Question

2.If the length of one of the diagonals of a square is p units, then what
is the perimeter of the square?​

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:Perimeter\:of\:square\:=\:\dfrac{4\:p}{\sqrt{2}}}}}$

Step-by-step-explanation:

NOTE: Refer to the attachment for the diagram.

We have given that,

The length of the diagonal of a square is "p" units.

We have to find the perimeter of the square.

In figure, □ABCD is a square.

AC is the diagonal of the square.

AC = p units

AB = BC = CD = AD = a units

Now, in △ABC, m∠B = 90°,

∴ AC² = AB² + BC² - - - [ Pythagoras theorem ]

⇒ ( p )² = ( a )² + ( a )²

⇒ p² = a² + a²

⇒ p² = 2a²

⇒ p = √2 a - - - [ Taking square roots ]

$\displaystyle{\implies\:\boxed{\pink{\sf\:a\:=\:\dfrac{p}{\sqrt{2}}\:}}}$

Now, we know that,

Perimeter of square = 4 * side

⇒ P ( □ABCD ) = 4 * AB

⇒ P ( □ABCD ) = 4 * a

$\displaystyle{\implies\sf\:P\:(\:\square\:ABCD\:)\:=\:4\:\times\:\dfrac{p}{\sqrt{2}}}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:Perimeter\:of\:square\:=\:\dfrac{4\:p}{\sqrt{2}}}}}}$

Answer 2

Question:-

• If the length of one of the diagonals of a square is p units, then what is the perimeter of the square?

To Find:-

• Find the perimeter of square.

Solution:-

Given ,

• AC is diagonal , AC is also p unit , AB = BC = CD = DA = a units.

Here ,

We have to use Pythagoras Theorem:-

$\tt\implies \: { AC }^{ 2 } = { AB}^{ 2 } + { BC }^{ 2 }$

$\tt\implies \: { p }^{ 2 } = { a }^{ 2 } + { a }^{ 2 }$

$\tt\implies \: p = \sqrt{ 2 } a$

$\tt\implies \: a = \dfrac { p } { \sqrt { 2 } }$

Now ,

We to multiply with 4 as square has four sides:-

$\tt\implies \: Perimeter = 4 × AB$

$\tt\implies \: Perimeter = 4a$

$\tt\implies \: Perimeter = \dfrac { 4p } { \sqrt { 2 } }$ ( Given , $\tt a = \dfrac { p } { \sqrt { 2 } }$

Hence ,

• Perimeter of square is $\tt \: \dfrac { 4p } { \sqrt { 2 } }$