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-BARSIC- [3]
2 years ago
10

Given a side of a Square is 3cm. Find area and perimeter of the square. ​

Mathematics
2 answers:
anastassius [24]2 years ago
8 0

Answer:

Area: 9cm²

Perimeter: 12 cm

Explanation:

Area is length x width

It's a square so length and width are the same

3 times 3 is 9

And to find the perimeter you just add the side lengths

Squares have 4 sides and the given side length is 3 so 3x4 OR 3+3+3+3 is how you get the perimeter.

Allisa [31]2 years ago
5 0

\star  \pink{\frak{Given}}

\\  \\

  • Side of a Square = 3cm

\\  \\

\star  \blue{\frak{To \: find :}}

\\  \\

  • Area of a Square

  • Perimeter of a Square

\\  \\

\star  \orange{\frak{Solution :}}

\\  \\

There are two parts in this question. In first part we have to find area and in second part we have to find perimeter of a Square.

\\  \\

\green{ \text{part \: 1}}

\\

<u>To find Area of a Square </u><u>:</u>

we know:

\bigstar \boxed{ \tt{Area \: of  \: a  \: Square =  }{ \tt{side}^{2} }}

\\

So :

\\

\dashrightarrow \sf{Area \: of  \: a  \: Square =  }{ \sf{side}^{2} }

\\  \\

\dashrightarrow \sf{Area \: of  \: a  \: Square =  }{ \sf{3}^{2} }

\\  \\

\dashrightarrow \sf{Area \: of  \: a  \: Square =  }{ \sf3 \times 3}

\\  \\

\dashrightarrow \sf{Area \: of  \: a  \: Square =  }{ \sf9cm {}^{2} }

\\  \\

\therefore \underline \textsf{ \textbf{Area \: of \: a \: square \:  =  \red{9 {cm} }}} \red{ \bf^{2}}

\\  \\

\green{ \text{part \: 2}}

\\  \\

<u>To find perimeter of square </u><u>:</u>

\\  \\

\bigstar \boxed{ \tt{Perimeter \: of  \: a  \: Square =  }{ \tt4 \times {side}}}

\\  \\

So:

\\  \\

\dashrightarrow \sf{Perimeter \: of  \: a  \: Square =  }{ \sf4 \times side} \\

\\  \\

\dashrightarrow \sf{Perimeter \: of  \: a  \: Square =  }{ \sf4 \times 3} \\

\\  \\

\dashrightarrow \sf{Perimeter \: of  \: a  \: Square =  }{ \sf12cm} \\

\\  \\

\therefore \underline{ \textsf{ \textbf{perimeter \: of \: a \: square \:  =  \red{12cm}}}}

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  1. Substitute in points [Distance Formula]:                                                       \displaystyle d = \sqrt{(5-11)^2+(8-4)^2}
  2. [√Radical] (Parenthesis) Subtract:                                                                 \displaystyle d = \sqrt{(-6)^2+(4)^2}
  3. [√Radical] Evaluate exponents:                                                                    \displaystyle d = \sqrt{36+16}
  4. [√Radical] Add:                                                                                               \displaystyle d = \sqrt{52}
  5. [√Radical] Simplify:                                                                                         \displaystyle d = 2\sqrt{13}
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