
Name four important towns in your district. Note the distance between them in km. Express these in centimeters and millimeters.
Answer
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Hint: Note down any four towns in your district and the distance between them. Since we need to write down the distance between each other we will get six cases. Write the distances in kilometers first and then convert them into centimeters and millimeters as asked in the question.
Complete step by step answer:
Now let us consider these four towns in some districts as A, B, C, D.
Since they asked us to find the distance between each of them, we will get,
Let us find the distance from A to B, C, D.
Let the distance between A and B in kilometers is $12.8km\;$ ,
Since $1km = 1000m = 100000cm = 1000000mm$
Which can also be written as,
$1km = 1000m = {10^5}cm = {10^6}mm$
Now the distance between A and B in centimeters is $1280000cm = 12.8 \times {10^5}cm$ and in millimeters is $12800000mm = 12.8 \times {10^6}mm$ .
The distance between A and C in kilometers is $16.2km\;$ ,
Now the distance between A and B in centimeters is $1620000cm = 16.2 \times {10^5}cm$ and in millimeters is $16200000mm = 16.2 \times {10^6}mm$ .
The distance between A and D in kilometers is $9.7km\;$ ,
Now the distance between A and B in centimeters is $970000cm = 9.7 \times {10^5}cm$ and in millimeters is $9700000mm = 9.7 \times {10^6}mm$ .
Let us find the distance from B to C, D.
Let the distance between B and C in kilometers is $6.5km\;$ ,
Now the distance between B and C in centimeters is $650000cm = 6.5 \times {10^5}cm$ and in millimeters is $6500000mm = 6.5 \times {10^6}mm$ .
The distance between B and D in kilometers is $1.2km\;$ ,
Now the distance between B and D in centimeters is $120000cm = 1.2 \times {10^5}cm$ and in millimeters is $1200000mm = 1.2 \times {10^6}mm$ .
Let us find the distance from C to D.
Let the distance between C and D in kilometers is $3.7km\;$ ,
Now the distance between C and D in centimeters is $370000cm = 3.7 \times {10^5}cm$ and in millimeters is $3700000mm = 3.7 \times {10^6}mm$ .
$\therefore$ $1km = 1000m = 100000cm = 1000000mm$
Note: Never forget to write the units of the given quantity after the whole evaluation. Units are what give quantity recognition and identity. Whenever there are more zeros in the end, we can write it in powers of $10\;$ to reduce space and time.
Complete step by step answer:
Now let us consider these four towns in some districts as A, B, C, D.
Since they asked us to find the distance between each of them, we will get,
Let us find the distance from A to B, C, D.
Let the distance between A and B in kilometers is $12.8km\;$ ,
Since $1km = 1000m = 100000cm = 1000000mm$
Which can also be written as,
$1km = 1000m = {10^5}cm = {10^6}mm$
Now the distance between A and B in centimeters is $1280000cm = 12.8 \times {10^5}cm$ and in millimeters is $12800000mm = 12.8 \times {10^6}mm$ .
The distance between A and C in kilometers is $16.2km\;$ ,
Now the distance between A and B in centimeters is $1620000cm = 16.2 \times {10^5}cm$ and in millimeters is $16200000mm = 16.2 \times {10^6}mm$ .
The distance between A and D in kilometers is $9.7km\;$ ,
Now the distance between A and B in centimeters is $970000cm = 9.7 \times {10^5}cm$ and in millimeters is $9700000mm = 9.7 \times {10^6}mm$ .
Let us find the distance from B to C, D.
Let the distance between B and C in kilometers is $6.5km\;$ ,
Now the distance between B and C in centimeters is $650000cm = 6.5 \times {10^5}cm$ and in millimeters is $6500000mm = 6.5 \times {10^6}mm$ .
The distance between B and D in kilometers is $1.2km\;$ ,
Now the distance between B and D in centimeters is $120000cm = 1.2 \times {10^5}cm$ and in millimeters is $1200000mm = 1.2 \times {10^6}mm$ .
Let us find the distance from C to D.
Let the distance between C and D in kilometers is $3.7km\;$ ,
Now the distance between C and D in centimeters is $370000cm = 3.7 \times {10^5}cm$ and in millimeters is $3700000mm = 3.7 \times {10^6}mm$ .
$\therefore$ $1km = 1000m = 100000cm = 1000000mm$
Note: Never forget to write the units of the given quantity after the whole evaluation. Units are what give quantity recognition and identity. Whenever there are more zeros in the end, we can write it in powers of $10\;$ to reduce space and time.
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