Question

The sum of the areas of two squares is 360 if the difference between their perimeter is 48 then find sides of square

Answer 1

let the side of the first square be x and second square be y

Given , 4x - 4y = 48 => x - y = 12 => x = 12 + y ----(1)

x^2 + y^2 = 360

=> (12+y)^2 + y^2 = 360 {∵ fromeq(1) }

{ (a+b)^2 = a^2 + b^2 + 2ab }

=> 144 + y^2 + 24y + y^2 = 360

=> 2y^2 + 24y -216 = 0

{ Factors of -216×2 whose sum or difference is 24 are -12 and 36}

=> 2y^2 - 12y + 36y - 216 = 0

=> 2y( y - 6 ) +36 ( y - 6 ) = 0

=> ( y -6 )( 2y + 36 ) =0

=> y = 6 (or) y = -18

side of square can't be negative so y is 6

from eq(1) ,

x = 12 + 6 = 18

Answer 2

$\red{\underline{\underline{Answer:}}}$

$\sf{The \ sides \ of \ square \ are \ 18 \ and}$

$\sf{6 \ units \ respectively.}$

$\orange{\underline{\underline{Given:}}}$

$\sf{\implies{The \ sum \ of \ the \ areas}}$

$\sf{of \ two \ squares \ is \ 360.}$

$\sf{\implies{Difference \ between \ the \ perimeter}}$

$\sf{is \ 48.}$

$\pink{\underline{\underline{To \ find:}}}$

$\sf{Sides \ of \ square.}$

$\green{\underline{\underline{Solution:}}}$

$\sf{Let \ the \ side \ of \ first \ square \ be}$

$\sf{x \ unit's \ and \ side \ of \ second \ square}$

$\sf{be \ y \ units.}$

$\sf{According \ to \ first \ condition.}$

$\sf{\implies{x^{2}+y^{2}=360...(1)}}$

$\sf{According \ to \ second \ condition.}$

$\sf{4x-4y=48}$

$\sf{4(x-y)=48}$

$\sf{x-y=\frac{48}{4}}$

$\sf{\implies{x-y=12...(2)}}$

$\sf{a^{2}+b^{2}=(x-y)^{2}+2ab}$

$\sf{According \ to \ identity}$

$\sf{x^{2}+y{2}=(x-y)^{2}+2xy}$

$\sf{from \ (1) \ and \ (2)}$

$\sf{360=12^{2}+2xy}$

$\sf{360=144+2xy}$

$\sf{2xy=360-144}$

$\sf{2xy=216}$

$\sf{xy=\frac{216}{2}}$

$\sf{\implies{xy=108...(3)}}$

$\sf{(a+b)^{2}=(a-b)^{2}+4ab}$

$\sf{According \ to \ identity}$

$\sf{(x+y)^{2}=144+4(108)}$

$\sf{(x+y)^{2}=144+432}$

$\sf{(x+y)^{2}=576}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\sf{\implies{x+y=24...(4)}}$

$\sf{Add \ equations \ (2) \ and \ (4)}$

$\sf{x-y=12}$

$\sf{+}$

$\sf{x+y=24}$

_______________

$\sf{2x=36}$

$\sf{x=\frac{36}{2}}$

$\sf{\implies{x=18}}$

$\sf{Substitute \ x=18 \ in \ equation \ (4)}$

$\sf{18+y=24}$

$\sf{y=24-18}$

$\sf{\implies{y=6}}$

$\sf\purple{\tt{\therefore{The \ sides \ of \ square \ are \ 18 \ and}}}$

$\sf\purple{\tt{6 \ units \ respectively.}}$