Question

find the diagonal of a square whose side is 10cm​

Answer 1

Step-by-step explanation:

The side of square is 10 cm

The side of square is 10 cmIn right angled △ABD

The side of square is 10 cmIn right angled △ABDAB A+AD

The side of square is 10 cmIn right angled △ABDAB A+AD2 =BD 2[according to Pythagoras theorem]

The side of square is 10 cmIn right angled △ABDAB A+AD2 =BD 2[according to Pythagoras theorem](10) 2 +(10) 2 =BD2$

The side of square is 10 cmIn right angled △ABDAB A+AD2 =BD 2[according to Pythagoras theorem](10) 2 +(10) 2 =BD2$100+100=BD 2BD

The side of square is 10 cmIn right angled △ABDAB A+AD2 =BD 2[according to Pythagoras theorem](10) 2 +(10) 2 =BD2$100+100=BD 2BD2 =200BD=102 cm

The side of square is 10 cmIn right angled △ABDAB A+AD2 =BD 2[according to Pythagoras theorem](10) 2 +(10) 2 =BD2$100+100=BD 2BD2 =200BD=102 cm∴ Diagonal of square =102 cm.

Answer 2

Given:-

• Side of the square is 10 cm.

⠀

To find:-

• Diagonal of the square.

⠀

Solution:-

• In right angled triangle ABD.

⠀

We know that,

• All sides of a square are equal.

⠀

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

⠀

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

⠀

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

⠀

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

⠀

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

⠀

Hence,

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

⠀

Diagram:-

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