Math, asked by hitesh1611, 1 year ago

Sum of the area of two squares is 468 m² . If the difference of their perimeter is 24 m. Find the sides of the two squares.

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Answers

Answered by TooFree
39

STEP 1: Define x

Let the length of one square be x and the other be y.

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STEP 2: Form the equations:

sum of the area of the two squares is 468 m²

x² + y² = 468

.

Difference in their perimeter is 24 cm

4x - 4y = 24

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STEP 3 : Solve x and y:

x² + y² = 468------------------ [ 1 ]

4x - 4y = 24 ------------------ [ 2 ]

.

From [ 2 ]:

4x - 4y = 24

Divide by 4 through:

x - y = 6

Add y to both sides:

x = 6 + y ------------------ [ 3 ]

.

Substitute [ 3 ] into [ 2 ]

(6 + y)² + y² = 468

36 + 12y + y² + y² = 468

2y² + 12y - 431 = 0

y² + 6y - 216 = 0

(y - 12) (y + 18) = 0

y = 12 or y = -18 (rejected, because length cannot be negative)

.

When y = 12 ------------------ Substitute into [ 3 ]

x = 6 + 12

x = 18

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Answer: The length of the two squares are 12 m and 18 m

Answered by Anonymous
11

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 .

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