Question

In trapezium $ABCD$, as shown, $AB\parallel DC,AD=DC=BC=20cm$ and $A={60}^{\circ }$. Find:
Distance between $AB$ and $DC$.

Answer 1

Draw two perpendicular to $AB$ from the point $D$ and $C$ respectively.

Since $AB\parallel CD,PMCD$ will be a rectangle.

From the right triangle $APD$ we have

$\sin {60}^{\circ }=\frac{PD}{20}$

$\frac{\sqrt{3}}{2}=\frac{PD}{20}$

$PD=10\sqrt{3}$

Therefore the distance between $AB$ and $CD$ is $10\sqrt{3}$.