Question

In a triangle,$ABC,AD,BE$, and $CF$ are the altitudes, and $R$ is the circumradius then the radius of the circle $DEF$ is

• A. $2R$
• B. $R$
• C. $\frac{R}{2}$
• D. None of these

Answer 1

The correct option is C

$\frac{R}{2}$

Explanation for the correct option:

Finding the radius of the circle:

Considering in $∆DEF$

$$\begin{gathered} EF=a\cos A \\ DE=c\cos C \\ DF=b\cos B \end{gathered}$$

Let $R_1$is the circumradius of$∆DEF.$

$R_1=\frac{\left(aC\mathrm{os}A\right)\left(bC\mathrm{os}B\right)\left(cC\mathrm{os}C\right)}{4\times {\displaystyle \frac{1}{2}}DF\times DE\sin \angle EDF}\left[\because R=\frac{abc}{4∆}\right]$

$\angle EDF=180–2A$

So

$\begin{array}{rcl}{R}_1& =& \frac{(abc\cos A\cos B\cos C)}{2b\cos Bc\sin C\sin (180–2A)}\\ & =& \frac{a\cos A}{2\sin 2A}\\ & =& \frac{a\cos A}{4\sin A\cos A}\end{array}$

Putting$a=2R\sin A$

$$\begin{gathered} R_1=\frac{2R\sin A}{4\sin A} \\ =R/2 \end{gathered}$$

Hence, the correct option is C.