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Consider the semicircle sits on an isosceles triangle.

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$\"A(\\theta)\"$ is the area of the semicircle.

$\"B(\\theta)\"$ is the area of the triangle $\"PQR\"$ .

Consider the radius of the semicircle as $\"r\"$ .

In the right angle triangle, $\"\\sin\\theta=\\frac{\\textrm{opposite$ .

From the figure,

$\"\\sin\\frac{\\theta}{2}=\\frac{r}{10}\"$

$\"r=10\\sin\\frac{\\theta}{2}\"$ .

Area of the semicircle is $\"A=\\frac{1}{2}\\pi$ .

Substitute $\"r=10\\sin\\frac{\\theta}{2}\"$ in the $\"A=\\frac{1}{2}\\pi$ .

$\"A(\\theta)=\\frac{1}{2}\\pi$

$\"\\\\=\\frac{1}{2}\\pi$

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Area of the right angle triangle is $\"A=\\frac{1}{2}\\times$ .

Here base of the right angle triangle is radius of the semi circle.

$\"r=10\\sin\\frac{\\theta}{2}\"$

Therefore $\"\\mathrm{base}=10\\sin\\frac{\\theta}{2}\"$ .

In the right angle triangle, $\"\\cos\\theta=\\frac{\\textrm{adjacent$ .

From the figure,

$\"\\\\\\cos\\frac{\\theta}{2}=\\frac{\\mathrm{height}}{10}\\\\$ .

Thus $\"\\mathrm{base}=10\\sin\\frac{\\theta}{2}\"$ and $\"\\\\{\\mathrm{height}=10\\cos$ .

Area of the triangle $\"PQR\"$ , $\"B(\\theta)=2\\times$ .

Substitute corresponding values in the above expression.

$\"\\\\B(\\theta)=2\\times$

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Determination of $\"\\lim_{\\theta\\rightarrow$ :

Substitute $\"\\\\A(\\theta)=50\\pi\\sin^2\\frac{\\theta}{2}\"$ and $\"\\\\B(\\theta)=100\\sin\\frac{\\theta}{2}\\cos\\frac{\\theta}{2}\"$ in the above expression.

$\"\\\\\\lim_{\\theta\\rightarrow$

$\"\\\\\\lim_{\\theta\\rightarrow$ .

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$\"\\\\\\lim_{\\theta\\rightarrow$ .