Question

The following figure shows a closed victory stand whose dimensions are given in cm. If the bottom of the stand is open, find its volume and total surface area.

• A. $1,92,000c{m}^3$ and $15,000c{m}^2$
• B. $1,62,000c{m}^3$ and $15,000c{m}^2$
• C. $1,52,000c{m}^3$ and $15,000c{m}^2$
• D. $1,32,000c{m}^3$ and $15,000c{m}^2$

Answer 1

The correct option is D $1,32,000c{m}^3$ and $15,000c{m}^2$

Volume of the stand $=$ Volume of part $2$ of the stand with dimension $30,40,20$ $+$ Volume of part $1$ of stand with dimensions $30,40,60$ $+$ Volume of part $3$ of stand with dimensions $30,40,30$

Since each part is cuboidal, their volume $=length\times breadth\times height$

Hence, Total Volume of the stand

$=(30\times 40\times 20)+(30\times 40\times 60)+(30\times 40\times 30)$

$=(30\times 40)\times \{20+60+30\}$

$=132000{cm}^3$

Total Surface area $=$ Sum of surface areas of each top face

Total Surface area $=\left[\right(40\times 20)+2\times (30\times 20)+(40\times 30\left)\right]+\left[\right(40\times 40)+(30\times 40)+(30\times 40)+2\times (30\times 60\left)\right]+\left[\right(30\times 40)+(30\times 40)+2\times (30\times 30\left)\right]$

$=800+1200+1200+1600+1200+1200+3600+1200+1200+1800$

$=15000{cm}^2$