Question

If non-parallel sides of an isosceles trapezium are prolonged, an equilateral triangle ABC 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

R.E.F. Image.

we know attitude is also a median in an equilateral triangle.

Therefore, BG = GC = 3 cm.

From $\Delta ABG$ by Pythagoras theorem.

$(AB{)}^2=(AG{)}^2+(BG{)}^2$

$(6{)}^2=(AG{)}^2+(3{)}^2$

$\left(AG\right)=\sqrt{36-9}$

$AG=\sqrt{27}=5.20cm.$

Height of trapezium = 2.60 cm $\to$ (given in question $\Rightarrow \frac{5.20}{2}=2.60cm$)

Parallel side of the trapezium: $a=BC=6cm$and $b=DE=\frac{BC}{2}=\frac{6}{2}=3cm$ {$\because$ its an equilateral triangle}

Therefore, Area of trapezium = $\frac{(a+b)\times h}{2}$

$=\frac{(6+3)\times 2.60}{2}=9\times 1.3=11.70c{m}^2.$