22. Given that \[ \frac{10-\sqrt{32}}{\sqrt{2}}=a+b \sqrt{2} \] where a and b are integer. Find the values of a and b . \[ \begin{array}{r} a=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\ b=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \end{array} \] 23. A shed has dimensions, in metres, of \[ \text { height }=\sqrt{5}, \text { width }=\sqrt{6} \text { and length }=\frac{9}{\sqrt{2}} \] Find the volume of the shed. Give your answer in the form \( \mathrm{a} \sqrt{ } 15 \), where a is an integer.

To solve the first problem, we start with the equation: \[ \frac{10-\sqrt{32}}{\sqrt{2}} = a + b \sqrt{2}. \] First, we simplify \(\sqrt{32}\) to \(4\sqrt{2}\): \[ 10 - \sqrt{32} = 10 - 4\sqrt{2}. \] Now we substitute this back into our equation: \[ \frac{10 - 4\sqrt{2}}{\sqrt{2}} = \frac{10}{\sqrt{2}} - 4. \] Next, rationalizing \(\frac{10}{\sqrt{2}}\): \[ \frac{10}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}. \] Thus: \[ \frac{10 - 4\sqrt{2}}{\sqrt{2}} = 5\sqrt{2} - 4. \] Now we equate this to \(a + b\sqrt{2}\): \[ a + b\sqrt{2} = -4 + 5\sqrt{2}. \] From this, we can see that: \[ a = -4 \quad \text{and} \quad b = 5. \] So, \[ \begin{array}{r} a = -4 \\ b = 5 \end{array} \] --- For the second problem, we find the volume of the shed using the formula for the volume of a rectangular prism: \[ \text{Volume} = \text{height} \times \text{width} \times \text{length}. \] Substituting in the dimensions: \[ \text{Volume} = \sqrt{5} \times \sqrt{6} \times \frac{9}{\sqrt{2}}. \] First, we simplify \(\sqrt{5} \times \sqrt{6}\) to \(\sqrt{30}\): \[ \text{Volume} = \sqrt{30} \times \frac{9}{\sqrt{2}} = 9 \times \frac{\sqrt{30}}{\sqrt{2}} = 9 \sqrt{\frac{30}{2}} = 9\sqrt{15}. \] Therefore, with \(a = 9\), \[ \text{Volume} = 9 \sqrt{15}. \] So, \[ a = 9. \]