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(B) Given $ABCD$ is a trapezium with $AB \parallel CD$.$AB=11, BC=4, CD=6, DA=3$.Draw $CE \parallel DA$ such that $E$ lies on $AB$. Then $AECD$ is a p

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$2.4$

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(B) Given $ABCD$ is a trapezium with $AB \parallel CD$.$AB=11, BC=4, CD=6, DA=3$.Draw $CE \parallel DA$ such that $E$ lies on $AB$. Then $AECD$ is a parallelogram.Thus,$AE=CD=6$ and $CE=DA=3$.Since $AB=11$ and $AE=6$,we have $EB = AB - AE = 11 - 6 = 5$.In $	riangle CEB$,the sides are $CE=3, BC=4, EB=5$.Since $3^2 + 4^2 = 9 + 16 = 25 = 5^2$,by the converse of the Pythagorean theorem,$	riangle CEB$ is a right-angled triangle with $\angle ECB = 90^\circ$.The area of $	riangle CEB = \frac{1}{2} 	imes CE 	imes BC = \frac{1}{2} 	imes 3 	imes 4 = 6$.Also,the area of $	riangle CEB = \frac{1}{2} 	imes EB 	imes h$,where $h$ is the altitude from $C$ to $EB$ (which is the distance between $AB$ and $CD$).$6 = \frac{1}{2} 	imes 5 	imes h \Rightarrow h = \frac{12}{5} = 2.4$.Therefore,the distance between $AB$ and $CD$ is $2.4$.

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

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