In the given figure, ABCABC is a right-angled triangle with AB=7 cmAB=7 cm and AC=9 cmAC=9 cm. A circle with center OO has been inscribed inside the triangle. Calculate the value of rr, the radius of the inscribed circle.
Answer:
2.3 cm2.3 cm
- Let us join OO to A,B,A,B, and CC and draw OD⊥ABOD⊥AB, OE⊥BCOE⊥BC and OF⊥CAOF⊥CA.
We see that OD,OE,OD,OE, and OFOF are the radius of the circle with center OO.
⟹OD=OE=OF=r cm⟹OD=OE=OF=r cm
Also, △ABC△ABC is a right-angled triangle. ⟹ Area of △ABC=12×AB×AC=12×7 cm×9 cm=31.5 cm2 - Let us now find the area of △ABC in terms of r. Area of △ABC=Area of △OAB+Area of △OBC+Area of △OCA=12×AB×OD+12×BC×OE+12×CA×OF=12×AB×r+12×BC×r+12×CA×r=12(AB+BC+CA)×r=12(Perimeter of △ABC)×r
- Comparing the area of △ABC obtained in step 1 and step 2, we have Area of △ABC=31.5 cm2=12×(AB+BC+CA)×r⟹31.5 cm2=12×(7+BC+9)×r…(i)
- Applying Pythagoras theorem in △ABC, we have BC2=AB2+AC2⟹BC=√(7)2+(9)2=11.4 cm
- Now, substituting the value of BC in eq (i), we have 31.5=12×(7+11.4+9)×r⟹31.5×2=27.4×r⟹r=31.5×227.4=2.3 cm
- Hence, the radius of the inscribed circle is 2.3 cm.