Length of the side of a similar triangle

Question

I am working on a problem and am a bit stuck.

We are given that triangle DCE is equilateral with side length $42$. As in $DE = EC = DC = 42$. We are then told that the area of the $\square DABC$ is equal to that of $\triangle AEB$. What is the length of $AB$?

I have determined that in order to solve this we need to understand the relationship between the area of a triangle and the area of a trapezoid. To do this we find:

$${1\over2} (B_1 + B_2)h = {1\over2}(b)(h)$$

$${1\over2} (x + 42)h = {1\over2}(x)(h)$$

Here is where I am a bit stuck. We now do not have an easy way to calculate the heights of either the trapezoid or the smaller triangle. What do I do?

Edit: I have reduced the problem to:

$${1\over8}(42 - x)^2(\sqrt{3}) = {1\over2}(42 + 42 - x)(h)$$

I guess the last missing piece is how do we find the height of the trapezoid?

Answer 1

Since none of the answers provided really explained the answer to this question - they mostly just gave an analytic solution - I have come back a year later to answer this question.

We are given that the triangle $\Delta$ABC is equilateral with side length = 42 and that the area of the parallelogram $\square$ADEC is half of the area of triangle $\Delta$ABC. (Shown by the picture below). We are then told to find the length of DE.

In order to solve this problem, you need to recognize that there are two similar triangles here (one inside of the other). Therefore we can conclude that triangle $\Delta$ABC is similar to triangle $\Delta$DEB. If we know that the length of the base of triangle $\Delta$ABC is 42, we can use this to find the length of the base of the other similar triangle.

First, as with working with any similarity problem, we need to find the linear scale factor between the two similar figures. In this case, we are given that the area of the parallelogram is $1\over2$ that of the entire figure so this is the linear scale factor. We then note that when using the similarity of figures, areas are related by $a^2 \over b^2$. As a result, we apply this transformation to the linear scale factor from our problem as such:

$$\sqrt{1\over2} = { { \sqrt{1} } \over { \sqrt{2}} } = {1 \over \sqrt{2}}$$

This is our new scale factor. Once we have this, we just simply multiply the scale factor by the side length to find the answer.

$$ {1 \over \sqrt{2}} \cdot 42 = {42 \over \sqrt{2}}$$

then

$$ {42 \over \sqrt{2}} \cdot {\sqrt{2} \over \sqrt{2}} = {42\sqrt{2} \over 2}$$

and then finally

$$21\sqrt{2}$$

Answer 2

Since triangle AEB is similar to triangle DEC we know that triangle AEB is also equilateral. We also know that area of AEB is half of area of DEC. Also, ratio of areas of similar triangles is ratio of square of corresponding sides. From that we get $$\ AB=21 \sqrt{2} $$ Hope it helps.

Answer 3

Let $FB=t$, then $EF$ will be $\sqrt 3t$ and $FG=21\sqrt 3-t\sqrt 3$ since $\angle EBF=\angle ECG=60°$.

So, you get $$\frac{1}{2}(2t)(t\sqrt 3)=\frac{1}{2}(2t+42)(21\sqrt 3-t\sqrt 3)$$

Answer 4

$\Delta AEB = \frac{1}{2} \Delta DEC$. Since $\Delta AEB \sim \Delta DEC$, by similarity, the side length of $AB$ is $\frac{1}{\sqrt2}$ that of $DC$. Therefore $AB = \frac{42}{\sqrt2} = \frac{42}{\sqrt2} \frac{\sqrt2}{\sqrt2} = 21\sqrt2$ directly.