Question

Sum of area of two squares is 468 metre square if the difference of their perimeter is 24 metre find the sides of the square​

Answer 1

Answer:

Side of one square is 6+6√6m and other square is 6√6m.

Step-by-step explanation:

If we consider the sides of the squares as x and y,

then area, a = x2+ y2 = 468 ---------1.

The difference between their perimeter, p = 4x-4y = 24.

4x-4y = 24

4x = 24+4y

x = (24+4y)/4

x = 6+y -------------2

Put x = 6+ y in equation 1,

x2+y2 = 468

(6+y)2 + y2 = 468

36+y2+y2 = 468

2y2 = 468-36

2y2 = 432

y2 = 432/2

y2 = 216

y = √216

y = √3×4×9×2

y = 2×3√3×2

y = 6√6 m.

Put y = 6√6 in equation 2,

x = 6+y

x = 6+6√6 m.

Verification:

x2+y2 = 468

(6+6√6)2 + (6√6)2 = 468

36+(36×6) + (36×6) = 468

36+216+216 = 468

468 = 468.

4x-4y = 24

4(6+6√6) - 4(6√6) = 24

24+24√6 - 24√6 = 24

24+0 = 24

24 = 24

Answer 2

${\huge{\sf{\bold{\boxed{\color{Pink}{Answer}}}}}}$

Sum of the areas of two squares is 468 m²

∵ x² + y² = 468 . ………..(1) .[ ∵ area of square = side²] → The difference of their perimeters is 24 m.

∵ 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),

∵ 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),

∵ y = x – 6

⇒ y = 18 – 6

∴ y = 12 m

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

Hope it's Helpful....:)