Question

Let $ABCD$ be a trapezium, in which $AB$ is parallel to $CD,AB=11,BC=4,CD=6$ and $DA=3$. The distance between $AB$ and $CD$ is

• A. $2$
• B. $2.4$
• C. $2.8$
• D. not determinable with the data

Answer 1

The correct option is B $2.4$

$AB=11,CD=6$, Draw $DF\perp AB$ and $CE\perp AB$. Then $EF=6$

Now, $AB=11$

$AF+FE=BE=11$

$\Rightarrow AF+6+BE=11\Rightarrow AF+BE=5\to 1$

In right angle triangle $△AFD$, using Pythagoras theorem

${AF}^2+{FD}^2=3^2\Rightarrow {AF}^2+h^2=9\to 2$

Similarly, $△CEB,h^2+{BE}^2=4^2=16\to 3$

Subtracting $3$ and $2$, we get

${BE}^2-{AF}^2=7\Rightarrow (BE-AF)(BE+AF)=7$

$\Rightarrow (BE-AF)\times 5=7\Rightarrow BE-AF=7.4\to 4$

Adding $1$ and $2$, we get

$2BE=6.4\Rightarrow BE=3.2$

Now, from $3$, $h^2+{\left(3.2\right)}^2=16\Rightarrow =\sqrt{16-10.24}$

$\Rightarrow h=2.4$

Hence, distance between AB and CD is $2.4$