Question

Solve :-

Sum of two area of square is 394 cm². And difference of perimeter is 8. Then find the sides
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Answer 1

Sol.1, Solving with Identity

Let $a$ and $b$($a>b$) be the two lengths of each square.

Sum of the area

= $a^2+b^2=394$ ...[Eqn. A]

Difference of perimeter

= $4(a-b)$ = 8

Hence $a-b=2$...[Eqn.B].

Let $\begin{cases} & a+b=u \\ & ab=v \end{cases}$

Then,

$(a-b)^2$

= $(a+b)^2-4ab$

= $u^2-4v$ ...[Eqn.1]

And,

$a^2+b^2$

= $(a+b)^2-2ab$

= $u^2-2v$ ...[Eqn.2]

Equating both equations,

$\begin{cases} & u^2-4v=4 \\ & u^2-2v=394 \end{cases}$

Hence, $u^2=784$ and $v=195$.

And hence,

$\begin{cases} & a+b=28 \\ & a-b=2 \end{cases}$

So,

$\begin{cases} & a=15 \\ & b= 13\end{cases}$

Hence, the two squares have sides 15 cm and 13 cm.

Sol. 2, Elimination of Variables

Let $a$ and $b$($a>b$) be the two lengths of each square.

Sum of the area

= $a^2+b^2=394$  ...[Eqn. A]

Difference of perimeter

= $4(a-b)=8$

Hence $a-b=2$, furthermore $b=a-2$...[Eqn.B]

Let us substitute.

$\rightarrow a^2+(a-2)^2=394$

$\rightarrow 2a^2-4a+4=394$

$\rightarrow 2a^2-4a-390=0$

$\rightarrow a^2-2a-195=0$

$\rightarrow (a-15)(a+13)=0$

Hence, $a=15$ cm is the greater length.

Another length of a square is $b=13$ cm.

Answer 2

Given,

• Sum of two area of square is 394 cm².
• The difference of perimeter is 8.

To Find,

• The Sides of Squares.

Solution,

Let's,

The Side of First Square = X

And,

The Side of Second Square = Y

The Area of First Square = Side × Side

→ The Area of First Square = X × X

→ The Area of First Square = X²

The Area of Second Square = Side × Side

→ The Area of Second Square = Y × Y

→ The Area of Second Square = Y²

The Sum of their Areas = 394cm²

X² + Y² = 394cm² •••(1st Equation)

The Perimeter of First Square = 4(Side)

→ The Perimeter of First Square = 4(X)

→ The Perimeter of First Square = 4X

The Perimeter of Second Square = 4(Side)

→ The Perimeter of Second Square = 4(Y)

→ The Perimeter of Second Square = 4Y

The Difference of Their Perimeters = 8cm

→ 4X - 4Y = 8

→ 4(X - Y) = 8

→ X - Y = 8/4

→ X - Y = 2 •••(2nd Equation)

X² + Y² = 394 •••(1st Equation)

X - Y = 2 •••(2nd Equation)

Finding The Value of Y in 2nd Equation;

X - Y = 2 •••(2nd Equation)

→ 2 = X - Y

→ Y = X - 2 •••(3rd Equation)

Substituting the Value of Y in 1st Equation;

X² + Y² = 394cm² •••(1st Equation)

→ X² + Y² = 394

→ X² + (X - 2)² = 394

→ X² + X² - 4X + 4 = 394

→ 2X² - 4X - 394 + 4 = 0

→ 2X² - 4X - 390 = 0

→ 2(X² - 2X - 195) = 0

→ X² - 2X - 195 = 0/2

→ X² - 2X - 195 = 0

→ X² - 15X + 13X - 195 = 0

→ X(X - 15) + 13(X - 15) = 0

→ (X - 15)(X + 13) = 0

X - 15 = 0

→ X = 15

Substituting the Value of X in 3rd Equation;

Y = X - 2 •••(3rd Equation)

→ Y = 15 - 2

→ Y = 13

Required Answer,

The Side of First Square = 15cm

The Side of Second Square = 13cm