Question

do it fast.......... ​

Answer 1

Given:-

• Side of the square is 10 cm.

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To find:-

• Diagonal of the square.

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Solution:-

• In right angled triangle ABD.

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We know that,

All sides of a square are equal.

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${\dag}\:{\underline{\boxed{\sf{\purple{Using\: Pythagoras\: theorem}}}}}$

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$\tt\longrightarrow{(10)^2 + (10)^2 = BD^2}$

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$\tt\longrightarrow{100 + 100 = BD^2}$

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$\tt\longrightarrow{BD^2 = 200}$

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$\bf\longrightarrow{\boxed{\orange{BD = 10\sqrt{2} \: cm}}}$

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Hence,

• the diagonal of the square is $\sf{10\sqrt{2} \: cm}$

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Diagram:-

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(1.6,-1.6)(1.6,-1.6)(1.6,1.6)\qbezier (-1.6,-1.6)(-1.6,-1.6)(-1.6,1.6)\qbezier (-1.6,-1.6)(-1.6,-1.6)(1.6,-1.6)\qbezier (-1.6,1.6)(-1.6,1.6)(1.6,1.6)\put (-2.1,1.8){\sf A}\put (1.8,1.75){\sf B}\put (1.9,-2){\sf C}\put (-2.1,-2.05){\sf D}\qbezier (-1.6,1.6)(-1.6,1.6)(1.6,-1.6)\put (-0.2,1.9){\bf 10 cm}\put (-0.2,-2){\bf 10cm}\put(1.9,-0.10){\bf\large 10cm}\put(-2.9,-0.10){\bf\large 10cm}\end{picture}$

Answer 2

Answer:

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$\tt\longrightarrow{(10)^2 + (10)^2 = BD^2}$

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$\tt\longrightarrow{100 + 100 = BD^2}$

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$\tt\longrightarrow{BD^2 = 200}$

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★ $\bf\longrightarrow{\boxed{\pink{BD = 10\sqrt{2} \: cm}}}$ ★

hope this helps you

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