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:
Step 1: Use similarity in the triangle
Since $DE \parallel BC$, triangles $\triangle ADE$ and $\triangle ABC$ are similar. Hence the corresponding sides are proportional:
\[
\frac{DE}{BC}=\frac{AD}{AB}.
\]
Step 2: Compute $AB$ and substitute
Given $AD=3\,\text{cm}$ and $DB=1.5\,\text{cm}$, so
\[
AB=AD+DB=3+1.5=4.5\,\text{cm}.
\]
Also $DE=4\,\text{cm}$. Using the ratio:
\[
\frac{4}{BC}=\frac{3}{4.5}\Rightarrow BC=\frac{4\times 4.5}{3}=4\times 1.5=6\,\text{cm}.
\]
Step 3: Conclusion
Therefore, the length of $BC$ is $6\,\text{cm}$.
The correct answer is option (D).
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: