Question

sum of the areas of two square is 468 m2.If the difference of their perimeter is 24 m, find the side of the two square.
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Answer 1

Answer:

the of first square is 12m and second square is 18m.

Step-by-step explanation:

Let side of first square be x and second square be y.

condition - i

P2 - P1 = 24m

or, 4y - 4x = 24

or, y - x = 6

or, y = 6 + x ......eqn i

condition - ii

A1 + A2 = 468

or, x² + y² = 468

or, x² + (6 + x)² = 468

or, x² + 36 + 12x + x² = 468

or, 2x² + 12x - 432 = 0

or, x² + 6x - 216 = 0

or, x² + (18 - 12)x - 216 = 0

or, x² + 18x - 12x - 216 = 0

or, x(x + 18) - 12(x - 18) = 0

or, (x + 18) (x - 12) = 0

either,

x + 18 = 0

therefore, x = -18 (rejected)

or,

x - 12 = 0

therefore, x = 12

again,

y = 6 + x

  = 6 + 12

  = 18

Answer 2

$\huge\sf\pink{Answer}$

☞ The sides of the squares are 12 m and 18 m

$\rule{110}1$

$\huge\sf\blue{Given}$

✭ Sum of the area of two squares is 468 m²

✭ Difference in their Perimeter is 24 m

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

◈ The side of the squares?

$\rule{110}1$

$\huge\sf\purple{Steps}$

◕ Sides of first square be x meter

◕ Sides of the other square be y meter

We know that Perimeter of a square is given by 4a, so then

➝ Perimeter of first square = 4x

➝ Perimeter of second square = 4y

➢ Perimeter of first square - Perimeter of second square = 24 m

➢ $\sf 4x-4y=24$

➢ $\sf x+y=6$

➢ $\sf x=y+6\quad -eq(1)$

Also Given that,

»» Area of first square = x² meter²

»» Area of the second square = y² meter²

$\bullet\:\underline{\textsf{As Per the Question}}$

➳ Area of First Square + Area of Second Square = 468 m²

➳ $\sf x^2+y^2 = 468^2$

From eq(1)

➳ $\sf(6+y^2)+y^2=468$

➳ $\sf 36+y^2+12y+y^2=468$

➳ $\sf 2y^2+12y+432 = 0$

➳ $\sf y^2+6y-216=0$

➳ $\sf y^2+18y-12y-216=0$

➳ $\sf y(y+18)-12(y+18)=0$

➳ $\sf (y+18)(y-12)=0$

➳ $\sf y=-18,12$

Ignoring Negatively the side of a square is 12 m

➳ $\sf\red{y=12}$

Substituting the value of y in eq(1)

➠ $\sf x = y+6$

➠ $\sf x=12+6$

➠ $\sf\orange{x=18 \ m}$

$\rule{170}3$