一个正方形的面积为 $(17-8\sqrt{2})\ \text{cm}^2$。该正方形的边长可以表示为 $(a+b\sqrt{2})\ \text{cm}$ 的形式，其中 $a$ and $b$ 是整数。证明 $2b^4-17b^2+16=0$。 [扇形面积为 $\frac{1}{2}r^2\theta$，扇形弧长为 $r\theta$。] 一个半径为 $r$ 的扇形，其弧长为 $(\sqrt{15}-\sqrt{3})\ \text{cm}$，面积为 $(3\sqrt{3}-\sqrt{15})\ \text{cm}^2$。证明 $r = \frac{6\sqrt{3}-2\sqrt{15}}{\sqrt{15}-\sqrt{3}}$ 并由此将 $r$ 表示为 $(p+q\sqrt{5})\ \text{cm}$ 的形式，其中 $p$ 和 $q$ 是整数。

1. Set up the equation relating the side length to the area of the square.

$(a+b\sqrt{2})^2 = 17-8\sqrt{2}$

2. Expand the left-hand side of the equation.

$a^2 + 2ab\sqrt{2} + 2b^2 = 17-8\sqrt{2}$
$(a^2 + 2b^2) + 2ab\sqrt{2} = 17-8\sqrt{2}$

3. Equate the rational and irrational parts to form simultaneous equations.

Since $a$ and $b$ are integers:
Rational parts: $a^2 + 2b^2 = 17$ --- (1)
Irrational parts: $2ab = -8 \implies ab = -4 \implies a = -\frac{4}{b}$ --- (2)

4. Substitute equation (2) into equation (1) to show the required result.

Substitute $a = -\frac{4}{b}$ into (1):
$\left(-\frac{4}{b}\right)^2 + 2b^2 = 17$
$\frac{16}{b^2} + 2b^2 = 17$
Multiply throughout by $b^2$:
$16 + 2b^4 = 17b^2$
$2b^4 - 17b^2 + 16 = 0$ (shown)

1. Set up equations for arc length and area

Arc length: $r\theta = \sqrt{15} - \sqrt{3}$
Area: $\frac{1}{2}r^2\theta = 3\sqrt{3} - \sqrt{15}$

2. Express area in terms of arc length to find $r$

$\frac{1}{2}r(r\theta) = 3\sqrt{3} - \sqrt{15}$
$\frac{1}{2}r(\sqrt{15} - \sqrt{3}) = 3\sqrt{3} - \sqrt{15}$

3. Rearrange to show the given expression for $r$

$r(\sqrt{15} - \sqrt{3}) = 6\sqrt{3} - 2\sqrt{15}$
$r = \frac{6\sqrt{3} - 2\sqrt{15}}{\sqrt{15} - \sqrt{3}}$

4. Rationalise the denominator

$r = \frac{6\sqrt{3} - 2\sqrt{15}}{\sqrt{15} - \sqrt{3}} \times \frac{\sqrt{15} + \sqrt{3}}{\sqrt{15} + \sqrt{3}}$
$r = \frac{6\sqrt{45} + 6(3) - 2(15) - 2\sqrt{45}}{15 - 3}$

5. Simplify the expression to the required form

$r = \frac{4\sqrt{45} + 18 - 30}{12}$
$r = \frac{4(3\sqrt{5}) - 12}{12}$
$r = \frac{12\sqrt{5} - 12}{12}$
$r = -1 + \sqrt{5}$