Math, asked by shuklaamarujala, 9 months ago

20. Find the length of the diagonal of a square whose area is 128 cm Also, find its perimeter.
21. The area of a square field is 8 hectares. How long would a man take to cross it digonally waking with speed of 4kmph?
22. The cost of harvesting a square field at 900 per hectare is 8100. Find
the cost of putting a fence around it at rupee18
find its perimeter.​

Answers

Answered by TakenName
14

20. The area of a square is one side squared.

Let us consider:-

  • A side is x cm.

The given area is 128 cm².

This implies that:-

\sf{\to x^2=128}

\sf{\to x=\sqrt{128}}

\sf{\to x=\sqrt{2^5}}

\sf{\therefore x=4\sqrt{2}} cm ...(1)

Here, the diagonal of a square is always x × √2.

\sf{\therefore \sqrt{2} x=8} cm (ANSWER)

Here, the perimeter of a square is always 4x.

\sf{\therefore 4x=16\sqrt{2} } cm (ANSWER)

21. The area of a square is one side squared.

Let us consider:-

  • A side is x m.

The given area is 80000 m².

This implies that:-

\sf{\to x^2=80000}

\sf{\to x=\sqrt{80000} }

\sf{\to x=\sqrt{8\times 10^4}}

\sf{\to x=200\sqrt{2} } m

\sf{\therefore x=\dfrac{\sqrt{2} }{5} } km ...(1)

Now

Here is the definition of speed.

\sf{\to s=\dfrac{d}{t} } (here, km/h)

As per the question:-

\sf{\to 4=\dfrac{\sqrt{2} }{5} \times \dfrac{1}{t} }

\sf{\to t=\dfrac{\sqrt{2} }{5\times 4}}

\sf{\therefore t=\dfrac{\sqrt{2} }{20}  } hours (ANSWER)

22. Given that:-

  • (Cost):(Field)=1:9
  • The cost of putting the fence is 18.
  • The field is a square.

This results in insufficient information, because there is no area of field.

This problem cannot be solved. (ANSWER)

Extra Information

  • Such problems with insufficient information, cannot be solved.

For example:-

"A captain owns 26 sheep and 10 goats. How old is the captain?"

  • A physist does not memorize any formula. (Quora Answer)

For example, we used the definition.

\sf{\to s=\dfrac{d}{t}}

Formulas are derived from definitions, they don't need to memorize but basics.

The derivation would look like this:

\sf{\to d=st\:\&\:t=\dfrac{d}{s} }

Thanks for reading!

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