
Find the area of trapezium \[ABCD\] as given in the figure in which \[ADCE\] is a rectangle.
Answer
530.4k+ views
Hint: A trapezium is a quadrilateral which is defined as a shape with four sides and one set of parallels side.
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
Complete step by step answer:
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
\[ABCD = 36c{m^2}\]
Note: Properties of trapezium
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.
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
Complete step by step answer:
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
\[ABCD = 36c{m^2}\]
Note: Properties of trapezium
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.
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