Exact form answers

3 Archimedes, about 2200 years ago, used regular polygons inside and outside circles to obtain approximations for \(\pi\).

1. Fig. 8.1 shows a regular 12 -sided polygon inscribed in a circle of radius 1 unit, centre \(\mathrm { O } . \mathrm { AB }\) is one of the sides of the polygon. C is the midpoint of AB . Archimedes used the fact that the circumference of the circle is greater than the perimeter of this polygon. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{c0fcd64b-8ca0-4309-9f58-c23cc4208f4d-2_457_422_457_936} \captionsetup{labelformat=empty} \caption{Fig. 8.1}
   \end{figure} (A) Show that \(\mathrm { AB } = 2 \sin 15 ^ { \circ }\).
   (B) Use a double angle formula to express \(\cos 30 ^ { \circ }\) in terms of \(\sin 15 ^ { \circ }\). Using the exact value of \(\cos 30 ^ { \circ }\), show that \(\sin 15 ^ { \circ } = \frac { 1 } { 2 } \sqrt { 2 - \sqrt { 3 } }\).
   (C) Use this result to find an exact expression for the perimeter of the polygon. Hence show that \(\pi > 6 \sqrt { 2 - \sqrt { 3 } }\).
2. In Fig. 8.2, a regular 12-sided polygon lies outside the circle of radius 1 unit, which touches each side of the polygon. F is the midpoint of DE. Archimedes used the fact that the circumference of the circle is less than the perimeter of this polygon. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{c0fcd64b-8ca0-4309-9f58-c23cc4208f4d-2_450_416_1562_940} \captionsetup{labelformat=empty} \caption{Fig. 8.2}
   \end{figure} (A) Show that \(\mathrm { DE } = 2 \tan 15 ^ { \circ }\).
   (B) Let \(t = \tan 15 ^ { \circ }\). Use a double angle formula to express \(\tan 30 ^ { \circ }\) in terms of \(t\). Hence show that \(t ^ { 2 } + 2 \sqrt { 3 } t - 1 = 0\).
   (C) Solve this equation, and hence show that \(\pi < 12 ( 2 - \sqrt { 3 } )\).
3. Use the results in parts (i)( \(C\) ) and (ii)( \(C\) ) to establish upper and lower bounds for the value of \(\pi\), giving your answers in decimal form.

16 On a unit circle, the inscribed regular polygon with 12 edges gives a lower bound for \(\pi\), and the escribed regular polygon with 12 edges gives an upper bound for \(\pi\). Calculate the values of these bounds for \(\pi\), giving your answers:

1. in surd form
2. correct to 2 decimal places. www.ocr.org.uk after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, The Triangle Building, Shaftesbury Road, Cambridge CB2 9 EA.
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\includegraphics{figure_8} The shape of a badge is a sector \(ABC\) of a circle with centre \(A\) and radius \(AB\), as shown in Fig 1. The triangle \(ABC\) is equilateral and has a perpendicular height 3 cm.

1. Find, in surd form, the length \(AB\). [2]
2. Find, in terms of \(\pi\), the area of the badge. [2]
3. Prove that the perimeter of the badge is \(\frac{2\sqrt{3}}{3}(\pi + 6)\) cm. [3]

\includegraphics{figure_1} The shape of a badge is a sector \(ABC\) of a circle with centre \(A\) and radius \(AB\), as shown in Fig 1. The triangle \(ABC\) is equilateral and has a perpendicular height 3 cm.

1. Find, in surd form, the length \(AB\). [2]
2. Find, in terms of \(\pi\), the area of the badge. [2]
3. Prove that the perimeter of the badge is \(\frac{2\sqrt{3}}{3}(\pi + 6)\) cm. [2]

\includegraphics{figure_1} The shape of a badge is a sector \(ABC\) of a circle with centre \(A\) and radius \(AB\), as shown in Fig 1. The triangle \(ABC\) is equilateral and has a perpendicular height 3 cm.

1. Find, in surd form, the length \(AB\). [2]
2. Find, in terms of \(\pi\), the area of the badge. [2]
3. Prove that the perimeter of the badge is \(\frac{2\sqrt{3}}{3}(\pi + 6)\) cm. [3]