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Trapezium is a four sided shape where two sides are parallel,side lengths and angles are not equal.

The parallel sides are called bases and the non parallel sides are called legs

Types of Trapezium

Isosceles Trapezium: “The legs or the not parallel sides are equal.”

Scalene Trapezium: “A trapezium with all the sides and angles of different measures’’

Right Trapezium: “ A right trapezium has at least two right angles”


1. In a trapezium at least two opposite sides are parallel.

2. The diagonals intersect each other .

3. The sides which are not parallel in a trapezium are not equal except in the Isosceles trapezium.

4.The line that joins the mid-points of the non-parallel sides is always parallel to the bases or parallel sides which is equal to half the sum of the parallel sides.

Area and Perimeter

The parallel sides on are called the bases. The one that is longer is called the big base ( B) and the other the small base (b) of the trapezium.

The area of the trapezium is equal to half the product of the sum of the bases with height.

The formula is written as:

$ \displaystyle A=\frac{h\left( B+b \right)}{2}$

The  perimeter is the total lengths of all sides, the sum of the two bases and the two other sides.

$ \displaystyle P=a+b+B+c$

Example 1: On the figure below we have given the sides of trapezium. Find the height$ \displaystyle DE$.


Solution: If we observe our isosceles trapezium we se that $ \displaystyle EFCD$ is a rectangle with $ \displaystyle EF=10$

From that we calculate $ \displaystyle AB-EF=22-10=12$

This value is to be shared equally for $ \displaystyle AE$ and $ \displaystyle QB$ because our two right angle triangles $ \displaystyle AED$ and $ \displaystyle BFC$ are congruent from rule 2 side-angle-side. ( check congruent triangles rules)

$ \displaystyle AE=FB=6$

To find the height $ \displaystyle DE$ we use the the Pythagorean theorem on triangle $ \displaystyle AED$

$ \displaystyle {{\left( {DE} \right)}^{2}}={{\left( {AD} \right)}^{2}}-{{\left( {AE} \right)}^{2}}$

$ \displaystyle {{\left( {DE} \right)}^{2}}={{\left( {10} \right)}^{2}}-{{\left( 6 \right)}^{2}}$

$ \displaystyle {{\left( {DE} \right)}^{2}}=100-36$

$ \displaystyle {{\left( {DE} \right)}^{2}}=64$

$ \displaystyle DE=8$

So the height is $ \displaystyle DE=CF=8$.

Example 2: On the figure below we are given an isosceles trapezium where $ \displaystyle DC=CF=5cm$ and $ \displaystyle \widehat{{ABC}}={{45}^{\circ }}$. Find the area.


Solution: Firstly we observe our figure and see that $ \displaystyle DEFC$ is a square since $ \displaystyle DC=EF$

Also $ \displaystyle DC$ and $ \displaystyle EF$ are heights of our trapezium.

The right angle triangle $ \displaystyle BFC$ is an isosceles right angle triangle because based on the property that all angles on a triangle add up to $ \displaystyle {{180}^{\circ }}$ we find that $ \displaystyle \widehat{{FCB}}={{180}^{\circ }}-({{45}^{\circ }}+{{90}^{\circ }})={{45}^{\circ }}$

Since our right angle triangle is an isosceles triangle then $ \displaystyle CF=FB=5cm$

Based on the same reasoning also $ \displaystyle DE=EA=5cm$

Now we find the big base: $\displaystyle AB=AE+EF+FB=$$\displaystyle 5+5+5=15cm$

The area is: $ \displaystyle A=\frac{{\left( {B+b} \right)\cdot h}}{2}$

$ \displaystyle A=\frac{{\left( {15+5} \right)\cdot 5}}{2}=50c{{m}^{2}}$

Example 3: Find the area of the trapezium with diagonals $ \displaystyle 8cm$ and $ \displaystyle 7cm$ and bases $ \displaystyle 6cm$ and $ \displaystyle 3cm$.


Solution: Firstly we build $ \displaystyle CE\bot AB$ and $ \displaystyle DF\bot AB$


We have $ \displaystyle EF=DC=3$

We note $ \displaystyle AF=x$

$ \displaystyle EB=AB-AF-FE$

$ \displaystyle EB=6-x-3=3-x$

On the triangles $ \displaystyle ACE$ and $ \displaystyle BDF$ we apply the Pythagorean theorem:

On triangle $ \displaystyle ACE$

$ \displaystyle {{\left( {CE} \right)}^{2}}={{\left( {AC} \right)}^{2}}-{{\left( {AE} \right)}^{2}}$

$ \displaystyle {{\left( {CE} \right)}^{2}}=49-{{(x+3)}^{2}}$

On triangle $ \displaystyle BDF$

$ \displaystyle {{\left( {DF} \right)}^{2}}={{\left( {BD} \right)}^{2}}-{{\left( {BF} \right)}^{2}}$

$ \displaystyle {{\left( {DF} \right)}^{2}}=64-{{(6-x)}^{2}}$

Since $ \displaystyle CE=DF$ we have:

$\displaystyle 49-{{(x+3)}^{2}}=$$ \displaystyle 64-{{(6-x)}^{2}}$

$\displaystyle 49-{{x}^{2}}-6x-9=$$\displaystyle 64-36+12x-{{x}^{2}}$

$\displaystyle 49-{{x}^{2}}-6x-9=$$\displaystyle 64-36+12x-{{x}^{2}}$

$ \displaystyle 40-6x=12x+28$

Check how to solve first degree equation

$ \displaystyle x=\frac{2}{3}$

$\displaystyle AE=AF+FE=$$\displaystyle \frac{2}{3}+3=\frac{{11}}{3}$

$\displaystyle {{\left( {CE} \right)}^{2}}=$$\displaystyle 49-{{(\frac{{11}}{3}+3)}^{2}}$


$ \displaystyle CE=\frac{{8\sqrt{5}}}{3}$

Now we find the area:

$ \displaystyle {{A}_{{ABCD}}}=\frac{{(AB+DC)\cdot CE}}{2}$

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