Question

In $\Delta ABC,AB=3,BC=4,CA=5.$ Semi circles are drawn outwardly on AB, BC as diameters. PQ is a common tangent to the semi circles. The length of PQ is

• A. $\frac{5}{2}$
• B. $\sqrt{6}$
• C. $\sqrt{5}$
• D. $\frac{7}{2}$.

Answer 1

The correct option is B $\sqrt{6}$

Here,

$\overline{PD}=1.5$ and $\overline{PD}\perp \overline{PQ}$

($\because \overline{PD}$ is the radius of the circle with diameter $\overline{AB}$ where $\overline{AB}=3$ and radius is perpendicular to tangent)

Similarly,

$\overline{EQ}=2$ and $\overline{EQ}\perp \overline{PQ}$

$\therefore$ PQFD is a rectangle and $△DFE$ is a right angled triangle

ie, $\overline{PQ}=\overline{DF}$

and ${\overline{BD}}^2+{\overline{BE}}^2{=\overline{DE}}^2{=\overline{DF}}^2+{\overline{FE}}^2$

($\because △BDE$and$△DEF$ are right angles with same hypotenuse)

ie, ${\overline{DF}}^2+{\overline{FE}}^2={1.5}^2+2^2\to \left(1\right)$

But, $\overline{FE}=\overline{QE}-\overline{PD}$

($\because$ PQFE is a rectangle)

$\therefore \overline{FE}=2-1.5=0.5$

Substituting above equation in (1),

${\overline{DF}}^2={1.5}^2+2^2-{0.5}^2=6$

ie, $\overline{DF}=\sqrt{6}$

As PQFD is a rectangle,

$\overline{DF}=\overline{PQ}=\sqrt{6}$

So, Option B is correct