In △abc a line segment de connects sides ab and bc so that de ∥ ac . the lengths of the sides of △dbe are four times shorter than the lengths of the sides of △abc. what is the area of △abc if the area of a trapezoid adec is 30cm^2?

  32 cm^2

Step-by-step explanation:

The sides of ∆DBE or 1/4 of the sides of ∆ABC, so its area is (1/4)^2 = 1/16 of the area of ∆ABC. The trapezoid's area is then 1 -1/16 = 15/16 of the total area of ∆ABC. Said another way, the area of ∆ABC is 16/15 times the area of the trapezoid.

  Area ∆ABC = (16/15)·(30 cm^2) = 32 cm^2

5/6

step-by-step explanation:

if i remember correctly with perpendicular stuff you just take away the negative

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3