represent the height of the rectangle: \( A=\frac{1}{b}+\frac{1}{2} \) \( \frac{4}{b^{2}}-1 \)

We are told that the area of a rectangle is obtained by multiplying its base and its height. In our problem the rectangle has - Base: \[ \frac{1}{b}+\frac{1}{2} \] - Height: \[ \frac{4}{b^2}-1 \] Thus, the area \( A \) is given by \[ A = \left(\frac{1}{b}+\frac{1}{2}\right)\left(\frac{4}{b^2}-1\right). \] If we need to represent the height in terms of the area and the base, we recall that \[ \text{Area} = \text{base} \times \text{height}. \] Therefore, the height \( h \) can be written as \[ h = \frac{A}{\text{base}}. \] In our situation, since \[ A = \frac{4}{b^2}-1 \quad \text{and} \quad \text{base} = \frac{1}{b}+\frac{1}{2}, \] we have \[ h = \frac{\frac{4}{b^2}-1}{\frac{1}{b}+\frac{1}{2}}. \] We now simplify this expression step by step. 1. Express the denominator with a common denominator: \[ \frac{1}{b}+\frac{1}{2} = \frac{2}{2b}+\frac{b}{2b} = \frac{2+b}{2b}. \] 2. Substitute back into the expression for \( h \): \[ h = \frac{\frac{4}{b^2}-1}{\frac{2+b}{2b}}. \] 3. Dividing by a fraction is equivalent to multiplying by its reciprocal: \[ h = \left(\frac{4}{b^2}-1\right) \cdot \frac{2b}{2+b}. \] 4. Write \(\frac{4}{b^2}-1\) as a single fraction: \[ \frac{4}{b^2}-1 = \frac{4-b^2}{b^2}. \] 5. Substitute this into the expression for \( h \): \[ h = \frac{4-b^2}{b^2} \cdot \frac{2b}{2+b}. \] 6. Simplify by canceling a factor of \( b \) in the numerator with one in the denominator: \[ h = \frac{2b(4-b^2)}{b^2(2+b)} = \frac{2(4-b^2)}{b(2+b)}. \] Thus, the height of the rectangle in terms of \( b \) is \[ h = \frac{2(4-b^2)}{b(2+b)}. \]