Question

The lengths of three unequal edges of a rectangular solid block are in G.P. . The volume of the block is $216\, cm ^3$ and the total surface area is $252 \,cm ^2$. 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 $a, a r$ and $a r^2$, respectively.
Now, volume $=216 \,cm ^3$
$\Rightarrow a(a r)\left(a r^2\right)-216$
$\left(\because \text { volume of cuboid } -f \times b \times h\right)$....(i)
$ \Rightarrow (a r)^3=(6)^3 $
$ \Rightarrow a r=6 cm \text { (taking cube root) ...(ii) }$
and total surface area$=252 cm ^2 $
$ 2\left[a(a r)+a r\left(a r^2\right)+a\left(a r^2\right)\right]=252$
$ \left[\because \text { surface area of cuboid } =2(l b+b h+h l)
\right] $
From Eq. (ii), we get
$ 2(6 a+36 r+36)=252$
$ \Rightarrow 12(a+6 r+6)=252 $
$\Rightarrow a+6 r=15 $ (divide both sides by $12)...(iii) $
$ \Rightarrow a+6 \times\left(\frac{6}{a}\right)=15 $[from Eq.(ii)]
$ \Rightarrow a^2-15 a+36=0 $
$ \Rightarrow (a-12)(a-3)=0 $
$ \Rightarrow a=3,12 $
From Eq. (iii), we get
when $a=3$, then $ 3+6 r=15 \Rightarrow r=2$
and when $a=12$, then $12+6 r=15 \Rightarrow r=\frac{1}{2}$
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 $12 cm$.