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Question

The incircle of ABC having integral side lengths and least perimeter, touches AB,BC,CA at R,P,Q respectively. If 2AR+5BP+5CQ=6r, where r is the radius of the incircle, then

A
ΔABC is isosceles
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B
ΔABC is equilateral
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C
Area of ABC is 27 sq. units
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D
Inradius of ABC is 4 units
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Solution

The correct option is D Inradius of ABC is 4 units
Let tanA2=x,tanB2=y,tanC2=z


x=rAR,y=rBP,z=rCQ
Putting the value of AR,BP and CQ in given relation, we get
2xr+5yr+5zr=6r2x+5y+5z=6 (1)

Using tanA2tanB2=1, we get
xy+yz+zx=1 (2)
If we interchange y and z in both equations (1) and (2), then they remain unchanged.
ABC is isosceles with B=Cy=z

Now, both the equation reduce to
x=35y2xy+y2=1
Solving both, we get
9y26y+1=0y=z=13x=43
Therefore,
AR=3r4, BP=3r, CQ=3r
Now, AB=AR+BR
AB=3r4+3r
AB=15r4=AC, BC=6r
Now, the perimeter of the triangle
=2(AR+BP+CQ)=27r2
For perimeter to be smallest integer and sides also integer, r must be 4.
=rs=4×27r4=108 sq. unit

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