In the diagram, right-angled triangles $A E D$ and $B F C$ are constructed inside rectangle $A B C D$ so that $F$ lies on $D E$. If $A E=21, E D=72$ and $B F=45$, what is the length of $A B$ ?
(A) 50
(B) 48
(C) 52
(D) 54
(E) 56

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-113.jpg?height=336&width=548&top_left_y=675&top_left_x=1187)
By the Pythagorean Theorem in $\triangle A E D, A D^{2}=A E^{2}+E D^{2}=21^{2}+72^{2}=5625$, so $A D=75$. Since $A B C D$ is a rectangle, $B C=A D=75$. Also, by the Pythagorean Theorem in $\triangle B F C$, $F C^{2}=B C^{2}-B F^{2}=75^{2}-45^{2}=3600$, so $F C=60$.

Draw a line through $F$ parallel to $A B$, meeting $A D$ at $X$ and $B C$ at $Y$.

To determine the length of $A B$, we can find the lengths of $F Y$ and $F X$.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-187.jpg?height=345&width=537&top_left_y=1527&top_left_x=848)

## Step 1: Calculate the length of $F Y$

The easiest method to do this is to calculate the area of $\triangle B F C$ in two different ways. We know that $\triangle B F C$ is right-angled at $F$, so its area is equal to $\frac{1}{2}(B F)(F C)$ or $\frac{1}{2}(45)(60)=1350$.

Also, we can think of $F Y$ as the height of $\triangle B F C$, so its area is equal to $\frac{1}{2}(F Y)(B C)$ or $\frac{1}{2}(F Y)(75)$.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-187.jpg?height=340&width=531&top_left_y=2267&top_left_x=859)

Therefore, $\frac{1}{2}(F Y)(75)=1350$, so $F Y=36$.

(We could have also approached this by letting $F Y=h, B Y=x$ and so $Y C=75-x$. We could have then used the Pythagorean Theorem twice in the two little triangles to create two equations in two unknowns.)

Since $F Y=36$, then by the Pythagorean Theorem,

$$
B Y^{2}=B F^{2}-F Y^{2}=45^{2}-36^{2}=729
$$

so $B Y=27$.

Thus, $Y C=B C-B Y=48$.

## Step 2: Calculate the length of $F X$

Method 1 - Similar triangles

Since $\triangle A E D$ and $\triangle F X D$ are right-angled at $E$ and $X$ respectively and share a common angle $D$, then they are similar.

Since $Y C=48$, then $X D=48$.

Since $\triangle A E D$ and $\triangle F X D$ are similar, then $\frac{F X}{X D}=\frac{A E}{E D}$ or $\frac{F X}{48}=\frac{21}{72}$ so $F X=14$.

Method 2 - Mimicking Step 1

Drop a perpendicular from $E$ to $A D$, meeting $A D$ at $Z$.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-188.jpg?height=325&width=708&top_left_y=1277&top_left_x=768)

We can use exactly the same argument from Step 1 to calculate that $E Z=\frac{504}{25}$ and that $Z D=\frac{1728}{25}$.

Since $D F$ is a straight line, then the ratio $F X: E Z$ equals the ratio $D X: D Z$, ie. $\frac{F X}{\frac{504}{25}}=\frac{48}{\frac{1728}{25}}$ or $\frac{F X}{504}=\frac{48}{1728}$ or $F X=14$.

Method 3 - Areas

Join $A$ to $F$. Let $F X=x$ and $E F=a$. Then $F D=72-a$.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-188.jpg?height=209&width=740&top_left_y=2031&top_left_x=752)

Since $A E=21$ and $E D=72$, then the area of $\triangle A E D$ is $\frac{1}{2}(21)(72)=756$.

Now, the area of $\triangle A E D$ is equal to the sum of the areas of $\triangle A E F$ and $\triangle A F D$, or

$$
756=\frac{1}{2}(21)(a)+\frac{1}{2}(75)(x)
$$

so $21 a+75 x=1512$ or $a+\frac{25}{7} x=72$.

Now in $\triangle F X D$, we have $F X=x, G D=48$ and $F D=72-a$.

By the Pythagorean Theorem, $x^{2}+48^{2}=(72-a)^{2}=\left(\frac{25}{7} x\right)^{2}=\frac{625}{49} x^{2}$.

Therefore, $48^{2}=\frac{576}{49} x^{2}$, or $x=14$.

Therefore, $A B=X Y=F X+F Y=36+14=50$.

ANSWER: (A)