Question

sum of the areas of two squares is 544 metre square if the difference of their perimeter is 32 find the sides of the two squares​

Answer 1

AnswEr:-

Sides of the two squares = 12 m & 20 m

Step-by-step-explanation:-

Let the sides of two squares be a & b m respectively.

As we know,

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

$\star\:\large{\boxed{\sf\blue{Perimeter \: of \: square = 4 \times Side }}}$

ATQ:-

➠ a² + b² = 544 ---------(Eq.1)

Secondly,

➠ 4a - 4b = 32

➠ 4(a - b) = 32

➠ a - b = 32/4

➠ a - b = 8 -----(Eq.2)

Now we go to (Eq.1):-

➠ (a - b)² + 2ab = 544

[Used identity: a² + b² = (a - b)² + 2ab]

➠ (8)² + 2ab = 544

➠ 2ab = 544 - 64

➠ 2ab = 480

➠ ab = 480/2

➠ ab = 240 -------(Eq.3)

Now we will use this (Eq.3) in (Eq.2):-

As we got (Eq.2):-

*Squaring both sides :-

➠ (a - b)² = 8²

➠ (a + b)² - 4ab = 64

[Used identity: (a - b)² = (a + b)² - 4ab]

➠ (a + b)² - 4 × 240 = 64

➠ (a + b)² = 64 + 960

➠ (a + b)² = 1024

➠ a + b = √1024

➠ a + b = 32 ------(Eq.4)

Now adding both the (Eq.2) & (Eq.4):-

➠ a - b + a + b = 8 + 32

➠ 2a = 40

➠ a = 40/2

➠ a = 20

Therefore,side of one square = 20 m

$\rule{200}{1}$

Now putting value of a in (Eq.2):-

➠ 20 - b = 8

➠ b = 20 - 8

➠ b = 12

Therefore,side of second square = 12 m

$\therefore\underline{\textsf{Sides \: of \: two \: squares = {\textbf{12\:m \: \& \: 20 \: m }}}}$

Answer 2

Answer:

Side of first square = 20 m

Side of second square = 12 m

Step-by-step explanation:

Let the sides of first and second square be X and Y .

Area of first square = (X)² And, Area of second square = (Y)²

According to question,

(X)² + (Y)² = 544 m² ------------(1).

Perimeter of first square = 4 × X, and,

Perimeter of second square = 4 × Y

According to question,

4X - 4Y = 32 -----------(2)

From equation (2) we get,

   4X - 4Y = 32

⇒ 4(X-Y) = 32

⇒ X - Y = 32/4

⇒ X - Y = 8

⇒ X = 8+Y ---------(3)

Putting the value of X in equation (1)

⇒ (X)² + (Y)² = 544

⇒ (8+Y)² + (Y)² = 544

⇒ (8)² + (Y)² + 2 × 8 × Y + (Y)² = 544

⇒ 64 + Y² + 16Y + Y² = 544

⇒ 2Y² + 16Y - 544 +64 = 0

⇒ 2Y² + 16Y -480 = 0

⇒ 2( Y² + 8Y - 240) = 0

⇒ Y² + 8Y - 240 = 0

⇒ Y² + 20Y - 12Y -240 = 0

⇒ Y(Y+20) - 12(Y+20) = 0

⇒ (Y+20) (Y-12) = 0

⇒ (Y+20) = 0 Or (Y-12) = 0

⇒ Y = -20 OR Y = 12

Putting Y = 12 in EQUATION (3)

⇒ X = 8+Y = 8+12 = 20

Therefore, Side of first square = X = 20 m

and,

Side of second square = Y = 12 m.