Question

If non-parallel sides of an isosceles trapezium are prolonged, an equilateral triangle with sides of 6cm would be formed. Knowing that the trapezium is half the height of the triangle, calculate the area of the trapezium.

Answer 1

Hint: In this question, we will first find the side BG and then in triangle ABG which is a right triangle, by using pythagoras theorem we will find the altitude AG. After this we will calculate FG and finally find the area of the trapezium using the formula, area = $\dfrac{1}{2} \times $ (sum of the parallel sides$ \times $height).

Complete step-by-step answer:
It is given that BDEC is a trapezium and ABC is an equilateral triangle formed by extending the sides BD and EC.
Each side of triangle = 6cm
Height of triangle = 2(Height of trapezium).
To find the area of the trapezium, we have to find its height.
So first of all, we will find the height of the triangle.
AG is the altitude of triangle ABC.
We know that the altitude of an equilateral triangle divides the triangle into two equal halves. So, AG is also the median of the triangle.
$\therefore $ BG = GC = $\dfrac{6}{2}$ =3cm
Now, in $\vartriangle $ABG, we will apply pythagoras theorem to find the length of AG.
On applying pythagoras theorem in $\vartriangle $ABG, we have:
$A{B^2} = A{G^2} + B{G^2}$
$ \Rightarrow A{G^2} = A{B^2} - B{G^2}$
$ \Rightarrow AG = \sqrt {A{B^2} - B{G^2}} $
Putting the value of AB and BG, we have:
$ \Rightarrow AG = \sqrt {{6^2} - {{(3)}^2}} = \sqrt {36 - 9} = \sqrt {27} $ =5.2cm
We know that F is the midpoint of AG.
$\therefore $ FG = $\dfrac{{AG}}{2} = \dfrac{{5.2}}{2}$ = 2.6cm
Also, we know that formula to find area of trapezium is given by:
Area of trapezium = $\dfrac{1}{2} \times $ (sum of the parallel sides$ \times $height).
Now, we will calculate length ‘DE’.
We know that the line joining the mid points of two sides is parallel to the third side and also half of it.
\[\therefore \] DE=$\dfrac{{BC}}{2} = \dfrac{6}{2}$ =3cm.
Putting the values in above equation, we get:
Area of trapezium = $\dfrac{1}{2} \times \left\{ {\left( {BC + DE} \right) \times FG} \right\} = \dfrac{1}{2}(6 + 3) \times 2.6 = \dfrac{1}{2} \times 9 \times 2.6 = 11.70c{m^2}$
Therefore, the area of the trapezium = 11.70$c{m^2}$

Note: In this type of question, you should remember the formula to find the area of trapezium. You should remember the properties related to an equilateral triangle. All sides of an equilateral triangle are equal and the altitude is also the median of the triangle. You must remember the pythagoras theorem which states that the sum of squares of two sides is equal to the square of its third largest side.