Question

Solve this Equation + Surface Area of Solid Based Question

Answer 1

The area of the rectangle is given by:

`A_(rectangle)=2x(x-1)=2x^2-2x`

The area of the trapezium is given by:

`A_(trapezium)=(x\lbrack(2x+2)+(x-1)\rbrack)/(2)=(3x^2+x)/(2)`

Then we have:

`\begin{gathered} A_(rectangle)+3=A_(trapezium) \\ 2x^2-2x+3=(3x^(2)+x)/(2) \\ 4x^2-4x+6=3x^2+x \\ x^2-5x+6=0 \\ x=(5\pm√((-5)^2-4\cdot6))/(2) \\ x_+=3 \\ x_-=2 \end{gathered}`

Then, the two possible values for the area of the rectangle are:

`\begin{gathered} 2\cdot3^2-2\cdot3=12\text{ cm}^2 \\ 2\cdot2^2-2\cdot2=4\text{ cm}^2 \end{gathered}`