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Incircle of Delta ABC touches AB, BC, CA at R, P, Q, respectively. If (2)/(AR)+(5)/(BP)+(5)/(CQ)=(6)/(r ) and the perimeter of the triangle is the smallest integer, then answer the following questions : The inradius of incircle of Delta ABC is |
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Answer» 4  LET `tan.(A)/(2)= x, tan.(B)/(2)=y, tan.(C )/(2)=z` `THEREFORE x=(r )/(AR), y = (r )/(BP), z=(r )/(CQ)` Put the VALUES of AR, BP and CQ in given relation, we get `(2x)/(r )+(5y)/(r )+(5z)/(r )=(6)/(r )` `rarr 2x+5y+5z=6` .....(1) Also in the triangle, `xy + yz + zx = 1`.....(2) It we interchange y and z, then both equations (1) and (2) remain unchanged `rArr ABC` is isoceles with `angle B = angle C rArr y =z` So form (1) and (2), we get x = 3 - 5 y.......(3) and `2xy + y^(2)=1` ....(4) Solving we get, `x = (4)/(3), y = z = (1)/(3)` `therefore AR = (3r)/(4), BP= 3r, C = 3r` Now perimeter `2s=2.AR+2.BP+2.CQ=(27r)/(2)` Given perimeter is smallest integer `rarr r =2` `therefore s=(27)/(2)` `therefore Delta = rs = 2xx(27)/(2)=27` sq. unit. |
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