geometry
The diagonals of a trapezoid are perpendicular and have lengths 8 and 10. Find the length of the median of the trapezoid.
I have no idea about this problem except assuming that the trpezoid is a rhombus. This is a duplicate, but the solution on The diagonals of a trapezoid are perpendicular and have lengths 8 and 10. Find the length of the median of the trapezoid. is a little bit too advanced for me. Can anyone solve this using a simpler method?
First Solution: Let our trapezoid be $ABCD$ as in the diagram supplied by pedja. Let the diagonals meet at $O$.
Note that $\triangle OAB$ and $\triangle OCD$ are similar. Indeed we know the scaling factor. Since $AB=20$ and $CD=7$, the sides of $\triangle OCD$ are $\frac{7}{20}$ times the corresponding sides of $\triangle OAB$.
That is very useful. We have $AC=13=AO+\frac{7}{20}AO$. It follows that $$AO=\frac{(20)(13)}{27}, \quad\text{and similarly,}\quad BO=\frac{(20)(5\sqrt{10})}{27}.$$
If we want to use the usual formula for the area of a trapezoid, all we need is the height of the trapezoid. That is $1+\frac{7}{20}$ times the height of $\triangle OAB$.
The height of $\triangle OAB$ can be found in various ways. For example, we can use the Heron Formula to find the area of $\triangle OAB$, since we know all three sides. Or else we can use trigonometry. The Cosine Law can be used to compute the cosine of $\angle OAB$. Then we can find an exact (or approximate) expression for the sine of that angle. From this we can find the height of $\triangle OAB$.
Second Solution: This is a variant of the first solution that uses somewhat more geometry. Let $\alpha$ be the area of $\triangle OAB$.
We first compute the area of $\triangle COB$. Triangles $OAB$ and $COB$ can be viewed as having bases $OA$ and $CO$ respectively, and the same height. But the ratio of $CO$ to $OA$ is $\frac{7}{20}$, so the area of $\triangle COB$ is $\frac{7}{20}\alpha$.
Since triangles $ABC$ and $ABD$ have the same area, by subtraction so do $\triangle COB$ and $\triangle DOA$. And since $\triangle OCD$ is $\triangle OAB$ scaled by the linear factor $\frac{7}{20}$, the area of $\triangle OCD$ is $\left(\frac{7}{20}\right)^2\alpha$. Putting things together, we find that the area of our trapezoid is $$\alpha +2\frac{7}{20}\alpha +\left(\frac{7}{20}\right)^2\alpha,\quad\text{that is,}\quad \left(\frac{27}{20}\right)^2\alpha.$$ Pretty! Finally, by the similarity argument of the first solution, we know the sides of $\triangle OAB$, so we can find $\alpha$ by using Heron's Formula.
Let $ABCD$ a trapezoid such that $FG$ is its median, $AC$ and $DB$ are perpendicular, $DC=w$, $AB =z$, $AC=8$ and $DB=10$.
Let $r$ such that $r \parallel AC$ and $D \in r$.
Let $s$ such that $A \in s$ and $B \in s$.
Let point $E$ such that $\{E\} =r \cap s$. See the figure below:
It follows that: $$DE=CA=8,$$ $$EA=w,$$ $$FG = \frac{w+z}{2} \quad (1)$$ and $$DE \perp DB.$$ Using the Pythagorean Theorem in $\triangle EDB$, we get: $$w+z= \sqrt{164} = 2 \sqrt{41}.$$ From $(1)$ we get: $$FG= \sqrt{41}.$$ Therefore it is possible to determine $FG$ without assuming a rhombus.
Best Answer
Assume that the diagonals cut themselves in four segments with lengths $a,b,c,d$, in such a way that $b+d=10,a+c=8$ and the right triangle with legs $a,b$ is similar to the right triangle with legs $c,d$, so that $\frac{b}{a}=\frac{d}{c}$ or, equivalently, $ad=bc$. That leads to the equation: $$ a(10-b)=b(8-a) $$ that is equivalent to $10a=8b$, hence $\frac{b}{a}=\frac{d}{c}=\frac{5}{4}$. So we have that all the trapezoids with perpendicular diagonals such that $a=x,\,b=\frac{5}{4}x,\,c=y,\,d=\frac{5}{4}y,\,x+y=8$ meet the given constraints. For such trapezoids, the bases have length $$ \frac{x}{4}\sqrt{41},\qquad \frac{y}{4}\sqrt{41} $$ by the Pythagorean theorem, hence the "median" has length $$ \frac{x+y}{8}\sqrt{41} = \color{red}{\sqrt{41}}.$$
In general, if a trapezoid has perpendicular diagonals with length $\ell_1,\ell_2$, the segment joining the midpoints of the diagonal sides has length $\frac{1}{4}\sqrt{\ell_1^2+\ell_2}$. Here it comes a proof (almost) without words: the red segments have the same length.