Question

A right triangle has legs 4 cm and 3cm. How far from the vertex must this triangle be cut by a line parallel to the longer diagonal so that (a) the area of the small right triangle will be equal to the area of the trapezoid formed?(b) the perimeter of the small triangle is equal to the perimeter of the trapezoid?

Answer 1

Let's just consider a referential system that will be set on the vertex not containing the height of this triangle:

So, we can now take \(A(0,0)\) as the vertex of the triangle and further deduce the other coordinates, AB will be the leg with length 4cm, therefore B must be the point \(B(4,0)\)
And the remaining one being \(C(4,3)\) and since the area of the small triangle defined will be equal to the trapezoid, let's then make these areas equal, by first defining a point "T" and a point "M" belonging to AC and AB respectively.
We would have to find the hypotenuse (longer diagonal) in order to proceed with this.

Answer 2

Ah, wait, Let's change that a little and make "T" belong to CB, since the diving line must be parallel to the hypotenuse.

Answer 3

to find AC you can use pythagoras theorem
\(AB^2+BC^2=AC^2\)

Answer 4

These will divide the triangle in a trapezoid and a triangle so, these two points will have their coordinates \(T(x_t,y_t)\) and \(M(x_m,y_m)\) let's define them with the condition that \[A _{ACTM}=A _{MBT}\]
\[\frac{ 3(AC+TM) }{ 2 }=\frac{ AB.MB }{ 2 }\]

Answer 5

The hypotenuse is 5

Answer 6

oh no, failed again:
\[\frac{ 3(AC+TM) }{ 2 }=\frac{ MB.BT }{ 2 }\]

Answer 7

Maybe I'm doing this wrong... I am stretching a variabe without end...