Question

ABCD is a rectangle with E as a point on CD and F is a point on BC such that `angle`AEF = `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?

Answer 1

`384` `cm^2`

Explanation:

Let's start by finding `AE` and `EF`. We know that:

`AE^2 + EF^2 = 25^2`

One possibility is the `7`-`24`-`25` triangle, where `AE = 24` and `EF = 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 `EC` and `CF`), 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*5=15` and `4*5=20`. Since the diagram depicts `AE` as being longer than `EF`, let's say that:

`AE = 20`
`EF = 15`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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 `AE = 20` as a hypotenuse:

`AD^2 + DE^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`.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The same thing can be done for the triangle with hypotenuse `EF = 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`.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
We have one final triangle to solve -- the one with `AF` as a hypotenuse and `AB` and `BF` 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 = 1/2(15)(20) + 1/2(12)(16) + 1/2(9)(12) + 1/2(7)(24)`

`A = 150 + 96 + 54 + 84`

`A = 384` or just `16xx24=384`

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

Final Answer