Question

if a area of a square is a² + b²/4 + c²/9 +ab+ b²/3 + 2/3ca, find the perimeter of the square.

pls give the correct answer whoever will answer it​

Answer 1

Appropriate Question :-

if a area of a square is a² + b²/4 + c²/9 +ab+ bc/3 + 2/3ca, find the perimeter of the square.

$\large\underline{\sf{Given- }}$

$\rm \:Area_{square}  =  \: {a}^{2} + \dfrac{ {b}^{2} }{4} + \dfrac{ {c}^{2} }{9} + ab+ \dfrac{ {b}c }{3} + \dfrac{2}{3} ca$

$\large\underline{\sf{To\:Find - }}$

$\boxed{ \rm{ Perimeter_{square}}}$

$\large\underline{\sf{Solution-}}$

Given area of square

$\rm \:  =  \:  \: {a}^{2} + \dfrac{ {b}^{2} }{4} + \dfrac{ {c}^{2} }{9} + ab+ \dfrac{ {b}c }{3} + \dfrac{2}{3} ca$

can be rewritten as

$\rm \:  =  \:  \: {a}^{2} + \dfrac{ {b}^{2} }{ {2}^{2} } + \dfrac{ {c}^{2} }{ {3}^{2} } + ab+ \dfrac{ {b}c }{3} + \dfrac{2}{3} ca$

$\rm \:  =  \:  \: {a}^{2} + {\bigg(\dfrac{b}{2} \bigg) }^{2} + {\bigg(\dfrac{c}{3} \bigg) }^{2} + 2 \times a \times \dfrac{b}{2} + 2 \times \dfrac{b}{2} \times \dfrac{c}{2} + 2 \times a \times \dfrac{c}{3}$

We know that

$\boxed{ \rm{ {a}^{2}+{b}^{2}+{c}^{2} + 2ab + 2bc + 2ca = {(a + b + c)}^{2}}}$

So, using this identity, we get

$\rm \:  =  \:  \: {\bigg(a + \dfrac{b}{2} + \dfrac{c}{3} \bigg) }^{2}$

So, we get

$\bf\implies \:Area_{square} = {\bigg(a + \dfrac{b}{2} + \dfrac{c}{3} \bigg)}^{2}$

We know that,

$\boxed{ \bf{ Area_{square} = {(side)}^{2}}}$

So, we get

$\rm :\longmapsto\: {(side)}^{2} = {\bigg(a + \dfrac{b}{2} + \dfrac{c}{3} \bigg)}^{2}$

$\bf\implies \:side = \bigg(a + \dfrac{b}{2} + \dfrac{c}{3} \bigg)$

Now, we have to find the Perimeter of square.

We know that,

$\red{\bf :\longmapsto\:Perimeter_{square} = 4 \times side}$

So,

$\rm :\longmapsto\:Perimeter_{square} = 4 \times \bigg(a + \dfrac{b}{2} + \dfrac{c}{3} \bigg)$

$\bf :\longmapsto\:Perimeter_{square} = 4a +2b + \dfrac{4c}{3} \: units$

Additional Information :-

$\boxed{ \rm{ Area_{rectangle} = length \times breadth}}$

$\boxed{ \rm{ Perimeter_{rectangle} =2( length + breadth)}}$

$\boxed{ \rm{ Area_{rhombus} = base \times height}}$

$\boxed{ \rm{ Area_{parallelogram} = base \times height}}$

$\boxed{ \rm{ Area_{circle} = \pi \: {r}^{2}}}$

$\boxed{ \rm{ Perimeter_{circle} =2 \: \pi \: {r}}}$