Question

$2$ cubes each of volume $64$ $c{m}^3$ are joined end to end. Find the total surface area of the resulting cuboid.

• A. $160c{m}^2$
• B. $180c{m}^2$
• C. $260c{m}^2$
• D. $190c{m}^2$

Answer 1

The correct option is A $160c{m}^2$

given cube

$a=b=c=l$

Volume $=a\times b\times c=l^3=64c{m}^3$

$l=4cm$

now

$a=2l,b=c=l$

T.S.A $=2\left[\right(length\times breadth)+(breadth\times height)+(height\times length\left)\right]$

$=2\times \left(\right(a\times b)+(b\times c)+(a\times c\left)\right)$

$=2\times (2l^2+l^2+2l^2)$

$=10l^2-10\times (4{)}^2-10\times 16=160c{m}^2$