Question

prove by vector method that the median of trapezium is parallel to the parallel sides and is half of the sum of parallel sides
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Answer 1

800 is answer of this question

Answer 2

Given:

A trapezium ABCD with side AD ║ side BC

To Find:

We have to prove that the median of the trapezium MN is parallel to the parallel sides and is half of the sum of parallel sides.

Solution:

Let  $\[\overline a \]$, $\[\overline b \]$, $\[\overline c \]$, and $\[\overline d \]$ be respectively the position vectors of the vertices A, B, C, and D of the ABCD trapezium.

Then the vectors AD and BC are parallel.

Therefore, there exists a scalar k, such that

                $\[\begin{gathered} \overline {AD} = k \times \overline {BC} \hfill \\ \therefore \overline {AD} + \overline {BC} = k \times \overline {BC} + \overline {BC} \hfill \\ = \left( {k + 1} \right)\overline {BC\,} \,\,\,\,\,\,.......(1) \hfill \\ \end{gathered} \]$

Let m and n be the position vectors of the midpoints M and N of the non-parallel sides AB and DC respectively. Then line segment MN is the median of the trapezium.

By the midpoint formula,

 m =  $\frac{a+b}{2}$ and n = $\frac{d+c}{2}$

       $\[\begin{gathered} \therefore \overline {MN} {\text{ = n - m}} \hfill \\ {\text{ = }}\frac{{d + c}}{2}{\text{ - }}\frac{{a + b}}{2} \hfill \\ {\text{ = }}\frac{1}{2}\left[ {d + c - a - b} \right] \hfill \\ \end{gathered} \]$

        $\[\begin{gathered} {\text{ = }}\frac{1}{2}\left[ {\left( {d - a} \right) + \left( {c - a} \right)} \right] \hfill \\ {\text{ = }}\frac{{\overline {AD} {\text{ + }}\overline {BC} }}{2} \hfill \\ {\text{ = }}\frac{{\left( {k + 1} \right) \times \overline {BC} }}{2}\,\,\,...................\left( {By\,Eqn.1} \right) \hfill \\ \end{gathered} \]$

∴ $\overline {MN}$ and $\overline {BC}$ are parallel vectors

∴$\overline {MN}$ ║$\overline {BC}$ where

∴ The median MN is parallel to the parallel sides BC and AD of the trapezium.

Now $\overline {AD}$ and  $\overline {BC}$ are collinear.

Now

             $\[\begin{gathered} \left| {\overline {AD} + \overline {BC} } \right| = AD + BC \hfill \\ \therefore MN = \frac{1}{2}\left( {AD + BC} \right) \hfill \\ \end{gathered} \]$

Hence we prove that the median of the trapezium is parallel to the parallel sides and is half of the sum of parallel sides.