please answer the question
Answers
Answer:
18 m and 12 m
Step-by-step explanation:
Let a and b be the sides of two squares respectively.
a² + b² = 468 ----- (1)
Given that the difference between their perimeters is 24 meters.
⇒ 4*a - 4*b=24
⇒ a - b = 6
⇒ a =b + 6 ----- (2)
On substituting the equation(2) in the equation (1),We get
⇒ (b+6)² + b² = 468
⇒ b² + 36 + 12b + b² = 468
⇒ 2b² + 12b - 432 = 0
⇒ b² + 6*b - 216 = 0
⇒ b² + 18b - 12b - 216 = 0
⇒ b(b + 18) - 12(b + 18) = 0
⇒ (b + 18)(b - 12) = 0
⇒ b + 18 = 0 or b - 12 = 0
⇒ b = -18 or b = 12
⇒ b = 12
Place b = 12 in (2), we get
⇒ a = b + 6
⇒ a = 18
Therefore ,
The sides of two squares is 18 m and 12 m respectively.
Hope it helps!
Answer :
18 m and 12 m
Note :
• Square : A quadrilateral whose all the sides are equal and each of whose angle is 90° is called a square .
• Perimeter of the square = 4a
• Area of the square = a²
Where a is the side of square .
Solution :
Let the sides of 1st and 2nd square (where 1st square is bigger than 2nd square) be a m and a' m respectively .
Then ,
Area of 1st square = a²
Area of 2nd square = a'²
Also ,
Perimeter of 1st square = 4a
Perimeter of 2nd square = 4a'
Now ,
According to the question , sum of areas of the two squares is 468 m² .
Thus ,
→ a² + a'² = 468 --------(1)
Also ,
The difference of the perimeters of the squares is 24 m .
Thus ,
→ 4a - 4a' = 24
→ 4(a - a') = 24
→ a - a' = 24/4
→ a - a' = 6
→ a = 6 + a' ---------(2)
Now ,
Putting a = 6 + a' in eq-(2) , we get ;
=> a² + a'² = 468
=> (6 + a')² + a'² = 468
=> 6² + a'² + 2×6×a' + a'² = 468
=> 36 + 2a'² + 12a' = 468
=> 2a'² + 12a' + 36 - 468 = 0
=> 2a'² + 12a' - 432 = 0
=> 2(a'² + 6a' - 216) = 0
=> a'² + 6a' - 216 = 0
=> a'² + 18a' - 12a' - 216 = 0
=> a'(a' + 18) - 12(a' + 18) = 0
=> (a' + 18)(a' - 12) = 0
=> a' = -18 , 12
=> a'= 12 (appropriate value)
( Note : a' = -18 is rejected value because side can't be negative )
Now ,
Putting a' = 12 in eq-(2) , we get ;
=> a = 6 + a'
=> a = 6 + 12
=> a = 18