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Question:
→ Sum of the areas of two squares is 486 m². If the difference of their perimeters is 12m, find the sides of the two squares .
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Answers
hii dude your answer is
Let the sides of first and second square be X and Y .Area of first square = (X)²And,Area of second square = (Y)²According to question,(X)² + (Y)² = 468 m² ------------(1).Perimeter of first square = 4 × Xand,Perimeter of second square = 4 × YAccording to question,4X - 4Y = 24 -----------(2)From equation (2) we get,4X - 4Y = 244(X-Y) = 24X - Y = 24/4 X - Y = 6X = 6+Y ---------(3)Putting the value of X in equation (1)(X)² + (Y)² = 468(6+Y)² + (Y)² = 468(6)² + (Y)² + 2 × 6 × Y + (Y)² = 46836 + Y² + 12Y + Y² = 4682Y² + 12Y - 468 +36 = 02Y² + 12Y -432 = 02( Y² + 6Y - 216) = 0Y² + 6Y - 216 = 0Y² + 18Y - 12Y -216 = 0Y(Y+18) - 12(Y+18) = 0(Y+18) (Y-12) = 0(Y+18) = 0 Or (Y-12) = 0Y = -18 OR Y = 12Putting Y = 12 in EQUATION (3)X = 6+Y = 6+12 = 18Side of first square = X = 18 mand,Side of second square = Y = 12 m.
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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 .