# Length change of diagonal in deformable triangle

Hi

I have a problem understanding a figure in my lecture notes. The figure is the following one

It shows the deformation of a triangular element from time $t$ to time $t+dt$: So at t it is a isosceles triangle and at t+dt it is deformed. According to my lecture notes (page 16, eq. 28), the increase in the diagonal AC is given by

$$\delta(AC) = \frac{a+d}{\sqrt{2}} + \frac{b+c}{\sqrt{2}}$$

It is not clear to me why that is the case. I would just have said it should be
$$\sqrt{(a+d)^2 + (b+c)^2}$$
but this seems not to be the case. Does anyone see how one arrives at the first expression? I'd be very happy to get some help, I am pretty stuck.

Best,
Niles.

Last edited:

Nugatory
Mentor
It is not clear to me why that is the case. I would just have said it should be
$$\sqrt{(a+d)^2 + (b+c)^2}$$
but this seems not to be the case.
That's the correct calculation for the distance that the vertex C moves. However, the vertex C is not moving directly along the original AC line, so that's not the amount by which the distance AC changes.

If I am five km to the north of you, and I move one km, will the distance between us change by exactly one km? Only if I move due north or due south.