Question

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

Answer 1

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

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

From right triangle $ADP$

$\cos {60}^{\circ }=\frac{AP}{AD}$

$\frac{1}{2}=\frac{AP}{20}$

$AP=10$

Similarly from the right triangle $BMC$, we get $BM=10cm$.

Now from the rectangle $PMCD$, we get $CD=PM=20cm$.

$AB=AP+PM+MB=10+20+10=40cm$.