Step 1: Prime factorization
- $12 = 2^2 \times 3$
- $15 = 3 \times 5$
- $21 = 3 \times 7$
Step 2: Take highest powers of all primes
\[
LCM = 2^2 \times 3 \times 5 \times 7
\]
\[
LCM = 4 \times 3 \times 5 \times 7 = 420
\]
Step 3: Re-check options
The $LCM(12, 15, 21) = 420$.
So, 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:
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: