
How do you convert $19.87\text{ k}{{\text{m}}^{2}}$ to ${{\text{m}}^{2}}$?
Answer
557.7k+ views
Hint: We will first see the conversion from km to m. Then we will square both sides of the equation that gives the conversion. Then we will obtain the conversion from $\text{k}{{\text{m}}^{2}}$ to ${{\text{m}}^{2}}$. Then we will use the unitary method to obtain the conversion from $19.87\text{ k}{{\text{m}}^{2}}$ to ${{\text{m}}^{2}}$. For this, we will multiply the number given by the conversion for $\text{1 k}{{\text{m}}^{2}}$ to ${{\text{m}}^{2}}$ and obtain the required solution.
Complete step-by-step solution:
We know that $1\text{ km}=1000\text{ m}$. This is the conversion from 1 km to m. We have to obtain the conversion from $\text{k}{{\text{m}}^{2}}$ to ${{\text{m}}^{2}}$. For this, we will square the equation $1\text{ km}=1000\text{ m}$. So, we have the following,
${{\left( 1\text{ km} \right)}^{2}}={{\left( 1000\text{ m} \right)}^{2}}$
Squaring both the sides, we obtain the following equation,
$\begin{align}
& {{1}^{2}}\text{ k}{{\text{m}}^{2}}=1000000\text{ }{{\text{m}}^{2}} \\
& \therefore 1\text{ k}{{\text{m}}^{2}}={{10}^{6}}\text{ }{{\text{m}}^{2}} \\
\end{align}$
The above equation gives us a relation between $\text{k}{{\text{m}}^{2}}$ and ${{\text{m}}^{2}}$. Now, we will see the definition of unitary method. The unitary method is a technique for solving problems by first finding the value of a single unit and then finding the necessary value by multiplying the single unit value. Now, we have obtained that
$1\text{ k}{{\text{m}}^{2}}={{10}^{6}}\text{ }{{\text{m}}^{2}}$
So, we have to find the following,
$19.87\text{ k}{{\text{m}}^{2}}=?\text{ }{{\text{m}}^{2}}$
So, we will now multiply the equation $1\text{ k}{{\text{m}}^{2}}={{10}^{6}}\text{ }{{\text{m}}^{2}}$ by the number 19.87 to obtain the required answer. We get the following,
$19.87\text{ k}{{\text{m}}^{2}}=19.87\times {{10}^{6}}\text{ }{{\text{m}}^{2}}$
Thus, we have converted $19.87\text{ k}{{\text{m}}^{2}}$ to ${{\text{m}}^{2}}$.
Note: We should be familiar with the metric system. We should know the standard conversions between units of a given measurement. We might need to use these conversions while solving word problems involving the measurements. We should be careful with the power of 10 used in conversion between units as there is a possibility of error in this step of the conversion.
Complete step-by-step solution:
We know that $1\text{ km}=1000\text{ m}$. This is the conversion from 1 km to m. We have to obtain the conversion from $\text{k}{{\text{m}}^{2}}$ to ${{\text{m}}^{2}}$. For this, we will square the equation $1\text{ km}=1000\text{ m}$. So, we have the following,
${{\left( 1\text{ km} \right)}^{2}}={{\left( 1000\text{ m} \right)}^{2}}$
Squaring both the sides, we obtain the following equation,
$\begin{align}
& {{1}^{2}}\text{ k}{{\text{m}}^{2}}=1000000\text{ }{{\text{m}}^{2}} \\
& \therefore 1\text{ k}{{\text{m}}^{2}}={{10}^{6}}\text{ }{{\text{m}}^{2}} \\
\end{align}$
The above equation gives us a relation between $\text{k}{{\text{m}}^{2}}$ and ${{\text{m}}^{2}}$. Now, we will see the definition of unitary method. The unitary method is a technique for solving problems by first finding the value of a single unit and then finding the necessary value by multiplying the single unit value. Now, we have obtained that
$1\text{ k}{{\text{m}}^{2}}={{10}^{6}}\text{ }{{\text{m}}^{2}}$
So, we have to find the following,
$19.87\text{ k}{{\text{m}}^{2}}=?\text{ }{{\text{m}}^{2}}$
So, we will now multiply the equation $1\text{ k}{{\text{m}}^{2}}={{10}^{6}}\text{ }{{\text{m}}^{2}}$ by the number 19.87 to obtain the required answer. We get the following,
$19.87\text{ k}{{\text{m}}^{2}}=19.87\times {{10}^{6}}\text{ }{{\text{m}}^{2}}$
Thus, we have converted $19.87\text{ k}{{\text{m}}^{2}}$ to ${{\text{m}}^{2}}$.
Note: We should be familiar with the metric system. We should know the standard conversions between units of a given measurement. We might need to use these conversions while solving word problems involving the measurements. We should be careful with the power of 10 used in conversion between units as there is a possibility of error in this step of the conversion.
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