Question

If $R\ge r>0$ and $d>0,$ then $0<\frac{d^2+R^2-r^2}{2dR}\le 1$

• A. is satisfied if $|d-R|\le r$
• B. is satisfied if $|d-R|\le 2r$
• C. is satisfied if $|d-R|\ge r$
• D. is not satisfied at all

Answer 1

Answer: (A)

Given $R\ge r>0$ and $d>0$
$\Rightarrow 0<\frac{d^2+R^2-r^2}{2dR}\le 1$
$\Rightarrow 0<\left(d+R-r\right)\left(d+R+r\right)\le 2dR;$ which is true iff
$\left(d^2+R^2-r^2\right)\le 2dR,$ which is true iff $d^2+R^2-2dR\le r^2$
$\Rightarrow {\left(d-R\right)}^2\le r^2$
$\left|d-R\right|\le r,$ which is also $-r\le \left(d-R\right)\le r$
$\left|d-R\right|\le ,$ which is also $-r\le \left(d-R\right)\le r$