# ABCD is a rectangle with E as a point on CD and F is a point on BC such that angleAEF = 90^@ and AF = 25cm. The length of DE, EC, CF, FB, AE and EF are positive integers. What is the area of rectangle ABCD, in cm2?

## Jul 10, 2017

$384$ $c {m}^{2}$

#### Explanation:

Let's start by finding $A E$ and $E F$. We know that:

$A {E}^{2} + E {F}^{2} = {25}^{2}$

One possibility is the $7$-$24$-$25$ triangle, where $A E = 24$ and $E F = 7$. However, if we were to then try to find right triangles with a hypotenuse of 7 (which we would have to do to find $E C$ and $C F$), we would find that there is no combination of perfect squares that add up to ${7}^{2} = 49$. So the $7$-$24$-$25$ triangle doesn't work.

The only other right triangles we can make are ones where everything is scaled up by a certain factor. In this case, 25 only has factors of 5, so the only other triangles we can make are ones with a hypotenuse of $5$, and then multiply every side by $5$ to get the correct size.

The only Pythagorean triple with a hypotenuse of $5$ is the $3$-$4$-$5$ triangle.

If we multiply all of the sides of this triangle by $5$, we get that the legs must be $3 \cdot 5 = 15$ and $4 \cdot 5 = 20$. Since the diagram depicts $A E$ as being longer than $E F$, let's say that:

$A E = 20$
$E F = 15$

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Now we've figured out our two leg lengths for the shaded triangle. As it turns out, these two legs are both a hypotenuse for another triangle which must have all integer side lengths. Let's start with the triangle with $A E = 20$ as a hypotenuse:

$A {D}^{2} + D {E}^{2} = {20}^{2}$

Using trial and error, or from memory, one can deduce that the only pair of numbers which produce a hypotenuse of 20 is (yet again) a $3$-$4$-$5$ triangle scaled up.

This time, we need to scale everything up by a factor of $4$. This means that our $3$-$4$-$5$ triangle turns into a $12$-$16$-$20$ triangle.

This means that the two legs of this triangle are $12$ and $16$.

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The same thing can be done for the triangle with hypotenuse $E F = 15$.

Again, we see that we must take a $3$-$4$-$5$ triangle and scale it up. This time, it is by a factor of 3.

This means our $3$-$4$-$5$ triangle becomes a $9$-$12$-$15$ triangle.

This means that the two legs of this triangle are $9$ and $12$.

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We have one final triangle to solve -- the one with $A F$ as a hypotenuse and $A B$ and $B F$ as legs.

The only other Pythagorean triple with $25$ as a hypotenuse is the $7$-$24$-$25$ triangle, and based on the diagram, this must be what the last triangle is (note how the two triangles with hypotenuse $25$ are noticeably different, and since one was the scaled $3$-$4$-$5$, the other must be the $7$-$24$-$25$).

This means that the two legs of this triangle are $7$ and $24$.

The dimensions in the figure now appeaar as shown below: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now, to find the area of the whole rectangle. Notice that we've found the leg lengths of all 4 triangles that this shape was divided into. This means that we can find the areas of all 4 triangles, and then simply add them all together to get the area of the rectangle.

$A = \frac{1}{2} \left(15\right) \left(20\right) + \frac{1}{2} \left(12\right) \left(16\right) + \frac{1}{2} \left(9\right) \left(12\right) + \frac{1}{2} \left(7\right) \left(24\right)$

$A = 150 + 96 + 54 + 84$

$A = 384$ or just $16 \times 24 = 384$

So the area of the rectangle is $384$ square centimeters.