Sum of areas of two squares is 468
Asked by admin @ in Math viewed by 334 People
sum of the areas of two squares is 468 m square if the difference of their perimeter is 24 m find the sides of the two square
Asked by admin @ in Math viewed by 334 People
sum of the areas of two squares is 468 m square if the difference of their perimeter is 24 m find the sides of the two square
Answered by admin @
→ 18m and 12 m .
Let the sides of two squares be x m and y m respectively .
→ Sum of the areas of two squares is 468 m² .
A/Q,
∵ x² + y² = 468 . ...........(1) .
[ ∵ area of square = side² . ]
→ 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) .
∵ 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 .
∵ y = x - 6 .
⇒ y = 18 - 6 .
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