Answer:
The answer is given below
Step-by-step explanation:
The question is not complete, we need the dimensions of the triangle and rectangle.
From the image attached, the height of the triangle is 2x and the base is x. For the rectangle the length is 3x - 2 and the width is x - 1.
The area of the triangle = [tex]\frac{1}{2}(base *height)=\frac{1}{2} (x*2x) = \frac{1}{2} (2x^2)= x^2[/tex]
Area of the rectangle = [tex]length * breadth=x-1(3x-2)= 3x^2-2x-3x+2=3x^2-5x+2[/tex]
the area of the rectangle is greater than the area of triangle. Therefore:
[tex]3x^2-5x+2>x^2\\3x^2-5x+2-x^2>0\\2x^2-5x+2>0\\2x^2-4x-x+2>0\\2x(x-2)-1(x-2)>0\\(2x-1)(x-2)>0\\2x-1>0\ or\ x-2>0\\x<\frac{1}{2}\ or\ x>2\\Therefore \ x>2[/tex]
