4. Find the area of a square, the length of diagonal is 2^2 m.​

2 thoughts on “4. Find the area of a square, the length of diagonal is 2^2 m.​”

1. [tex]\huge{\underline{\underline{\red{\bf{Solution}}}}}[/tex]

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   Diagonal divides the square in two equal right angled triangles. Let the each side of the square be x.

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   Given that,

   • Diagonal = 2²m
   • Diagonal = 4m

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   Now, using pythagoras theorem

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   → (Diagonal)² = (side)² + (side)²

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   → (4)² = x² + x²

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   → 16 = 2x²

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   → x² = 8

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   → x = √8

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   → x = 2√2

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   We get x = 2√2, that means each side of the square is 2√2m.

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   • Area of a square is given as

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   ★ Area = (side)²

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   → Area = (2√2)²

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   → Area = 4 × 2

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   → Area = 8 m²

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   Hence,

   • Area of the square is 8 m².

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2. Given :

   • Length of diagonal is 2² m. Means 4 m.

   To find :

   • Area of Square.

   Solution :

   [tex] \tt Here, \: length \: of \: diagonal\: given. [/tex]

   [tex] \tt All \: angles \: of \: square \: are \: 90\degree [/tex]

   [tex] \tt By \: Pythagoras\: theorem :[/tex]

   [tex] \tt \leadsto Side^{2} + Side^{2} = Diagonal^{2}[/tex]

   [tex] \tt \leadsto 2 \: side^{2} = (4)^{2} [/tex]

   [tex] \tt \leadsto 2 \: Side^{2} = 16 [/tex]

   [tex] \tt \leadsto Side^{2} = \dfrac{16}{2} [/tex]

   [tex] \tt \leadsto Side^{2} = 8[/tex]

   As we know,

   [tex] \boxed{\pmb{\tt Area \: of \: square = Side^{2}}} [/tex]

   [tex] \tt We \: have \: find\: side^{2} \: be \: 8 [/tex]

   Thus,

   Area of Square is 8 m².

   Reply