Question

21. Find the length of the diagonal of a square of area 128 cm². Also find the
perimeter of the square, correct to two decimal places.​

Answer 1

Given:-

• Area of a square is 128 cm².

⠀

To find:-

• Length of the diagonal of a square.
• Perimeter of the square.

⠀

Solution:-

Let,

• the Length of the diagonal be d.

⠀

⇛ Area of square = 1/2 × d²

⇛ 128 = d²/2

⇛ d² = 128 × 2

⇛ d = √256

⇛ d = 16 cm

⠀

Hence,

• the Length of the diagonal is 16 cm.

⠀

Now,

☆Area of square = a²

⇛ 128 = a²

⇛ a = √128

⇛ a = 11.31 cm

⠀

Then,

☆Perimeter of square = 4 × side

⇛ 4 × 11.31

⇛ 45.24 cm

⠀

Hence,

• the perimeter of the square is 45.24 cm.

Answer 2

$\huge \fbox{Answer}$

$\large \underline{ \frak{ \color{red}Given:}}$

$\mapsto \sf{Area \: of \: square=128cm²}$

$\large \underline{ \frak{ \color{rd}To \: find:}}$

$\leadsto \sf{Length \: of \: diagonal \: of \: square.}$

$\leadsto \sf{Perimeter \: of \: square.}$

$\large \underline{ \frak{ \color{blue}Formula \: required::}}$

$\underline{ \boxed{ \rm{Area \: of \: square = \frac{1}{2} \times( {Diagonal})^{2} }}}$

$\underline{ \fbox{Perimeter of square=4×side}}$

$\large \underline{ \frak{ \pink{According \: to \: Question:}}}$

${ \sf{:: \implies{Area \: of \: square = \frac{1}{2} \times({Diagonal})^{2} }}}$

${ \sf{:: \implies{128 {cm}^{2} = \frac{1}{2} \times( {Diagonal})^{2} }}}$

${ \sf{:: \implies{( {Diagonal})^{2} = 128 {cm}^{2} \times 2 }}}$

${ \sf{:: \implies{( {Diagonal})^{2} =256 {cm}^{2}}}}$

${ \sf{::\implies{Diagonal= \sqrt{256 {cm}^{2} } }}}$

${ \sf{ ::\implies{Diagonal= 16cm}}}$

${:: \implies{ \sf{Area \: of \: rectangle= {side}^{2} }}}$

${ ::\implies{ \sf{Side = \sqrt{128 {cm}^{2} } }}}$

${ ::\implies{ \sf{Side = 11. 31cm} }}$

$\bf{Then, }$

${:: \implies{ \sf{Perimeter \: of \: rectangle=4×side}}}$

${:: \implies{ \sf{Perimeter \: of \: rectangle=4×11.31cm}}}$

${:: \implies{ \sf{Perimeter \: of \: rectangle=45.25cm}}}$

$\bf{Hence, }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: { \underline{ \fbox{Perimeter \: of \: rectangle=45.25cm}}}$