centre is I. Now consider, ar(ΔAIB + ΔBIC + ΔCIA) = [(1/2) × AB × r] + [(1/2) × BC × r] + [(1/2) ×AC × r] ( 1/2 ) × r × ( 8 + 15 + 17 ) = ( 1/2 ) × r × (40) = 20r But, ar(ΔAIB + ΔBIC + ΔCIA) = ar(ΔABC )= 60 sq cm Hence 20r = 60 r = 3 cm