Step 1: Recall the distance formula
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Step 2: Substitute values
Points are $(x_1,y_1)=(2,3)$ and $(x_2,y_2)=(4,1)$.
\[
d = \sqrt{(4-2)^2 + (1-3)^2}
\]
\[
= \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8}
\]
Step 3: Simplify
\[
\sqrt{8} = 2\sqrt{2}
\]
\[
\boxed{d = 2\sqrt{2}}
\]
The product of $\sqrt{2}$ and $(2-\sqrt{2})$ will be:
If a tangent $PQ$ at a point $P$ of a circle of radius $5 \,\text{cm}$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12 \,\text{cm}$, then length of $PQ$ will be:
In the figure $DE \parallel BC$. If $AD = 3\,\text{cm}$, $DE = 4\,\text{cm}$ and $DB = 1.5\,\text{cm}$, then the measure of $BC$ will be: