Question

*If the sum of the areas of two squares is 468 m² and the difference of their perimeter is 24 m, then the sides of the two squares will be:*

1️⃣ 18m , 16m
2️⃣ 12m , 18m
3️⃣ 18m , 18m
4️⃣ 16m , 12m​

Answer 1

☯ Let's consider that the first side be a m and second side be A m.

⠀⠀⠀

Therefore,

• Area of first square is a m².
• Area of second square is A m².

Given that,

• The difference of their perimeter is 24 m.

⠀⠀⠀

$\star\boxed{\sf{\purple{Perimeter = 4 \times (Side)}}}$

⠀⠀⠀

So,

$:\implies\sf 4a - a = 24 \qquad\qquad\bigg\lgroup\bf eq\: (I)\bigg\rgroup$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━

⠀⠀⠀

$\underline{\bigstar\:\boldsymbol{According\:to\:the\:question\::}}$

• The sum of the areas of two squares is 468 m².

$:\implies\sf A^2 + a^2 = 468 \qquad\qquad\bigg\lgroup\bf eq\: (II)\bigg\rgroup$

$:\implies\sf 4a - A = 24 \\\\\\:\implies\sf a - A = \cancel\dfrac{24}{4} \\\\\\:\implies\sf a - A = 6 \\\\\\:\implies\sf A = a + 6$

⠀⠀⠀

⠀⠀⠀⠀$\dag\;{\underline{\frak{Substituting \ the \ Value \ of \ A \ in \ equation \ (2) \: :}}}$

⠀⠀⠀

$:\implies\sf A^2 + a^2 = 468 \\\\\\:\implies\sf (a + 6)^2 + a^2 = 468 \\\\\\:\implies\sf a^2 + 12a + 36 + a^2 = 468 \\\\\\:\implies\sf 2a^2 + 12a + 36 = 468 \\\\\\:\implies\sf 2(a^2 + 6a + 18 = 234) \\\\\\:\implies\sf a^2 + 6a - 216 = 0 \\\\\\:\implies\sf a^2 + 18a - 12a - 216 = 0 \\\\\\:\implies\sf a(a + 18) -12(a + 18) = 0 \\\\\\:\implies\sf(a - 12) (a + 18) = 0 \\\\\\:\implies\sf a = 12 \: \& \: a = - 18$

As,

Side of the square can't be negative. so, x is 12.

⠀⠀⠀

Hence,

• Side of first square, a = 12 m
• Side of second square, A = a + 6 = 12 + 6 = 18 m.

⠀⠀⠀

$\therefore\:{\underline{\sf{Option \ (b), \ is \ correct.}}}$

⠀⠀⠀

⠀⠀⠀⠀$\therefore\:{\underline{\sf{The\: sides \; of \ the \ two \ squares\ will \ be\: \bf{ 12, 18 \ m}.}}}$

Answer 2

$\: \: \: \: \: \:{\large{\bold{\sf{\underbrace{\underline{Let's \; understand \; the \; question \; 1^{st}}}}}}}$

➙ This question is from a very interesting chapter of mathematics = Quadratic equation and the question says that if the sum of the area of 2 squares are 468 m² and the difference of their perimeter is 24 m, then the sides of the two squares will be what ? Some options are given below –

➝ 18m , 16m

➝ 12m , 18m

➝ 18m , 18m

➝ 16m , 12m

${\large{\bold{\sf{\underline{Given \; that}}}}}$

➙ The sum of the area of 2 squares are 468 m²

➙ The difference of their perimeter is 24 m

${\large{\bold{\sf{\underline{To \; find}}}}}$

➙ The sides of the two squares.

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

➙ The sides of the two squares = 12 m and 18 m ( Option 2 )

Diagram of 1st square

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf D}\put(-0.5,4.2){\bf A}\put(4.2,-0.5){\bf C}\put(4.2,4.2){\bf B}\put(1.5,-0.6){\bf\large 12\ m}\put(4.4,2){\bf\large 12\ m}\end{picture}$

Diagram of 2nd square

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf D}\put(-0.5,4.2){\bf A}\put(4.2,-0.5){\bf C}\put(4.2,4.2){\bf B}\put(1.5,-0.6){\bf\large 18\ m}\put(4.4,2){\bf\large 18\ m}\end{picture}$

${\large{\bold{\sf{\underline{Assumptions}}}}}$

➙ Let a m is 1st square side

➙ Let A m is 2nd square side

$\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;{\sf{Henceforth,}}$

➙ a m² is the area of 1st square

➙ A m² is the area of 2nd square

${\large{\bold{\sf{\underline{Using \; concept}}}}}$

➙ Perimeter of square formula

${\large{\bold{\sf{\underline{Using \; formula}}}}}$

➙ Perimeter of square = 4 × sides

${\large{\bold{\sf{\underline{Full \; Solution}}}}}$

$\rule{150}{1}$

➨ Perimeter of square = 4 × side

➨ 24 = 4a - A

➨ 4a - A = 24 .....[ Equation 1 ]

$\rule{150}{1}$

~ Now let's work according to the question,

${\bullet}$ As it's given that the sum of the area of 2 squares are 468 m²

Henceforth,

➨ a m² + A m² = 468 m² .....[ Equation 2 ]

➨ 4a - A = 24

➨ a - A = ${\sf{\dfrac{24}{4}}}$

➨ a - A = 6

➨ a = 6 + A

$\rule{150}{1}$

~ Now let's substitute the value,

➨ a² + A² = 468

➨ (a+6)² + A² = 468

➨ a² + 12a + 36 + A² = 468

➨ 2a²+ 12a + 36 = 468

[ Let's cancel digits with 2 ]

➨ a² + 6a + 18 = 234

➨ a² + 6a - 216 = 0

➨ a² + 18a - 12a - 216 = 0

➨ a(a+18) - 12(a+18) = 0

➨ (a-12) (a+18) = 0.

➨ Henceforth, a = 12 and -18

[ As we already know that the side of square never be negative so we take it as positive ]

➨ Therefore, 12 m and 18 m are the sides of the two squares ( Option 2 )

${\large{\bold{\sf{\underline{Additional \; knowledge}}}}}$

Diagram of a square -

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf D}\put(-0.5,4.2){\bf A}\put(4.2,-0.5){\bf C}\put(4.2,4.2){\bf B}\put(1.5,-0.6){\bf\large x\ cm}\put(4.4,2){\bf\large x\ cm}\end{picture}$

Quadratic equation definition -

Any equation containing one term in which unknown is squared (²) & none term in which it is raised to a higher power solve for x in the quadratic equation!

Linear equation knowledge -

❶ Linear equation = Here we will have to deal with linear expressions in just one variable. Such equations are known to be “linear equation in one variable”

❷ An algebraic equation in an equality involving variable. It has an equality sign. The expression on the left of equality sign is LHS. The expression on the right of equality sign is RHS like in expression 2x - 3 = 7

↦ 2x is variable

↦ = is the sign of equality

↦ 7 is equation

↦ 2x - 3 is LHS

↦ 7 is RHS

❸ In an equation the values, of the expression on LHS and RHS are equal. This happen to be true ! for certain values of that variable. The values are the solution of that equation like,

↦ x = 5 is the solution of the equation

↦ 2x - 3 = 7 [ x = 5 ]

↦ LHS = 2 × 5 - 3 = 7 = RHS

↦ On the other hand, x = 10 is nor a solution of the equation [ x = 10 ]

↦ LHS = 2 × 10 - 3 = 17

↦ This isn't equal to the RHS.

❹ How to find solution for equation?

We have to assume that the 2 sides of the equation are in a balance. We have to perform the same mathematical operation on both sides of the equation so that the balance isn't disturbed. A few such steps give you your solution always...!