1.

Let ABCD is a rectangle with AB=a & BC=b & circle is drawn passing through A & B and touching side CD. Another circle is drawn passing through B & C and touching side AD. Let r_(1) & r_(2) be the radii of these two circle respectively. Minimum value of (r_(1)+r_(2) equals

Answer»

`5/8 (a-b)`
`5/8 (a+b)`
`3/8 (a-b)`
`3/8 (a+b)`

Solution :`r_(1)=b-x_(1)=OP=OA`
`AP_(1)=a//2`
`r_(1)^(2)=x_(1)^(2)+(a/2)^(2)=(b-x_(1))^(2)`
`x_(1)^(2)+(a^(2))/4=b^(2)+x_(1)^(2)-2bx_(1)`
`x_(1)=(4b^(2)-a^(2))/(8b)`
`r_(1)=b-x_(1)=(4b^(2)+a^(2))/(8b)`
SIMILARLY `r_(2)=(4A^(2)+b^(2))/(8a)`

`r_(1)+r_(2)=(a^(3)+b^(3)+4ab(a+b))/(8AB)`
`implies((a+b)(a^(2)+3ab+b^(2)))/(8ab)`
`=((a+b)/8)([(a-b)^(2)+5ab])/(AB)`
But `(a-b)^(2)ge0`
`r_(1)+r_(2)GE((a+b))/8.(5ab)/(ab)`
`impliesr_(1)+r_(2)ge(5(a+b))/8`


Discussion

No Comment Found

Related InterviewSolutions