Let $\AA$ denote the area of the shaded region in the following diagram: <!-- thumbNewton's Approximation of Pi - from Journey Through Genius--> :500px Consider the semicircle embedded in the cartesian plane: :whose radius is $\dfrac 1 2$ and :whose center is the point $\tuple {\dfrac 1 2, 0}$.  We have: {{begin-eqn}} {{eqn | l = \paren {x - \frac 1 2}^2 + \paren {y - 0}^2       | r = \frac 1 2 ^2       | c = Equation of Circle }} {{eqn | ll= \leadsto       | l = y       | r = \sqrt {\paren {x - x^2} } }} {{eqn | r = \sqrt x \sqrt {\paren {1 - x} } }} {{eqn | r = \sqrt x \paren {1 - \frac 1 2 x - \frac 1 8 x^2 - \frac 1 {16} x^3 - \frac 5 {128} x^4 - \frac 7 {256} x^5 - \cdots}       | c = General Binomial Theorem }} {{eqn | r = \paren {x^{\frac 1 2} - \dfrac 1 2 x^{\frac 3 2} - \dfrac 1 8 x^{\frac 5 2} - \dfrac 1 {16} x^{\frac 7 2} - \dfrac 5 {128} x^{\frac 9 2} - \dfrac 7 {256} x^{\frac {11} 2} - \cdots} }} {{end-eqn}} We calculate $\AA$ in $2$ different ways. First we calculate it as the definite integral between $x = 0$ and $x = \dfrac 1 4$: {{begin-eqn}} {{eqn | l = \AA       | r = \int_0^\frac 1 4 \paren {x^{\frac 1 2} - \dfrac 1 2 x^{\frac 3 2} - \dfrac 1 8 x^{\frac 5 2} - \dfrac 1 {16} x^{\frac 7 2} - \dfrac 5 {128} x^{\frac 9 2} - \dfrac 7 {256} x^{\frac {11} 2} - \cdots} \rd x       | c =  }} {{eqn | r = \intlimits {\paren {\frac 2 3 x^{\frac 3 2} - \frac 1 5 x^{\frac 5 2} - \frac 1 {28} x^{\frac 7 2} - \frac 1 {72} x^{\frac 9 2} - \frac 5 {704} x^{\frac {11} 2} - \frac 7 {1664} x^{\frac {13} 2} - \cdots } } {x \mathop = 0} {x \mathop = \frac 1 4}       | c =  }} {{eqn | n = 1       | r = \paren {\dfrac 2 {3 \times 2^3} - \dfrac 1 {5 \times 2^5} - \dfrac 1 {28 \times 2^7} - \dfrac 1 {72 \times 2^9} - \dfrac 5 {704 \times 2^{11} } - \dfrac 7 {1664 \times 2^{13} } - \cdots}       | c =  }} {{end-eqn}} {{qed|lemma}} Then we calculate $\AA$ using the techniques of plane geometry. From the construction, we have that: :$AC = CD$ :$AB = BC$ :$BD$ is common so by Triangle Side-Side-Side Equality: :$\triangle ABD = \triangle CBD$ Thus: :$AD = AC = CD$ and so $\triangle ACD$ is equilateral. Thus we have that $\triangle BCD$ has angles $30 \degrees$, $60 \degrees$ and $90 \degrees$. Hence by Pythagoras's Theorem: {{begin-eqn}} {{eqn | l = BD^2 + BC^2       | r = CD^2       | c =  }} {{eqn | ll= \leadsto       | l = BD       | r = \sqrt {CD^2 - BC^2}       | c =  }} {{eqn | r = \sqrt {\paren {\dfrac 1 2}^2 - \paren {\dfrac 1 4}^2}       | c =  }} {{eqn | r = \dfrac 1 4 \sqrt {4 - 1}       | c =  }} {{eqn | r = \dfrac {\sqrt 3} 4       | c =  }} {{end-eqn}} Then we observe that $\AA$ is: :the area $\AA_S$ of the sector of the semicircle whose central angle is $\dfrac \pi 3$ minus: :the area $\AA_T$ of the right triangle $\triangle BCD$. $\AA_S$ is $\dfrac 1 6$ of the area of the circle whose radius is $\dfrac 1 2$. Thus from Area of Circle: :$\AA_S = \dfrac 1 6 \paren {\pi \paren {\dfrac 1 2}^2}$ $\AA_T$ is given from Area of Triangle in Terms of Side and Altitude as: :$\AA_T = \dfrac 1 2 \paren {\dfrac 1 4} \paren {\dfrac {\sqrt 3} 4}$ Hence: {{begin-eqn}} {{eqn | l = \AA       | r = \AA_S - AA_T       | c =  }} {{eqn | r = \dfrac 1 6 \paren {\pi \paren {\dfrac 1 2}^2} - \dfrac 1 2 \paren {\dfrac 1 4} \paren {\dfrac {\sqrt 3} 4}       | c =  }} {{eqn | n = 2       | r = \dfrac {\pi} {24} - \dfrac {\sqrt 3} {32}       | c =  }} {{end-eqn}} {{qed|lemma}} Finally, setting $(1)$ equal to $(2)$: {{begin-eqn}} {{eqn | l = \dfrac {\pi} {24} - \dfrac {\sqrt 3} {32}       | r = \paren {\dfrac 2 {3 \times 2^3} - \dfrac 1 {5 \times 2^5} - \dfrac 1 {28 \times 2^7} - \dfrac 1 {72 \times 2^9} - \dfrac 5 {704 \times 2^{11} } - \dfrac 7 {1664 \times 2^{13} } - \cdots}       | c =  }} {{eqn | ll= \leadsto       | l = \pi       | r = \dfrac {3 \sqrt 3} 4 + 24 \paren {\dfrac 2 {3 \times 2^3} - \dfrac 1 {5 \times 2^5} - \dfrac 1 {28 \times 2^7} - \dfrac 1 {72 \times 2^9} - \dfrac 5 {704 \times 2^{11} } - \dfrac 7 {1664 \times 2^{13} } - \cdots}       | c =  }} {{end-eqn}} {{qed}} {{Namedfor|Isaac Newton|cat = Newton}}