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
ΔABCis isosceles
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
ΔABCis equilateral
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Area of △ABCis 27 sq. units
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Inradius of △ABCis 4 units
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is DInradius of △ABCis 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=6r⇒2x+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=∠C⇒y=z
Now, both the equation reduce to x=3–5y2xy+y2=1
Solving both, we get 9y2−6y+1=0⇒y=z=13⇒x=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=108sq. unit