Question

What is the area of trapezoid ABCD? A(-2,2) B(2,5) C(11,-7) D(1,-2)

Answer 1

Answer:

50 sq. units

Step-by-step explanation:

Join one diagonal of the trapezoid ABCD.

Assume that AC is the diagonal.

So, Area of the trapezoid = Area of Δ ABC + Area of Δ CAD.  

Now, coordinates of A(-2,2), B(2,5), C(11,-7) and D(1,-2) are given.

We know, that three vertices of a triangle are $(x_{1},y_{1} )$, $(x_{2},y_{2} )$ and $(x_{3},y_{3} )$ respectively, then the area of the triangle will be given by  

$\frac{1}{2} |x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+ x_{3}(y_{1}-y_{2} ) |$

Hence, area of Δ ABC will be  

$\frac{1}{2} |-2(5-(-7))+2(-7-2)+11(2-5)|= 37.5$ sq units.

Again, area of Δ CAD will be  

$\frac{1}{2} |11(2-(-2))-2(-2-(-7))+1(-7-2)|= 12.5$ sq units.

Therefore, the area of the trapezoid ABCD will be (37.5 + 12.5) = 50 sq, units. (Answer)