Calculating a section of a sector of a circle when given the arc length and angle of the sector of the circle. This question requires the use of the formulas to find the area of a sector of a circle and to find the area of a triangle.
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", "advice": "To find the area of the minor segment cut off by $AB$, we need to calculate the area of the whole sector, and the triangle $OAB$.
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\nFirstly, we need to calculate the radius $r$ using the formula
\n\\[ s = r \\theta \\implies r = \\frac{s}{\\theta}. \\]
\nSince $s=\\var{s}$ cm and $\\theta=\\var{t}$ rad.,
\n\\[ \\begin{split} r &\\,= \\frac{\\var{s}}{\\var{t}} \\text{ cm} \\\\ &\\,= \\var{r2} \\text{ cm (2 d.p.)}. \\end{split} \\]
\nThe equation to find the area of a sector of a circle is $A_{sector} = \\frac{1}{2}\\theta r^2$:
\n\\[ \\begin{split} A_{sector} &\\,= \\frac{1}{2} \\times \\var{t} \\times \\var{r2}^2 \\\\ &\\,= \\var{sar} \\text{ cm$^2$ (2 d.p.)}. \\end{split} \\]
\nTo find the area of the triangle $OAB$ we want to use the formula $A_{triangle}=\\frac{1}{2}ab\\sin(\\theta)$, where $a$ and $b$ are vertices $OA$ and $OB$. Since these are both equal to the radius of the circle, $r=\\var{r2} \\text{ cm},$
\n\\[ \\begin{split} A_{triangle} &\\,= \\frac{1}{2} \\times \\var{r2}^2 \\times \\sin(\\var{t}), \\\\ &\\,= \\var{tar} \\text{ cm$^2$ (2.d.p.).} \\end{split} \\]
\nTherefore, the minor area cut off by $AB$ is the difference between the area of the sector and the area of the triangle:
\n\\[ \\begin{split} A_{minor} &\\,= A_{sector} -A_{triangle} \\\\ &\\,= \\var{sar}\\text{ cm$^2$}-\\var{tar}\\text{ cm$^2$} \\\\ &\\,= \\var{mar} \\text{ cm$^2$ (2.d.p.)}. \\end{split} \\]
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\n(Give your answer to 2 decimal places where necessary.)
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