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Incircle of Delta ABC touches AB, BC, CA...

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

A

4

B

3

C

2

D

1

Text Solution

Verified by Experts

The correct Answer is:
C

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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