Question

sum of the areas of two squares is 468sqm. If the difference of their perimeters is 24m. Find the sides of the two squares.​

Answer 1

AnswEr :

• Let the Side of Square₁ be a and, Side of Square₂ be b.
• Sum of Areas = 468 m²
• Difference of Perimeter = 24 m

• Difference of the Perimeter :

⇒ Perimeter (Square₁ – Square₂) = 24 m

⇒ 4a – 4b = 24 m

⇒ 4(a – b) = 24 m

• Dividing both term by 4

⇒ (a – b) = 6 m

⇒ a = (6 + b) ⠀⠀ ⠀—eq.( I )

_______________________________

• Sum of Areas of Square :

↠ Area (Square₁ + Square₂) = 468 m²

↠ (a)² + (b)² = 468 m²

• Putting the value of a from eq.( I )

↠ (6 + b)² + (b)² = 468

↠ (36 + b² + 12b) + b² = 468

↠ 36 – 468 + b² + 12b + b² = 0

↠ 2b² + 12b – 432 = 0

↠ b² + 6b – 216 = 0

• By Splitting Middle term

↠ b² + 18b – 12b – 216 = 0

↠ b(b + 18) – 12(b + 18) = 0

↠ (b – 12)(b + 18) = 0

↠ b = 12 ⠀⠀or, ⠀⠀b = – 18

Side of Square can't be Negative, that's why we will take value of b = 12

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• S I D E ⠀O F ⠀S Q U A R E S :

◗ Square₁ = a = (b + 6) = (12 + 6) = 18 m

◗ Square₂ = b = 12 m

∴ Side of Squares will be 18 m & 12 m.

Answer 2

$\bf{\Huge{\boxed{\tt{\red{ANSWER\::}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}$

Sum of the areas of two squares is 468m². If the difference of their perimeter is 24m.

$\bf{\Large{\underline{\bf{To\:find\::}}}}$

The sides of the two squares.

$\bf{\Large{\underline{\tt{\blue{Explanation\::}}}}}$

Let the side of the one square be R.

Let the side of the other square be M.

Their area is know as:

$\leadsto\sf{\large{\boxed{\tt{Area\:of\:square\:=\:(Side)^{2} }}}}}$

$\bf{We\:have}\begin{cases}\rm{Area\:of\:one\:square\:=\:(R)^{2} }\\ \rm{Area\:of\:other\:square\:=\:(M)^{2} \end{cases}}$

According to the question:

$\bf{\Large{\boxed{\it{First\:Case\::}}}}$

$\longmapsto\sf{(R)^{2} +(M)^{2} \:=\:468.....................(1)}$

$\bf{\Large{\boxed{\it{Second\:Case\::}}}}$

$\longmapsto\sf{4R\:-\:4M\:=\:24}$

$\longmapsto\sf{4(R\:-\:M)\:=\:24}$

$\longmapsto\sf{R\:-\:M\:=\:\cancel{\frac{24}{4} }}$

$\longmapsto\sf{R\:-\:M\:=\:6.......................(2)}$

or

$\longmapsto\sf{\red{R\:=\:6+M}}$

The value of R using in equation (1), we get;

$\mapsto\sf{(6+M)^{2} \:+\:(M)^{2} \:=\:468}$

$\mapsto\sf{(6)^{2} \:+\:(M)^{2} \:+\:2*6*M\:+\:(M)^{2} =\:468}$

$\mapsto\sf{36\:+\;(M)^{2} \:+\:12M\:+(M)^{2} =\:468}$

$\mapsto\sf{2M^{2} \:+\:12M\:+\:36\:=\:468}$

$\mapsto\sf{2(M^{2} \:+\:6M\:+\:18)\:=\:468}$

$\mapsto\sf{M^{2} \:+\:6M\:+\:18\:=\cancel{\frac{468}{2} }}$

$\mapsto\sf{M^{2} \:+\:6M\:+\:18\:=\:234}$

$\mapsto\sf{M^{2} \:+\:6M+18\:-\:234\:=\:0}$

$\mapsto\sf{M^{2} \:+\:6M\:-\:216\:=\:0}$

$\mapsto\sf{M^{2} \:+\:18M\:-\:12M\:-\:216\:=\:0}$

$\mapsto\sf{M(M\:+\:18)\:-\:12(M+18)\:=\:0}$

$\mapsto\sf{(M+18)(M-12)\:=\:0}$

$\mapsto\sf{(M\:+\:18)\:=0\:\:\:\:\:\:\:or\:\:\:\:\:\:(M-12)\:=\:0}$

$\mapsto\sf{\red{M\:=\:-18\:\:\:\:\:\:\:\:\:\:or\:\:\:\:\:\:\:\:M\:=\:12}}$

We know that negative value is not acceptable.

∴ $\mapsto\sf{M\:=\:12m}$

Putting the value of M in place of R, we get;

$\longmapsto\sf{R\:=\:6+12}$

$\longmapsto\sf{\red{R\:=\:18m}}$

Thus,

$\bf{\Large{\boxed{\rm{The\:side\:of\:the\:square\:is\:\:18m\:\&\:12m}}}}$