The surface area of a cube is $18\dfrac{3}{8}{{m}^{2}}$. Find its volume.
Answer
571.5k+ views
Hint: In this problem we have to find the volume of the cube with the given surface area which is$18\dfrac{3}{8}{{m}^{2}}$. It is in mixed fraction, so we will convert it into normal fraction form. Now we will assume the side of the given cube as $a$ meters. We have the formula for the surface area of the cube which is having side length $a$ as $6{{a}^{2}}$. Now we will equate both the values and simplify the equation, then we will get the side of the cube which is $a$. From the value of side length of the cube we can simply find the volume by using the formula $v={{a}^{3}}$.
Formula use:
1.Surface area of the cube is $6{{a}^{2}}$.
2.Volume of cube is ${{a}^{3}}$.
Complete step by step answer:
Given that, the surface area of the cube is $18\dfrac{3}{8}{{m}^{2}}$.
We can observe that the above value is a mixed fraction, converting it into normal fraction, then we will have
$\begin{align}
& s=18\dfrac{3}{8}{{m}^{2}} \\
& \Rightarrow s=\dfrac{18\times 8+3}{8} \\
& \Rightarrow s=\dfrac{147}{8} \\
\end{align}$
Let us assume the side length of the cube as $a$.
We know that the surface area of the cube which is having side length $a$ is $6{{a}^{2}}$, equating the both the values, then we will get
$6{{a}^{2}}=\dfrac{147}{8}$
Dividing the above equation with $6$ on both sides, then we will have
$\begin{align}
& \Rightarrow \dfrac{6{{a}^{2}}}{6}=\dfrac{147}{6\times 8} \\
& \Rightarrow {{a}^{2}}=\dfrac{147}{48} \\
\end{align}$
Cancelling the common factor $3$ in both numerator and denominator, then we will get
\[\Rightarrow {{a}^{2}}=\dfrac{49}{16}\]
Taking square root on both sides of the above equation, then we will get
$\therefore a=\dfrac{7}{4}$
Now the volume of the cube which is having the side length $a$ is ${{a}^{3}}$, hence
$\begin{align}
& v={{\left( \dfrac{7}{4} \right)}^{3}} \\
& \Rightarrow v=\left( \dfrac{343}{64} \right){{m}^{3}} \\
\end{align}$
Note:
In this problem we have the three-dimensional shape cube. If they have mentioned the shape as a sphere, then we will use the surface area formula $s=4\pi {{r}^{2}}$. From this we will calculate the value of radius $r$ and use the volume formula $v=\dfrac{4}{3}\pi {{r}^{3}}$.
Formula use:
1.Surface area of the cube is $6{{a}^{2}}$.
2.Volume of cube is ${{a}^{3}}$.
Complete step by step answer:
Given that, the surface area of the cube is $18\dfrac{3}{8}{{m}^{2}}$.
We can observe that the above value is a mixed fraction, converting it into normal fraction, then we will have
$\begin{align}
& s=18\dfrac{3}{8}{{m}^{2}} \\
& \Rightarrow s=\dfrac{18\times 8+3}{8} \\
& \Rightarrow s=\dfrac{147}{8} \\
\end{align}$
Let us assume the side length of the cube as $a$.
We know that the surface area of the cube which is having side length $a$ is $6{{a}^{2}}$, equating the both the values, then we will get
$6{{a}^{2}}=\dfrac{147}{8}$
Dividing the above equation with $6$ on both sides, then we will have
$\begin{align}
& \Rightarrow \dfrac{6{{a}^{2}}}{6}=\dfrac{147}{6\times 8} \\
& \Rightarrow {{a}^{2}}=\dfrac{147}{48} \\
\end{align}$
Cancelling the common factor $3$ in both numerator and denominator, then we will get
\[\Rightarrow {{a}^{2}}=\dfrac{49}{16}\]
Taking square root on both sides of the above equation, then we will get
$\therefore a=\dfrac{7}{4}$
Now the volume of the cube which is having the side length $a$ is ${{a}^{3}}$, hence
$\begin{align}
& v={{\left( \dfrac{7}{4} \right)}^{3}} \\
& \Rightarrow v=\left( \dfrac{343}{64} \right){{m}^{3}} \\
\end{align}$
Note:
In this problem we have the three-dimensional shape cube. If they have mentioned the shape as a sphere, then we will use the surface area formula $s=4\pi {{r}^{2}}$. From this we will calculate the value of radius $r$ and use the volume formula $v=\dfrac{4}{3}\pi {{r}^{3}}$.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

In cricket, what is the term for a bowler taking five wickets in an innings?

Who Won 36 Oscar Awards? Record Holder Revealed

What is the name of Japan Parliament?

What is the median of the first 10 natural numbers class 10 maths CBSE

Why is it 530 pm in india when it is 1200 afternoon class 10 social science CBSE

