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The line segment connecting the midpoints of the non-parallel sides of a trapezoid is called the mid-segment.

If we drew a line segment, between the two non-parallels sides. From the mid-point of both sides, the trapezium will be divided into two unequal parts.

Area\[ = \dfrac{k}{2}(AB + CD)\]

We know that

Area of trapezoid \[ = \]area of triangle (1)\[ + \] area of rectangle \[ + \]area of triangle (2)

That means

\[A = \dfrac{{ah}}{2} + bh + \dfrac{{ch}}{2}\]

\[A = \dfrac{{ah + 2{b_1}h + cb}}{2}\]

Simplifying the equation, rearranging the term and factoring result to

\[A = \dfrac{h}{2}\left[ {{b_1}\_(a + {b_1}\_c)} \right].........(1)\]

We assume the longer base of the trapezium be \[{b_2}\] then

\[{b_2} = a + {b_1} + c...........(2)\]

Substituting (2) in equation (1)

\[A = \dfrac{h}{2}\left( {{b_1} + {b_2}} \right)\]

Therefore, the arc of trapezium with b bases \[{b_1} + {b_2}\] and altitude h is

\[A = \dfrac{h}{2}\left( {{b_1} + {b_2}} \right)\]

Therefore

Given image= where ADCE is a rectangle

Given

\[BE = 3cm\] \[AB = 3 + 3 = 6cm\]

\[EA = 3cm\] \[CD = 3cm\]

\[AD = 8cm\] \[AD = 8cm\]

Area of trapezium \[ABCD = \dfrac{1}{2} \times \] parallel sides \[ \times \]height

\[ = \dfrac{1}{2} \times (AB + CD) \times AD\]

\[ = \dfrac{1}{2} \times (6 + 3) \times 8\] {BY putting the value of AB, CD and Ad in the formula}

\[ = \dfrac{1}{2} \times 9 \times 8 = 36c{m^2}\]

Hence,

The area of the trapezium

\[ABCD = 36c{m^2}\]

The sum of all the four angles of the trapezium is equal to \[{360^0}\]

It has two parallels and two non-parallels sides

The diagonals of regular trapezium bisect each other

The length of the mid-segment is equal to half the sum of the parallel’s bases, in trapezium

Two Paris of adjacent angles of trapezium formed between the parallel’s sides and one of the non-parallels side, add up to 180 degrees.