wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
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
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
ΔABC is equilateral
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Area of ABC is 27 sq. units
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Inradius of ABC is 4 units
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
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

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon