In triangle ABC, D is a point on line AB and E is a point on

[Fouad1013]

In triangle ABC, D is a point on line AB and E is a point on line AC so that line DE is parallel to line BC. If line BC = 20 and the area of trapezoid DBCE is one-fourth the area of triangle ABC, then find the length of line DE.

 

[stapel]

You have formulas for the areas of triangles and trapezoids, based on their bases and heights. I would suggest trying them.

Draw the picture, if you haven't already. Draw the vertical line from point A to side BC, perpendicular to BC (and thus also to DE), which indicates the height "n" of the triangle ABC.

Label the portion of the height line within the new triangle AED as "h₁" and the height line within the trapezoid as "h₂". What equations can you create with this information? How far can you get toward solving the exercise?

Please reply showing all of your steps. Thank you.

Eliz.

 

[Fouad1013]

I got the formula for the areas
The area for the trapezoid will be h2(20+DE)
The area for triangle ABC will be (h2+h1)(20)/2
The area for triangle ADE will be h1(DE)/2

What do i do after this please reply
Thank You

 

[Fouad1013]

i got that DE=-80+10 square root 61

pleese tell me if this is right

 

[stapel]

i got that DE=-80+10 square root 61

pleese tell me if this is right

Since this says that the length of DE is a negative value, it is unlikely that this is correct.

Please reply showing all of your steps. Thank you.

Eliz.

 

[Denis]

Fouad, use a right triangle: make AB the hypotenuse;
triangle ABC will then be similar to triangle ADE;
you should have no problems: fairly easy.
You will already have ABC's height, which is AC.

 

[Fouad1013]

I think i got it this time.
Well we will draw the altitude from A. The section in triangle ADE will be labeled h1 and the section the trapazoid will be h2. 4 times the area of the trapazoid equals the area of the large triangle. And the area of the small triangle + the area of the trapazoid also equals the area of the large triangle. Then we can say that the area of the small triangle is 3 times the area of the trapazoid. The equation will be (h1)DE/2 = 3{h2(DE+20)/2}. When we solve for h1 we will get h1=(3h2DE + 60h2)/DE.
We also now that the area of the large triangle is (h1+h2)20/2. And that that equals 4 times the area of the trapazoid 4(h2{DE+20}/2). We set those equal to eachother and we simplify we get DEsquared - 60DE - 1200 = 0 when we use the quadratic formula we get DE=30+10square root 21

Please tell if this is right
Thank you

 

[Denis]

.....we get DE=30+10square root 21

Fouad, THINK a bit: since BC = 20, is your solution possible?

I told you to also use SIMILAR triangles ABC and ADE.

Let d = DE, e = height of trapezoid, f = height of triangle AED

From similar triangles:
f / d = (f + e) / 20
e = (20f - df) / d

Using ABC = 4(DBCE):
20(e + f) / 2 = 4[e(d + 20) / 2]
e = 5f / (d + 15)

So: (20f - df) / d = 5f / (d + 15)
Solve that for d (the f's will cancel out)

AND (as I told you earlier), using a right triangle would be much easier :idea:

 

[Fouad1013]

Thank you very much