Find the area of ​the shaded parabola segment

Question

For reference: Figure P1, drawn below, is called a parabola segment. It is bounded by a parabola arc and a line segment perpendicular to the axis of symmetry of the parabola. The height is 1 cm and the base is 2 cm On the line segment that joins the lower left vertex of the line segment to the vertex of the parabola, a parabola segment P2 was constructed, similar to P1, as shown in the figure below. What is the area of ​​the shaded part in the figure below? (Note: the shaded area corresponds to points in P2 that do not belong to P1).

I think I need to use the theorem: The area of ​​the parabolic segment is four-thirds the area of ​​a triangle with the same base and same height

$SP_1 = \frac{4}{3}(\frac{2.1}{2})=\frac{4}{3}$

$ \triangle P_2 \sim \triangle P_1: \frac{h}{1}=\frac{\sqrt2}{2}\implies h = \frac{\sqrt2}{2}$

$SP_2 = \frac{4}{3}(\frac{\sqrt2.\frac{\sqrt2}{2}}{2})=\frac{1}{3}$

I am not able to calculate the shaded area Obs: You cannot use integrals in this exercise.

Answer 1

Just improving upon Arthur's answer.

$A(x)$ represents area of $x$. And $x$ is a curved shape unless specified by $\triangle$. Let the area of shaded region be S. Notice that $$A(\triangle ABC) + A(P_2) +A(ACD) = A(P_1) + S\\ \frac{1}{2} + \frac{1}{3} + \frac{1}{2} \times \frac{4}{3} = \frac{4}{3} + S$$ Now solve the equation to get your answer.

Answer 2

Look at your figure. In the left half of $P_1$ you have drawn an isosceles right triangle with leg lengths 1. Calculate the area of the wedge that lies between that triangle and the shaded area (this is half the area of $P_1$, minus the area of the triangle).

Now the area of the shaded area is the area of $P_2$, minus the area of that wedge.