Solution:
Let the length, breadth and height of rectangular solid block be$\frac{\text{a}}{\text{r}},$ a and ar, respectively.
$\therefore\ \text{Volume}=\frac{\text{a}}{\text{r}}\times\text{a}\times\text{ar}=216\text{cm}^3$
$\Rightarrow\text{a}^3=216=6^3\Rightarrow\text{a}=6$
Also, Surface area $=2\Big(\frac{\text{a}}{\text{r}}.\text{a}+\text{a}.\text{ar}+\frac{\text{a}}{\text{r}}.\text{ar}\Big)=252$
$\Rightarrow2\text{a}^2\Big(\frac{1}{\text{r}}+\text{r}+1\Big)=252$
$\Rightarrow2\times36\Big(\frac{1+\text{r}^2+\text{r}}{\text{r}}\Big)=252$
$\Rightarrow2(1+\text{r}^2+\text{r})=7\text{r}$
$\Rightarrow2\text{r}^2-5\text{r}+2=0$
$\Rightarrow(2\text{r}-1)(\text{r}-2)=0$
$\therefore\ \text{r}=\frac{1}{2},2$
For $\text{r}=\frac{1}{2}:$ Length $=\frac{\text{a}}{\text{r}}=\frac{6\times2}{1}=12,$ Breadth = a = 6
Height $=\text{ar}=6\times\frac{1}{2}=3$
For r = 2: Length $=\frac{\text{a}}{\text{r}}=\frac{6}{2}=3,$ Breadth = a = 6
Height = ar = 6 × 2 = 12
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Angle made by line with measured anticlockwise is called inclination of the line:
The middle term in the expansion of $\Big(\frac{2\text{x}}{3}=\frac{3}{2\text{x}^{2}}\Big)^{2\text{n}}$ is:
${^\text{2n}}\text{C}_{\text{n}}$
$(-1)\ {^\text{2n}}\text{C}_{\text{n}}\ \text{x}^{-\text{n}}$
${^\text{2n}}\text{C}_{\text{n}}\ \text{x}^{-\text{n}}$
None of these.