Question

The lengths of three unequal edges of a rectangular solid block are in G.P. . The volume of the block is $216c{m}^3$ and the total surface area is $252c{m}^2$. The length of the longest edge is

• A. 12 cm
• B. 6 cm
• C. 18 cm
• D. 3 cm

Answer 1

Answer: (A)

Let the edges of rectangular block are $a,ar$ and $a{r}^2$, respectively.
Now, volume $=216c{m}^3$
$\Rightarrow a(ar)\left(a{r}^2\right)-216$
$\left(\because \text{ volume of cuboid }-f\times b\times h\right)$....(i)
$\Rightarrow (ar{)}^3=(6{)}^3$
$\Rightarrow ar=6cm\text{ (taking cube root) ...(ii) }$
and total surface area$=252c{m}^2$
$2\left[a(ar)+ar\left(a{r}^2\right)+a\left(a{r}^2\right)\right]=252$
$\left[\because \text{ surface area of cuboid }=2(lb+bh+hl)<br/>\right]$
From Eq. (ii), we get
$2(6a+36r+36)=252$
$\Rightarrow 12(a+6r+6)=252$
$\Rightarrow a+6r=15$ (divide both sides by $12)...(iii)$
$\Rightarrow a+6\times \left(\frac{6}{a}\right)=15$[from Eq.(ii)]
$\Rightarrow a^2-15a+36=0$
$\Rightarrow (a-12)(a-3)=0$
$\Rightarrow a=3,12$
From Eq. (iii), we get
when $a=3$, then $3+6r=15\Rightarrow r=2$
and when $a=12$, then $12+6r=15\Rightarrow r=\frac{1}{2}$
Now, edges are $3,3\times 2,3\times (2{)}^2$ or $12,12\times \left(\frac{1}{2}\right),12\times {\left(\frac{1}{2}\right)}^2$
i.e., $3,6,12$ or $12,6,3$.
Hence, the length of the longest edge is $12cm$.