The problem requires us to draw an accurate triangle using a ruler and protractor based on the given dimensions (sides of 9 cm and 10 cm, and one angle of 30 degrees). After drawing the triangle, we need to measure the length of side $a$ and provide the answer to one decimal place.

1. Problem Description

The problem requires us to draw an accurate triangle using a ruler and protractor based on the given dimensions (sides of 9 cm and 10 cm, and one angle of 30 degrees). After drawing the triangle, we need to measure the length of side $a$ and provide the answer to one decimal place.

2. Solution Steps

Since I can't use a ruler and protractor to draw, I'll use the Law of Cosines to calculate the length of side $a$.

The Law of Cosines states:

$c^2 = a^2 + b^2 - 2ab \cos(C)$

In our case, let the side with length 9 cm be $b$, the side with length 10 cm be $c$, and the angle of 30 degrees be $C$. We want to find the length of side $a$.

Using the Law of Cosines, we have:

$c^2 = a^2 + b^2 - 2ab \cos(C)$

$10^2 = a^2 + 9^2 - 2 \cdot a \cdot 9 \cdot \cos(30^\circ)$

$100 = a^2 + 81 - 18a \cos(30^\circ)$

$100 = a^2 + 81 - 18a (\frac{\sqrt{3}}{2})$

$100 = a^2 + 81 - 9\sqrt{3} a$

$a^2 - 9\sqrt{3} a - 19 = 0$

We have a quadratic equation in the form of $Ax^2 + Bx + C = 0$, where $A=1$, $B = -9\sqrt{3}$, and $C = -19$.

We can use the quadratic formula to solve for $a$:

$a = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

$a = \frac{9\sqrt{3} \pm \sqrt{(-9\sqrt{3})^2 - 4(1)(-19)}}{2(1)}$

$a = \frac{9\sqrt{3} \pm \sqrt{243 + 76}}{2}$

$a = \frac{9\sqrt{3} \pm \sqrt{319}}{2}$

$a = \frac{9\sqrt{3} \pm \sqrt{319}}{2}$

$a = \frac{15.588 \pm 17.861}{2}$

We have two possible solutions for $a$:

$a_1 = \frac{15.588 + 17.861}{2} = \frac{33.449}{2} = 16.7245$

$a_2 = \frac{15.588 - 17.861}{2} = \frac{-2.273}{2} = -1.1365$

Since the length of a side must be positive, we take the positive solution:

$a \approx 16.7245$

Rounded to one decimal place, $a \approx 16.7$ cm.

3. Final Answer

16.7 cm