The Great Pyramid of Giza in Egypt is a right pyramid with a square base. The pyramid is made of solid stone. The sides of the base are \(230\,{\text{m}}\) long. The diagram below represents this pyramid, labelled \({\text{VABCD}}\).

\({\text{V}}\) is the vertex of the pyramid. \({\text{O}}\) is the centre of the base, \({\text{ABCD}}\) . \({\text{M}}\) is the midpoint of \({\text{AB}}\). Angle \({\text{ABV}} = 58.3^\circ \) .

Show that the length of \({\text{VM}}\) is \(186\) metres, correct to three significant figures.

Calculate the height of the pyramid, \({\text{VO}}\) .

Find the volume of the pyramid.

Write down your answer to part (c) in the form \(a \times {10^k}\)  where \(1 \leqslant a < 10\) and \(k \in \mathbb{Z}\) .

Ahmad is a tour guide at the Great Pyramid of Giza. He claims that the amount of stone used to build the pyramid could build a wall \(5\) metres high and \(1\) metre wide stretching from Paris to Amsterdam, which are \(430\,{\text{km}}\) apart.

Determine whether Ahmad’s claim is correct. Give a reason.

Ahmad and his friends like to sit in the pyramid’s shadow, \({\text{ABW}}\), to cool down.
At mid-afternoon, \({\text{BW}} = 160\,{\text{m}}\)  and angle \({\text{ABW}} = 15^\circ .\)

i)     Calculate the length of \({\text{AW}}\) at mid-afternoon.

ii)    Calculate the area of the shadow, \({\text{ABW}}\), at mid-afternoon.

\(\tan \,(58.3) = \frac{{{\text{VM}}}}{{115}}\)   OR \(115 \times \tan \,(58.3^\circ )\)              (A1)(M1)

Note: Award (A1) for \(115\,\,\left( {ie\,\frac{{230}}{2}} \right)\)   seen, (M1) for correct substitution into trig formula.

\(\left( {{\text{VM}} = } \right)\,\,186.200\,({\text{m}})\)        (A1)

\(\left( {{\text{VM}} = } \right)\,\,186\,({\text{m}})\)           (AG)

Note: Both the rounded and unrounded answer must be seen for the final (A1) to be awarded.

\({\text{V}}{{\text{O}}^2} + {115^2} = {186^2}\)  OR \(\sqrt {{{186}^2} - {{115}^2}} \)       (M1)

Note: Award (M1) for correct substitution into Pythagoras formula. Accept alternative methods.

\({\text{(VO}} = )\,\,146\,({\text{m}})\,\,(146.188...)\)       (A1)(G2)

Note: Use of full calculator display for \({\text{VM}}\) gives \(146.443...\,{\text{(m)}}\).

Units are required in part (c)

\(\frac{1}{3}({230^2} \times 146.188...)\)       (M1)

Note: Award (M1) for correct substitution in volume formula. Follow through from part (b).

\( = 2\,580\,000\,{{\text{m}}^3}\,\,(2\,577\,785...\,{{\text{m}}^3})\)       (A1)(ft)(G2)

Note: The answer is \(2\,580\,000\,{{\text{m}}^3}\) , the units are required. Use of \({\text{OV}} = 146.442\) gives  \(2582271...\,{{\text{m}}^3}\)

Use of \({\text{OV}} = 146\) gives  \(2574466...\,{{\text{m}}^3}.\)

\(2.58 \times {10^6}\,({{\text{m}}^3})\)       (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for \(2.58\) and (A1)(ft) for \( \times {10^6}.\,\)

Award (A0)(A0) for answers of the type: \(2.58 \times {10^5}\,({{\text{m}}^3}).\)

Follow through from part (c).

the volume of a wall would be \(430\,000 \times 5 \times 1\)       (M1)

Note: Award (M1) for correct substitution into volume formula.

\(2150000\,({{\text{m}}^3})\)       (A1)(G2)

which is less than the volume of the pyramid       (R1)(ft)

Ahmad is correct.       (A1)(ft)

OR

the length of the wall would be \(\frac{{{\text{their part (c)}}}}{{5 \times 1 \times 1000}}\)       (M1)

Note: Award (M1) for dividing their part (c) by \(5000.\)

\(516\,({\text{km}})\)          (A1)(ft)(G2)

which is more than the distance from Paris to Amsterdam       (R1)(ft)

Ahmad is correct.       (A1)(ft)

Note: Do not award final (A1) without an explicit comparison. Follow through from part (c) or part (d). Award (R1) for reasoning that is consistent with their working in part (e); comparing two volumes, or comparing two lengths.

Units are required in part (f)(ii).

i)     \({\text{A}}{{\text{W}}^2} = {160^2} + {230^2} - 2 \times 160 \times 230 \times \cos \,(15^\circ )\)       (M1)(A1)

Note: Award (M1) for substitution into cosine rule formula, (A1) for correct substitution.

\({\text{AW}} = 86.1\,({\text{m}})\,\,\,(86.0689...)\)       (A1)(G2)

Note: Award (M0)(A0)(A0) if \({\text{BAW}}\) or \({\text{AWB}}\)  is considered to be a right angled triangle.

ii)    \({\text{area}} = \frac{1}{2} \times 230 \times 160 \times \sin \,(15^\circ )\)         (M1)(A1)

Note: Award (M1) for substitution into area formula, (A1) for correct substitutions.

