Question

Which is the area of quadrilateral ABDE?

Answer 1

Can you upload it as an image?

Answer 2

I attached one, I'm not sure if I did it right though @Owlcoffee

Answer 3

I cannot open it, I'm sorry :(

Answer 4

It's okay, thank you for trying :)

Answer 5

if the vertices's are (x1, y1)......(x4,y4) then area of abcd is=(1/2){x1y2-x2y1+x2y3-x3y2+x3y4-x4y3+x4y1-x1y4}

Answer 6

I drew one the best I could @Owlcoffee @Er.Mohd.AMIR

Answer 7

AB=40 SO CD=40 AC=32 DE=10 BD=32 BREAK FIG IN RECTANGLE ABCD AND TRIANGLE BED SO AREA OF RECTANGLE =40*32=1280 AND AREA OF TRIANGLE=1/2(32*10)=160 TOTAL AREA= 1440

Answer 8

Before we begin handling numbers we have to "prepare" everything, in the triangle, we need to know the catet, we know that the whole measure, and the rectangle composed, have a "mayor length" of 40 units, so therefore, the triangle has the low catet of 10 units.

I will mention that our objective is to find the area of the rectangle and then the triangle.
the whole area will be the sum of those areas:
\[(A) rect.ABDE+(A)tr.BDF=(A)cuad. ABFE\]

Answer 9

looks complex, but all the equation says is that the area of the rectangle and the area of the triangle compose the area of the whole cuadrilateral.

Answer 10

Thank you @Er.Mohd.AMIR @Owlcoffee

Answer 11

So, as a review, you calculate the area of a triangle and a rectangle with the following equations:
\[(A)rect.= b.h\]
\[(A)tri=\frac{ b.h }{ 2 }\]
So let's recognize the bases of each geometric body.
In ABED the base is ED, wich is 40m, and in BDF the base is DF wich is 10m.
and both heigths are shares and is 32m, so therefore:
\[b _{ABDE}.h+\frac{ b _{BDF}.h }{2 }=(A)cuad.ABFE\]
replacing:
\[(32)(40) + \frac{ (10)(32) }{ 2 } = (A)cuad.ABFE\]

Answer 12

Thank you, that helped a lot :3