sum of the area of two squares is 468m² if the difference of their perimeters is 24m , formulate the quadra equation to find the sides of the two squares.
Answers
Answer:
18 m and 12 m
Step-by-step explanation:
Let the first square be X and the other be Y
SO
Equation 1:
x² + y² = 468
Equation 2:
The difference between their perimeters is 24m
So
4x-4y=24
Divide all the equation by 4 so you get
x-y=6
SO y=x-6
(multiply by 4 because perimeter=4L)
SUBSTITUTE
x² + ( x - 6 )² = 468
x² + x² - 12x + 36 = 468
2x² - 12x + 36 - 468 = 0
x( x - 18 ) + 12( x - 18 ) = 0
(x-18)(x+12)
x=18 and x=-12(length cant be negative so -12 is wrong)
X=18m
So put the value of x in Y=X-6
Y=18-6=12m
So the sides are 18m and 12m
:))
S O L U T I O N :
Given :
Sum of the area of two squares is 468m², if the difference if their Perimeter is 24m.
Explanation :
Let the First side of Square be x & the second side be y.
As we know that formula of the area of Square & their Perimeter :
- Area = Side × Side
- Perimeter = 4 × side
According to the question :
➛ Perimeter = 4 × side
➛ 4x - 4y = 24
➛ 4(x - y) = 24
➛ x - y = 24/4
➛ x - y = 6
➛ x = 6 + y.............(1)
&
➛ Area of Square = Side × Side
➛ (x)² + (y)² = 468
➛ (6+y)² + (y)² = 468 [from eq.(1)]
➛ (6)² + (y)² + 2 × 6 × y + y² = 468
➛ 36 + y² + 12y + y² = 468
➛ 2y² + 12y = 468 - 36
➛ 2y² + 12y = 432
➛ 2(y² + 6y) = 432
➛ y² + 6y = 432/2
➛ y² + 6y = 216
➛ y² + 6y - 216 = 0
By using quadratic equation method :
As we know that given quadratic polynomial compared with ax² + bx + c :
- a = 1
- b = 6
- c = -216
Now,
➛ x = -b ± √b² - 4ac/2a
➛ x = -6 ± √(6)² - 4 × 1 × -216/2 × 1
➛ x = -6 ± √36 + 864/2
➛ x = -6 ±√900/2
➛ x = -6 ± 30/2
➛ x = -6 + 30/2 Or x = -6 - 30/2
➛ x = 24/2 Or x = -36/2
➛ x = 12 cm Or x = -18 cm
Therefore, as we knoW that negative value isn't acceptable.
Thus,
The first side of the square will be 12 cm & the second side of the square will be (12 + 6) cm = 18 cm .