Question

11. Sum of the areas of two squares is 468 m. If the difference of their perne given eq
find the sides of the two squares by quadratic equation
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Answer 1

Correct Question:-

The sum of areas of two different squares is 468 m². If the difference of their perimeters is 24 m, find the sides of two squares by quadratic equation.

Answer:-

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

Given:

Sum of their areas = 468 m².

We know that,

Area of a square = a² [ where a is the length of its side ]

Area of the 1st square = x²

Area of the second square = y².

→ x² + y² = 468 -- equation (1)

And,

Difference of their perimeters = 24 m

Perimeter of a square = 4 * length of its side.

Perimeter of first square = 4*x = 4x

Perimeter of the second square = 4*y = 4y

Hence,

→ 4x - 4y = 24

→ 4(x - y) = 24

→ x - y = 24/4

→ x - y = 6

→ x = 6 + y -- equation (2)

Substitute x = 6 + y in equation (1).

→ (6 + y)² + y² = 468

Using the formula (a + b)² = a² + b² + 2ab in LHS we get,

→ 36 + y² + 12y + y² = 468

→ 2y² + 12y + 36 = 468

→ 2(y² + 6y + 18) = 468

→ y² + 6y + 18 = 468/2

→ y² + 6y + 18 = 234

→ y² + 6y + 18 - 234 = 0

→ y² + 6y - 216 = 0

→ y² - 12y + 18y - 216 = 0

→ y (y - 12) + 18 (y - 12) = 0

→ (y + 18) * (y - 12) = 0

→ y + 18 = 0

→ y = - 18

→ y - 12 = 0

→ y = 12

Side of a square cannot be negative. So Positive value is taken. i.e., y = 12 m.

Substitute the value of y in equation (2).

→ x = 6 + y

→ x = 6 + 12

→ x = 18

Therefore,

• Length of side of 1st square = 18 m.

• Length of side of 2nd square = 12 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

Hope it's Helpful....:)