Answer:
AB = 40 cm
Distance = [tex]10\sqrt{3}[/tex] cm
Step-by-step explanation:
Draw two heights DE and CF. These heights together with bases of trapezoid form rectangle DEFC. In this rectangle, DC=EF=20 cm.
Consider right triangle ADE. In this triangle, angle ADE has the measure of 30° (the sum of the measures of all interior angles is always 180°, so m∠ADE=180°-60°-90°=30°)
Leg opposite to 30° angle is equal to half of the hypotenuse, so
[tex]AE=\dfrac{1}{2}AD=\dfrac{1}{2}\cdot 20=10\ cm[/tex]
Similarly, in triangle CBF, BF=10 cm.
Hence,
[tex]AB=AE+EF+FB=10+20+10=40\ cm[/tex]
The height of the trapezoid (the second leg of triangle ADE) is the distance between DC and AB. By the Pythagorean theorem,
[tex]AD^2=AE^2+DE^2\\ \\DE^2=20^2-10^2=400-100=300\\ \\DE=\sqrt{300}=10\sqrt{3}\ cm[/tex]
