Question

1. Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24m,find the sides of two squares
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Answer 1

Answer:

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Step-by-step explanation:

Sum of the areas of two squares = 468 m2

Let a and b be the sides of the two squares.

⇒a2 + b2 = 468…(1)

Also given that,

the difference of their perimeters = 24m

⇒4a - 4b = 24

⇒a - b = 6

⇒a = b + 6…(2)

We need to find the sides of the two squares.

Substituting the value of a from equation (2) in equation (1), we have,

(b + 6)2 + b2 = 468

⇒b2 + 62 + 2 × b × 6 + b2 = 468

⇒2b2 + 36 + 12b = 468

⇒2b2 + 36 + 12b - 468 = 0

⇒2b2 + 12b - 432 = 0

⇒b2 + 6b - 216 = 0

⇒b2 + 18b - 12b - 216 = 0

⇒b(b + 18) - 12(b + 18) = 0

⇒(b + 18)(b - 12) = 0

⇒b + 18 = 0 or b - 12 = 0

⇒b = -18 = 0 or b = 12

Side cannot be negative and hence b = 12 m.

Therefore, a = b + 6 = 12 + 6 = 18 m.

Answer 2

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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

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