Question

1. solve for x.

Explanation:

Step1: Determine the type of triangle

Since two sides of $\triangle DEF$ are equal (marked with the same tick - marks), $\triangle DEF$ is an isosceles triangle. So, $\angle D=\angle F$.

Step2: Use the property of the exterior angle of a triangle

The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. The exterior angle is $(11x - 18)^{\circ}$, and the non - adjacent interior angle is $\angle E = 40^{\circ}$. Also, because $\angle D=\angle F$, and in an isosceles triangle $\angle D=\angle F=\frac{180^{\circ}-\angle E}{2}=\frac{180 - 40}{2}=70^{\circ}$.
Using the exterior angle formula: $(11x-18)^{\circ}=\angle D+\angle F$. Since $\angle D = \angle F=70^{\circ}$, we have the equation $11x-18=70 + 70$.

Step3: Solve the equation for \(x\)

First, simplify the right - hand side of the equation: $11x-18 = 140$.
Then, add 18 to both sides of the equation: $11x=140 + 18$, so $11x=158$.
Finally, divide both sides by 11: $x=\frac{158}{11}\approx14.36$. Wait, no, we made a mistake above. Let's start from the exterior angle formula correctly.

Since $\triangle DEF$ is isosceles with $\angle D=\angle F$. The exterior angle $(11x - 18)^{\circ}$ is equal to $\angle D+\angle F$. And also, using the angle - sum property of a triangle, $\angle D=\angle F=\frac{180-\angle E}{2}$. But we can also use the exterior angle formula directly.

The exterior angle formula: $(11x-18)= \frac{180 - 40}{2}+\frac{180 - 40}{2}$.

Another way: The base angles of the isosceles triangle $\angle D=\angle F$. The exterior angle $(11x - 18)$ is equal to $2\angle D$. And $\angle D=\frac{180-\angle E}{2}=\frac{180 - 40}{2}=70$.

Or use the exterior angle formula: $(11x-18)=2\times\frac{180 - 40}{2}$.

Correctly, using the exterior angle formula:

$$
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$$

Answer:

\(x = 14\)