Construct a tangent to a circle of radius $4 \, cm$ from a point on the concentric circle of radius $6 \, cm$ and measure its length. Also,provide the justification for the construction and verify the measurement by actual calculation.

(N/A) Tangents on the given circle can be drawn as follows:
$1.$ Draw a circle of $4 \, cm$ radius with centre as $O$ on the given plane.
$2.$ Draw a circle of $6 \, cm$ radius taking $O$ as its centre. Locate a point $P$ on this circle and join $OP.$
$3.$ Bisect $OP.$ Let $M$ be the mid-point of $PO.$
$4.$ Taking $M$ as its centre and $MO$ as its radius,draw a circle. Let it intersect the given circle at the points $Q$ and $R$.
$5.$ Join $PQ$ and $PR.$ $PQ$ and $PR$ are the required tangents.
It can be observed that $PQ$ and $PR$ are of length $4.47 \, cm$ each.
In $\triangle PQO$:
Since $PQ$ is a tangent,$\angle PQO = 90^{\circ}$.
$PO = 6 \, cm$ (hypotenuse),
$QO = 4 \, cm$ (radius).
Applying the Pythagoras theorem in $\triangle PQO$,we obtain:
$PQ^2 + QO^2 = PO^2$
$PQ^2 + (4)^2 = (6)^2$
$PQ^2 + 16 = 36$
$PQ^2 = 36 - 16$
$PQ^2 = 20$
$PQ = \sqrt{20} = 2\sqrt{5} \approx 4.47 \, cm$.
Justification:
The construction can be justified by proving that $PQ$ and $PR$ are the tangents to the circle (whose centre is $O$ and radius is $4 \, cm$). For this,join $OQ$ and $OR$.
Since $\angle OQP$ is an angle in a semi-circle,$\angle OQP = 90^{\circ}$.
$\Rightarrow OQ \perp PQ$.
Since $OQ$ is the radius of the circle,$PQ$ must be a tangent to the circle. Similarly,$PR$ is a tangent to the circle.