\( = 4760\,{{\text{m}}^2}\,\,\,(4762.27...\,{{\text{m}}^2})\)       (A1)(G2)

Note: The answer is \(4760\,{{\text{m}}^2}\) , the units are required.

Question 4: Trigonometry, volume and area.
Many were able to write a correct trig ratio for part (a). The most common error was not to write the unrounded or the rounded answer. Some incorrectly used the given value of 186 in their proof. Part (b) was mostly answered correctly, with only a few candidates using Pythagoras’ Theorem incorrectly. Most candidates used the correct formula to calculate the volume of the pyramid, but some did not find the correct area for the base of the pyramid. Some lost a mark for missing or for incorrect units. Even with an incorrect answer for part (c), candidates did very well on part (d). In part (e) some excellent justifications were given. However, many struggled to convert kilometres to metres, others were confused and compared surface area instead of volume. Some thought the volumes needed to be the same. For part (f) candidates often assumed a right angle at BAW or BWA. When they used the sine and cosine rule, this was mostly done correctly.

Question 4: Trigonometry, volume and area.
Many were able to write a correct trig ratio for part (a). The most common error was not to write the unrounded or the rounded answer. Some incorrectly used the given value of 186 in their proof. Part (b) was mostly answered correctly, with only a few candidates using Pythagoras’ Theorem incorrectly. Most candidates used the correct formula to calculate the volume of the pyramid, but some did not find the correct area for the base of the pyramid. Some lost a mark for missing or for incorrect units. Even with an incorrect answer for part (c), candidates did very well on part (d). In part (e) some excellent justifications were given. However, many struggled to convert kilometres to metres, others were confused and compared surface area instead of volume. Some thought the volumes needed to be the same. For part (f) candidates often assumed a right angle at BAW or BWA. When they used the sine and cosine rule, this was mostly done correctly.

Question 4: Trigonometry, volume and area.
Many were able to write a correct trig ratio for part (a). The most common error was not to write the unrounded or the rounded answer. Some incorrectly used the given value of 186 in their proof. Part (b) was mostly answered correctly, with only a few candidates using Pythagoras’ Theorem incorrectly. Most candidates used the correct formula to calculate the volume of the pyramid, but some did not find the correct area for the base of the pyramid. Some lost a mark for missing or for incorrect units. Even with an incorrect answer for part (c), candidates did very well on part (d). In part (e) some excellent justifications were given. However, many struggled to convert kilometres to metres, others were confused and compared surface area instead of volume. Some thought the volumes needed to be the same. For part (f) candidates often assumed a right angle at BAW or BWA. When they used the sine and cosine rule, this was mostly done correctly.

Question 4: Trigonometry, volume and area.
Many were able to write a correct trig ratio for part (a). The most common error was not to write the unrounded or the rounded answer. Some incorrectly used the given value of 186 in their proof. Part (b) was mostly answered correctly, with only a few candidates using Pythagoras’ Theorem incorrectly. Most candidates used the correct formula to calculate the volume of the pyramid, but some did not find the correct area for the base of the pyramid. Some lost a mark for missing or for incorrect units. Even with an incorrect answer for part (c), candidates did very well on part (d). In part (e) some excellent justifications were given. However, many struggled to convert kilometres to metres, others were confused and compared surface area instead of volume. Some thought the volumes needed to be the same. For part (f) candidates often assumed a right angle at BAW or BWA. When they used the sine and cosine rule, this was mostly done correctly.

Question 4: Trigonometry, volume and area.
Many were able to write a correct trig ratio for part (a). The most common error was not to write the unrounded or the rounded answer. Some incorrectly used the given value of 186 in their proof. Part (b) was mostly answered correctly, with only a few candidates using Pythagoras’ Theorem incorrectly. Most candidates used the correct formula to calculate the volume of the pyramid, but some did not find the correct area for the base of the pyramid. Some lost a mark for missing or for incorrect units. Even with an incorrect answer for part (c), candidates did very well on part (d). In part (e) some excellent justifications were given. However, many struggled to convert kilometres to metres, others were confused and compared surface area instead of volume. Some thought the volumes needed to be the same. For part (f) candidates often assumed a right angle at BAW or BWA. When they used the sine and cosine rule, this was mostly done correctly.

Question 4: Trigonometry, volume and area.
Many were able to write a correct trig ratio for part (a). The most common error was not to write the unrounded or the rounded answer. Some incorrectly used the given value of 186 in their proof. Part (b) was mostly answered correctly, with only a few candidates using Pythagoras’ Theorem incorrectly. Most candidates used the correct formula to calculate the volume of the pyramid, but some did not find the correct area for the base of the pyramid. Some lost a mark for missing or for incorrect units. Even with an incorrect answer for part (c), candidates did very well on part (d). In part (e) some excellent justifications were given. However, many struggled to convert kilometres to metres, others were confused and compared surface area instead of volume. Some thought the volumes needed to be the same. For part (f) candidates often assumed a right angle at BAW or BWA. When they used the sine and cosine rule, this was mostly done correctly.