Answer:
Therefore,
Dimensions of the Original Square is
![x =11\ m](https://tex.z-dn.net/?f=x%20%3D11%5C%20m)
Step-by-step explanation:
Given:
Let the side of Square be "x" meter
Then the Dimensions of a Rectangle is formed by adding 6 m to one side and 3 m to the other side wil be.
![Length = x+6\\\\Width=x+3](https://tex.z-dn.net/?f=Length%20%3D%20x%2B6%5C%5C%5C%5CWidth%3Dx%2B3)
Area of Rectangle =238 m²
To Find:
x = ? (Dimension of Original Square)
Solution:
Area of Rectangle is given by
![\textrm{Area of Rectangle}=Length\times Width](https://tex.z-dn.net/?f=%5Ctextrm%7BArea%20of%20Rectangle%7D%3DLength%5Ctimes%20Width)
Substituting the values we get
![238=(x+6)\times (x+3)](https://tex.z-dn.net/?f=238%3D%28x%2B6%29%5Ctimes%20%28x%2B3%29)
Opening the Parenthesis we get
......Which is Quadratic equation
On Factorizing and Splitting the middle term we get
![x^{2}+20x-11x-220=0\\\\x(x+20)-11(x+20)=0\\\\(x+20)(x-11)=0\\\\x+20=0\ or\ x-11=0\\\\x=-20\ or\ x=11](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B20x-11x-220%3D0%5C%5C%5C%5Cx%28x%2B20%29-11%28x%2B20%29%3D0%5C%5C%5C%5C%28x%2B20%29%28x-11%29%3D0%5C%5C%5C%5Cx%2B20%3D0%5C%20or%5C%20x-11%3D0%5C%5C%5C%5Cx%3D-20%5C%20or%5C%20x%3D11)
As distance cannot be negative therefore,
![x =11](https://tex.z-dn.net/?f=x%20%3D11)
Therefore,
Dimensions of the Original Square is
![x =11\ m](https://tex.z-dn.net/?f=x%20%3D11%5C%20m)