Question
A cube whose volume is $\frac18$ cubic centimeter is placed on a cube whose volume is $1\ cm^3$. The two cubes are then placed on top a third cube whose volume is $8\ cm^3$. The heigght of the stacked cubes is:
✓
Answer
Let,
$V_1, V_2, V_3$
$\rightarrow$ Volume of the three cubes
$a_1, a_2, a_3$
$\rightarrow$ Side of the three cubes
We know that,
$a^3 = V$
$So,$
$\text{a}^3_1=\text{V}$
$\text{a}^3_1=\frac18$
$\text{a}_1=\frac12\text{ cm}$
Similarly,
$\text{a}^3_2=1$
$\text{a}_2=1\text{ cm}$
And;
$\text{a}^3_3=8$
$\text{a}_2=2\text{ cm}$
So the height of the resulting structure,
$=\frac12+1+2$
$= 0.5 +1 + 2$
$= 3.5\ cm$
The height of the structure is $3.5\ cm.$
$V_1, V_2, V_3$
$\rightarrow$ Volume of the three cubes
$a_1, a_2, a_3$
$\rightarrow$ Side of the three cubes
We know that,
$a^3 = V$
$So,$
$\text{a}^3_1=\text{V}$
$\text{a}^3_1=\frac18$
$\text{a}_1=\frac12\text{ cm}$
Similarly,
$\text{a}^3_2=1$
$\text{a}_2=1\text{ cm}$
And;
$\text{a}^3_3=8$
$\text{a}_2=2\text{ cm}$
So the height of the resulting structure,
$=\frac12+1+2$
$= 0.5 +1 + 2$
$= 3.5\ cm$
The height of the structure is $3.5\ cm.$
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