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(C) $ABCD$ is a trapezium in which $AB \parallel DC$. Let the height of the trapezium $ABCD$ be $H$. Since $E$ and $F$ are the mid-points of the non-p

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$(3a + b) : (a + 3b)$

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(C) $ABCD$ is a trapezium in which $AB \parallel DC$. Let the height of the trapezium $ABCD$ be $H$. Since $E$ and $F$ are the mid-points of the non-parallel sides $AD$ and $BC$,the line segment $EF$ is parallel to $AB$ and $DC$,and its length is $EF = \frac{1}{2}(a + b)$.The height of the smaller trapeziums $ABFE$ and $EFCD$ will be $h = \frac{H}{2}$.The area of a trapezium is given by $\frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$.$\operatorname{ar}(ABFE) = \frac{1}{2} 	imes (AB + EF) 	imes h = \frac{1}{2} 	imes \left(a + \frac{a+b}{2}ight) 	imes \frac{H}{2} = \frac{1}{2} 	imes \left(\frac{3a+b}{2}ight) 	imes \frac{H}{2} = \frac{H(3a+b)}{8}$.$\operatorname{ar}(EFCD) = \frac{1}{2} 	imes (EF + DC) 	imes h = \frac{1}{2} 	imes \left(\frac{a+b}{2} + bight) 	imes \frac{H}{2} = \frac{1}{2} 	imes \left(\frac{a+3b}{2}ight) 	imes \frac{H}{2} = \frac{H(a+3b)}{8}$.Therefore,the ratio is $\frac{\operatorname{ar}(ABFE)}{\operatorname{ar}(EFCD)} = \frac{\frac{H(3a+b)}{8}}{\frac{H(a+3b)}{8}} = \frac{3a+b}{a+3b}$.Thus,the required ratio is $(3a+b) : (a+3b)$.

$ABCD$ is a trapezium with parallel sides $AB = a \text{ cm}$ and $DC = b \text{ cm}$. $E$ and $F$ are the mid-points of the non-parallel sides. The ratio of $\operatorname{ar}(ABFE)$ and $\operatorname{ar}(EFCD)$ is

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