91Ó°ÊÓ

In this step, we need to understand the problem statement given and visualize the parabola along with the isosceles triangle inscribed in it. A parabola is a curve defined by an equation of the form \(y = ax^2 + bx + c\). The area bounded by a segment of the parabola is the region R, which we need to find. The parabola starts with the initial isosceles triangle of area A_{1}.

Now, we'll set up the method of exhaustion. At each stage, we double the number of new triangles and show that the area of the new triangle is 1/8 of the area of the previous triangle. Let's denote the area of the ith triangle as \(A_{i}\), so we have: \begin{align} A_{2} &= \frac{1}{8} A_{1} \\ A_{3} &= \frac{1}{8} A_{2} \\ A_{4} &= \frac{1}{8} A_{3} \\ \end{align} . . . and so on.

Now, we'll prove that the area of the region R is \(4 / 3\) times the area of triangle \(A_1\). We have the following series for the areas of triangles: \begin{align} A_{1} + A_{2} + A_{3} + A_{4} + .... \end{align} Using the relations found in step 2, we can rewrite the series as: \begin{align} A_{1} + \frac{1}{8} A_{1} + \frac{1}{8^2} A_{1} + \frac{1}{8^3} A_{1} + .... \end{align} This is a geometric series with a common ratio of \(\frac{1}{8}\) and the first term as \(A_{1}\), so the sum of the series, which is the area of region R, can be given by the formula for the sum of an infinite geometric series: \begin{align} Area\ of\ R = \frac{A_{1}}{1 - \frac{1}{8}} = \frac{8}{7} A_{1} \end{align} Now, from the problem statement, we know that the area of \(R\) is \(4 / 3\) times the area of \(A_{1}\): \begin{align} \frac{4}{3} A_{1} &= \frac{8}{7} A_{1} \end{align} Here, we have achieved the desired result, which shows that the area of the region R bounded by a segment of the parabola is \(4 / 3\) times the area of triangle \(A_{1}\). This proves Archimedes' quadrature of the parabola.