Question

1. solve for x
2. find the measures of angle 1 and 2

Explanation:

Step1: Recall mid - segment formula for trapezoid

For a trapezoid with bases \(b_1\) and \(b_2\) and mid - segment \(m\), the formula is \(m=\frac{b_1 + b_2}{2}\). Here, \(b_1 = 11\), \(b_2=17\) and \(m=-x + 21\). So, \(-x + 21=\frac{11 + 17}{2}\).

Step2: Simplify the right - hand side

Calculate \(\frac{11+17}{2}=\frac{28}{2}=14\). So the equation becomes \(-x + 21=14\).

Step3: Solve for \(x\)

Subtract 21 from both sides: \(-x=14 - 21=-7\). Multiply both sides by - 1 to get \(x = 7\).

Step4: For the angle problem in question 16

Since the two triangles are isosceles (indicated by the equal - side marks), in the isosceles triangle with the \(70^{\circ}\) angle, if the \(70^{\circ}\) angle is a base angle, then the other base angle is also \(70^{\circ}\), and the vertex angle is \(180-(70 + 70)=40^{\circ}\). Angle 1 and the \(70^{\circ}\) angle are vertical angles, so \(\angle1 = 70^{\circ}\). Angle 2 is the vertex angle of the other isosceles triangle, and since the two isosceles triangles are symmetrically related, \(\angle2=40^{\circ}\).

Answer:

1. \(x = 7\)
2. \(\angle1 = 70^{\circ}\), \(\angle2 = 40^{\circ}\)