Question

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

A) \[1,72,000c{m^3}\; \text{and}\; 15,000c{m^2}\]
B) \[1,42,000c{m^3}\; \text{and}\; 15,000c{m^2}\]
C) \[1,62,000c{m^3}\; \text{and}\; 15,000c{m^2}\]
D) \[1,32,000c{m^3}\; \text{and}\; 15,000c{m^2}\]

Answer 1

Hint:
We’ll break down the whole structure into cube and cuboid because we know their formulas for volume and surface area. After calculating surface area and volume for these segments we’ll simply aff them to get the final answer.

Complete step by step solution:
Formula used:
1. Volume of cuboid=\[length \times breadth \times height = l \times b \times h\]
2. Area of rectangle=\[length \times breadth = l \times b\]

To find volume:
For victory stand number 3:
\[l = 40,b = 30,b = 20\]
\[ \Rightarrow V\left( {stand3} \right) = l \times b \times h = 40 \times 30 \times 20 = 24000c{m^3}\]

For victory stand number 1:
\[
  l = 40,b = 30,b = 60 \\
   \Rightarrow V\left( {stand1} \right) = l \times b \times h = 40 \times 30 \times 60 = 72000c{m^3} \\
\]

For victory stand number 2:
\[
  l = 40,b = 30,b = 30 \\
   \Rightarrow V\left( {stand2} \right) = l \times b \times h = 40 \times 30 \times 30 = 36000c{m^3} \\
\]

Total volume of victory stand
\[
   = V\left( {stand3} \right) + V\left( {stand1} \right) + V\left( {stand2} \right) \\
   \Rightarrow 24000 + 72000 + 36000 \\
   \Rightarrow 1,32,000c{m^3} \\
\]

To find total surface area:
For this, we will split that each cuboid into 6 rectangles.
We will not consider the bottom of the stand because it is open.

For victory stand number 3:
Total surface area \[ = \]front face \[ + \]back face \[ + \]side face \[ + \]top face
Since front and back are same,
\[
   \Rightarrow 2(30 \times 20) + 40 \times 30 + 40 \times 20 \\
   \Rightarrow 1200 + 1200 + 800 \\
   \Rightarrow 3200c{m^2} \\
\]

For victory stand number 2:
Total surface area \[ = \]front face \[ + \]back face \[ + \]side face \[ + \]top face
Since front and back are same,
\[
   \Rightarrow 2\left( {30 \times 30} \right) + 40 \times 30 + 40 \times 30 \\
   \Rightarrow 1800 + 1200 + 1200 \\
   \Rightarrow 4200c{m^2} \\
\]

For victory stand number 1:
For let face of this stand 1 length is \[60 - 20 = 40\]
For right face of stand 1length is \[60 - 30 = 30\]
Total surface area \[ = \]front face \[ + \]back face \[ + \]left face \[ + \]right face \[ + \] top face
Since front and back are same,
\[
   \Rightarrow 2\left( {30 \times 60} \right) + 40 \times 40 + 40 \times 30 + 40 \times 30 \\
   \Rightarrow 3600 + 1600 + 2400 \\
   \Rightarrow 7600c{m^2} \\
\]
Total surface area:
area stand 1 \[ + \] area stand 2 \[ + \]area stand 3
\[
   \Rightarrow 7600 + 4200 + 3200 \\
   \Rightarrow 15000c{m^2} \\
\]
Total volume is \[ \Rightarrow 1,32,000c{m^3}\]
Total surface area \[ \Rightarrow 15000c{m^2}\]

So, the correct option is D.

Note:
When we are finding the surface area we need to be cautious about the surfaces. We need to count all the surfaces and make sure that we are not taking lower surface into consideration. Do not consider the bottom of the stand. So, in a nutshell, we have to find areas with visible parts only.