Question

find the values of x that would make the area of the rectangle greater than the area of the triangle.can you help me pls​

Answer 1

Answer:

$x>3$

Step-by-step explanation:

$The\ area\ of\ the\ rectangle\ in\ terms\ of\ x=6x\\The\ area\ of\ the\ triangle\ in\ terms\ of\ x=\frac{4(x+6)}{2}=2(x+6) \\Given\ condition:\\The\ area\ of\ the\ rectangle>The\ area\ of\ the\ triangle\\Hence,\\6x>2(x+6)\\Now, as\ practically\ any\ of\ the\ side\ of\ the\ polygons\ cannot\ be\ negative.\\\\Hence,\\6x>2x+12\\6x-2x>2x+12-2x\\4x>12\\\frac{4x}{4} >\frac{12}{4} \\x>3$

$Hence,\\x\ could\ be\ any\ real\ number\ greater\ than\ 3.$