# The area of a triangle with base 24 cm is trisected by two segments parallel to the base. What is the length of the longer of these segments?

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### 1 Answer

Draw triangle ABC with base 24. Draw DE and FG parallel to BC. By assumption, the area of triangle ABC is trisected by DE and FG, so the area of triangle ADE is `1/3` the area of triangle ABC, and the area of triangle AFG is `2/3` the area of triangle ABC.

Thus Area ADE:Area ABC = 1:3

AndÂ Area AFG:Area ABC = 2:3

(1) Note that `Delta ABC ` ~ `Delta ADE` ~`Delta AFG` ;since the lines are parallel we have AA` ` ~.

(2) Since the triangles are similar the sides are in proportion. This proportion is the scale factor.

(3) We use the fact that for two similar figures, the ratio of the areas is the ratio of the scale factor squared.

With FG>DE we have:

` `Area `Delta AFG` :Area `Delta ABC` =`2:3` implies that the scale factor is the ratio `sqrt(2):sqrt(3)` .

Thus the ratio of FG:BC=`sqrt(2):sqrt(3)` or:

`FG:24=sqrt(2):sqrt(3)` ==> `(FG)/24=(sqrt(2))/sqrt(3)` ==> `FG=(24sqrt(2))/sqrt(3)=8sqrt(6)`

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**The length of the longer area trisector is `8sqrt(6)` **

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**Similar reasoning will lead you to DE=`8sqrt(3)` . If `h_1` is the height of `Delta ADE` and `h_2` the height of `Delta AFG` , and h the height of `Delta ABC` we see that the ratio of the heights is `h_1:h=1:sqrt(3)` and `h_2:h=sqrt(2):sqrt(3)` .

Consider a specific case: let h=1. Now, DEGF is a trapezoid whose area is `1/2(b_1+b_2)h` ; if h=1 then the area of `Delta ABC=12` ,`h_1/1=1/sqrt(3)==>h_1=1/sqrt(3),sqrt(2)/(sqrt(3))=h_2/1` and the area of the trapezoid will be 4. Now plugging in what we know we find:

`A=1/2(8sqrt(3)+8sqrt(6))(sqrt(2)/sqrt(3)-1/sqrt(3))=4` as required. (The height of the trapezoid is `h_2-h_1` )

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