Question

The sum of areas of two squares is 468m² If the difference of their perimeters is 24Cm, Find the sides of the two squares.​

Answer 1

THANKS 621

Reported

SachinKrMbMaths AryaBhatta

Answer:

→ 18m and 12 m .

Step-by-step explanation:

Let the sides of two squares be x m and y m respectively .

Case 1 .

→ Sum of the areas of two squares is 468 m² .

A/Q,

∵ x² + y² = 468 . ...........(1) .

[ ∵ area of square = side² . ]

Case 2 .

→ The difference of their perimeters is 24 m .

A/Q,

∵ 4x - 4y = 24 .

[ ∵ Perimeter of square = 4 × side . ]

⇒ 4( x - y ) = 24 .

⇒ x - y = 24/4 .

⇒ x - y = 6 .

∴ y = x - 6 ..........(2) .

From equation (1) and (2) , we get

∵ x² + ( x - 6 )² = 468 .

⇒ x² + x² - 12x + 36 = 468 .

⇒ 2x² - 12x + 36 - 468 = 0 .

⇒ 2x² - 12x - 432 = 0 .

⇒ 2( x² - 6x - 216 ) = 0 .

⇒ x² - 6x - 216 = 0 .

⇒ x² - 18x + 12x - 216 = 0 .

⇒ x( x - 18 ) + 12( x - 18 ) = 0 .

⇒ ( x + 12 ) ( x - 18 ) = 0 .

⇒ x + 12 = 0 and x - 18 = 0 .

⇒ x = - 12m [ rejected ] . and x = 18m .

∴ x = 18 m .

Put the value of 'x' in equation (2), we get

∵ y = x - 6 .

⇒ y = 18 - 6 .

∴ y = 12 m .

Hence, sides of two squares are 18m and 12m respectively

Answer 2

$\huge\underline\mathbb{SOLUTION:-}$

$\mathsf {Let, \: the \: side \: of \: the \: smaller \: square \: be \: y.}$

$\mathsf {Let, \: the \: side \: of \: the \: larger \: square \: be \: x.}$

$\mathsf {ATQ: \: x {}^{2} + y {}^{2} = 468}$

$\mathsf {Cond.ll \: 4x - 4y = 24}$

$\implies \mathsf {x - y = 6}$

$\implies \mathsf {x = 6 + y}$

$\mathsf {x {}^{2} + y {}^{2} = 468}$

$\implies \mathsf {(6 + y) {}^{2} + y {}^{2} = 468}$

$\mathsf {On \: solving \: we \: get \: y = 12}$

$\implies \mathsf {x = (12 + 6) = 18 \: m}$

$\therefore \mathsf \blue {Sides \: are \: 18m \: and \: 12m.}$