Question 1210512: Shown below is rectangle $EFGH$. Its diagonals meet at $Y$. Let $X$ be the foot if an altitude is dropped from $E$ to $\overline{FH}$. If $FY = 24$ and $HX = 28$ and $EF = 10$, find the perimeter of rectangle $EFGH$.
Answer by CPhill(2138)   (Show Source):
You can put this solution on YOUR website! This is a challenging geometry problem that requires using properties of rectangles, the altitude theorem in right triangles, and the Pythagorean theorem.
Let's first establish the properties and known values.
## 1. Establish Known Values and Properties
| Property | Value | Reason |
| :---: | :--- | :--- |
| **$FY$** | 24 | Given |
| **$EF$** | 10 | Given |
| **$HX$** | 28 | Given |
| **$EY$** | 24 | Diagonals of a rectangle are equal and bisect each other. (All four segments $EY, FY, GY, HY$ are equal). |
| **$HY$** | 24 | Diagonals of a rectangle are equal and bisect each other. |
| **$FH$** | 48 | $FH = FY + HY = 24 + 24$. $FH$ is the diagonal. |
## 2. Find the Length of $FX$
The segment $FX$ is part of the diagonal $FH$.
$$FX = FH - HX$$
$$FX = 48 - 28$$
$$\mathbf{FX = 20}$$
## 3. Find the Length of Altitude $EX$
The triangle $\triangle EFH$ is a right triangle since $\angle FEH = 90^\circ$ (property of a rectangle).
**Wait:** The angle $\angle HEF$ is $90^{\circ}$, but $\overline{EX}$ is the altitude dropped from $E$ to the hypotenuse $\overline{FH}$.
* $EF$ is a leg of the right triangle $\triangle EFH$.
We can use the **Altitude Theorem** (specifically, the Geometric Mean Theorem for a leg) on the right triangle $\triangle EFH$ with the altitude $\overline{EX}$.
The relationship for the leg $\overline{EF}$ is:
$$EF^2 = FX \cdot FH$$
Let's check if the given values are consistent:
$$10^2 = 20 \cdot 48$$
$$100 = 960$$
**The equation is FALSE.** $100 \neq 960$.
This means the problem statement contains **contradictory geometric constraints**. It is impossible for a rectangle $EFGH$ to simultaneously have $FY=24$, $HX=28$, and $EF=10$.
***
## 💡 How to Proceed (Assuming a Typo in $EF$)
Since a perimeter is requested, we must assume there was a typo and proceed by assuming **$EF^2 = FX \cdot FH$ is the correct theorem** and using the given values for $FY$, $HX$, and the properties of the rectangle to find $EF$.
### Recalculating $EF$ (Assuming other values are correct)
If $FY=24, HX=28$ are correct:
* $FH = 48$
* $FX = 20$
* $EF^2 = FX \cdot FH = 20 \cdot 48 = 960$
* $EF = \sqrt{960} = \sqrt{64 \cdot 15} = 8\sqrt{15}$
### 4. Find the Length of $EH$ (The other side)
We use the Pythagorean theorem on the right triangle $\triangle EFH$:
$$EF^2 + EH^2 = FH^2$$
Using $EF^2 = 960$ and $FH^2 = 48^2 = 2304$:
$$960 + EH^2 = 2304$$
$$EH^2 = 2304 - 960$$
$$EH^2 = 1344$$
$$EH = \sqrt{1344} = \sqrt{64 \cdot 21} = 8\sqrt{21}$$
### 5. Calculate the Perimeter
The perimeter $P$ of rectangle $EFGH$ is $2(EF + EH)$.
$$P = 2(8\sqrt{15} + 8\sqrt{21})$$
$$P = 16(\sqrt{15} + \sqrt{21})$$
***
## Final Conclusion
Due to the geometric contradiction ($100 \neq 960$), this problem has no solution as written.
**If we assume the intent was to find the perimeter given $FY=24$ and $HX=28$ (forcing $EF=8\sqrt{15}$):**
$$\text{Perimeter} = \mathbf{16(\sqrt{15} + \sqrt{21})}$$
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