How this triangle subtraction works?

Here is a triangle, whose area is $$\frac{1}{2}(V-V_o)t$$, where $$V_o$$ and $$V$$ are $$y$$-coordinates.

$$\frac{1}{2}(V-V_o)t$$ is also $$\frac{1}{2}Vt - \frac{1}{2}V_ot$$ i.e., difference between two other triangles.

While this is true, how to geometrically see how this works?

• The triangle in blue and the triangle with the dashed sides have the same altitude and the same base. – Bernard Oct 30 '19 at 19:32
• this may help – AgentS Oct 30 '19 at 19:33
• Because the area of a triangle is base times the height and divide by 2. The base is in both cases $V-V_0$ and the height is in both cases $t$: $A=\frac{(V-V_0)\cdot t}{2}$ – callculus Oct 30 '19 at 19:33
• mathworld.wolfram.com/CavalierisPrinciple.html – Xander Henderson Oct 30 '19 at 19:35
• thanks all. Didn't know to see this way. And those links and you guys are very helpful. – Saran Oct 30 '19 at 19:37