fractions - Let $a = frac{9+sqrt{45}}{2}$. Find $frac{1}{a}$

I've been wrapping my head around this question lately:

Let

$$a = \frac{9+\sqrt{45}}{2}$$

Find the value of

$$\frac{1}{a}$$

I've done it like this:

$$\frac{1}{a}=\frac{2}{9+\sqrt{45}}$$

I rationalize the denominator like this:

$$\frac{1}{a}=\frac{2}{9+\sqrt{45}} \times (\frac{9-\sqrt{45}}{9-\sqrt{45}})$$

This is what I should get:

$$\frac{1}{a} = \frac{2(9-\sqrt{45})}{81-45} \rightarrow \frac{1}{a}=\frac{18-2\sqrt{45}}{36})$$

Which I can simplify to:

$$\frac{1}{a}=\frac{2(9-\sqrt{45})}{36}\rightarrow\frac{1}{a}=\frac{9-\sqrt{45}}{18}$$

However, this answer can't be found in my multiple choice question here:

Any hints on what I'm doing wrong?

Answer

$$\frac { 1 }{ a } =\frac { 9-\sqrt { 45 } }{ 18 } =\frac { 3\left( 3-\sqrt { 5 } \right) }{ 18 } =\frac { 3-\sqrt { 5 } }{ 6 } $$