Sum of the areas of two squares is 463 m^2. If the difference of their perimeters is 24 m.
find the sides of the two squares.
Answers
Answer:
Let the length of each side of a square be x metres. Then, its perimeter is 4x.
It is given that the difference of the perimeters of two squares is 24m
Perimter of second square=24+4x metres
Length of each side of second square =
4
24+4x
metres=(6+x)metres
It is given that the sum of the areas of two squares is 468m
2
∴x
2
+(6+x)
2
=468⇒x
2
+(36+12x+x
2
)=468⇒2x
2
+12x−432=0⇒x
2
+6x−216=0
This is the required equation.
Correct 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:
The sides of the squares are 18 m and 12 m
Step-by-step explanation:
Given :
- Sum of the areas of two squares is 468 m²
- The difference of their perimeters is 24 m.
To find :
the sides of the squares
Solution :
Let 'a' and 'b' are the sides of the two squares.
- Area of the square = (side)²
Sum of the areas of two squares = 468 m²
a² + b² = 468 ➟ [1]
- Perimeter of the square = 4(side)
Difference of the perimeters = 24 m
4a - 4b = 24
4(a - b) = 24
a - b = 24/4
a - b = 6
a = 6 + b
Substitute the value of a in equation [1]
a² + b² = 468
(6 + b)² + b² = 468
6² + b² + 2(6)(b) + b² = 468
36 + b² + 12b + b² = 468
2b² + 12b = 468 - 36
2b² + 12b = 432
2b² + 12b - 432 = 0
2(b² + 6b - 216) = 0
b² + 6b - 216 = 0
Factorizing it,
b² + 18b - 12b - 216 = 0
b(b + 18) - 12(b + 18) = 0
(b + 18) (b - 12) = 0
⇒ b = -18, + 12
b can not be negative. Hence, b = 12
Then,
a = 6 + b
a = 6 + 12
a = 18
Therefore, the sides of the two squares are 18 m and 12 m