Cutting a rectangular piece from a right triangle

Question

A piece of sheet metal has the shape of an acute right-angled triangle AB with side BC and height AA' each having 12 cm. From it, a rectangle with two vertices is cut on the base BC and the other two on AB, respectively AC. Calculate the perimeter of this rectangle.

MY DRAWING

Okey, so, we know that MNPQ a rectangle(I named the points that made the rectangle M,N,P,Q). MN is parallel with AB and applying the the fundamental theorem of similarity, wecan say that the traingle MAQ is similar with CAB.

What should I do next??? Any idea is welcome! Thank you!

Answer 1

Mark point $X$ as the intersection of $AA'$ and $MQ$.

$MQ$ is parallel to $CB$ so $ABC$ and $AQM$ are similar triangles (as stated in the question).

$AA'=BC$, hence $MQ=AX$ by similar triangles. Clearly $XA'=QP$ so we have $AA=AX+XA'=MQ+QP=12$cm.

Since $MQ+QP$ is half the perimeter of $MNPQ$ the full perimeter is $24$cm.

Answer 2

Here are some proportions obtained from similar triangles:

$$\frac{MN}{AA'} = \frac{CM}{CA} \\ \frac{MQ}{CB} = \frac{AM}{CA}$$

Add up the LHS above and get $$\frac{MN}{AA'} + \frac{MQ}{CB} = \frac{CM+ AM}{CA}=1$$

In your case $AA'= CB$, so we get the perimeter of the rectangle constant, no matter how is situated.

$\bf{Added:}$ Say we introduce some coordinates on the height $AA'$. Then the perimeter of the rectangle corresponding to a point on the height is an affine function. But note that it takes equal values at $A$ and at $A'$. It follows that it is a constant function.