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Plz answer my question ,
Sum of the areas of two squares is 468 m² . If the difference of their perimeters is 24 m , find the sides of the squares .
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Answers
Answer:
18m, 12m
Step-by-step explanation:
Let the sides of the squares be x and y. {x > y}
(i) Area of the larger square = x².
Then,perimeter of larger square = 4x.
(ii) Area of the smaller square = y²
Then,perimeter of smaller square = 4y.
Now,
(1) Sum of areas:
Sum of areas of two squares is 468.
x² + y² = 468
(ii) Difference of their perimeters is 24.
4x - 4y = 24
x - y = 6
x = 6 + y
Substitute x = 6 + y in (i), we get
⇒ (6 + y)² + y² = 468
⇒ 36 + y² + 12y + y² = 468
⇒ 2y² + 12y - 432 = 0
⇒ y² + 6y - 216 = 0
⇒ y² + 18y - 12y - 216 = 0
⇒ y(y + 18) - 12(y + 18) = 0
⇒ (y - 12)(y + 18) = 0
⇒ y = 12,-18{Length of the side of the square cannot be negative}.
⇒ y = 12.
Then:
x = 6 + y
x = 18
Therefore:
Length of the sides of the two squares is 18m, 12m.
Hope it helps!
Question:
→ Sum of the areas of two squares is 468 m² . If the difference of their perimeters is 24 m , find the sides of the two squares .
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 .