In trapezium ABCD,AB || CD. Points P and Q ar | vedclass

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Question

(D) In a trapezium,the line segment joining the midpoints of the non-parallel sides is parallel to the parallel sides and its length is equal to half

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) In a trapezium,the line segment joining the midpoints of the non-parallel sides is parallel to the parallel sides and its length is equal to half the sum of the lengths of the parallel sides.Mathematically,for a trapezium $ABCD$ with $AB || CD$,where $P$ and $Q$ are midpoints of $AD$ and $BC$ respectively,the formula is:$PQ = \frac{1}{2} (AB + CD)$Given $AB = 18 \, cm$ and $PQ = 15 \, cm$,we substitute these values into the formula:$15 = \frac{1}{2} (18 + CD)$Multiply both sides by $2$:$30 = 18 + CD$Subtract $18$ from both sides:$CD = 30 - 18 = 12 \, cm$.Therefore,the length of $CD$ is $12 \, cm$.

In trapezium $ABCD$,$AB || CD$. Points $P$ and $Q$ are the midpoints of $AD$ and $BC$ respectively. If $AB = 18 \, cm$ and $PQ = 15 \, cm$,then $CD = \dots \, cm$.

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