Question

In $\Delta ABC;AD,BE$ and $CF$ are the altitudes and $R$ is the circumradius then the radius of the circle through $DEF$ is

• A. $2R$
• B. $R$
• C. $\frac{R}{2}$
• D. $\frac{3R}{2}$

Answer 1

The correct option is C $\frac{R}{2}$

In $\Delta DEF$
$EF=a\cos A,DE=b\cos B$ and $DF=c\cos C$
If circumradius of $\Delta DEF$ is $R_1$
Then $R_1=\frac{\left(a\cos A\right)\left(b\cos B\right)\left(c\cos C\right)}{4.\frac{1}{2}DF.DE\sin \left(\angle EDF\right)}$
Here $(\angle EDF={180}^0-2A)$
$=\frac{abc\cos A\cos B\cos C}{4.\frac{1}{2}.b\cos B.\cos C.\sin ({180}^0-2A)}$
$=\frac{a\cos A}{2\sin 2A}=\frac{2R\sin A\cos A}{4\sin A\cos A}\therefore R_1=\frac{R}{2}$