2026-08-31T11:35:08.257777 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL Figure 1 shows a cuboid \(ABCDEFGH\) in which the...

Every part of this question rests on one habit: locate a right-angled triangle inside the solid, then use Pythagoras for a length or a trigonometric ratio for an angle. In a cuboid every vertical edge is perpendicular to the horizontal base, and every face is a rectangle, which is what supplies the right angles.

(a) Show that \(AC = 15\,\mathrm{cm}\) [2]

\(AC\) is a diagonal of the rectangle \(ABCD\), and in a rectangle angle \(ABC = 90^{\circ}\), so Pythagoras applies in triangle \(ABC\):

\[AC^{2} = AB^{2} + BC^{2} = 12^{2} + 9^{2} = 144 + 81 = 225 \quad \textbf{[M1]}\] \[AC = \sqrt{225} = 15\,\mathrm{cm} \quad \textbf{[A1]}\]

(b) Show that \(AG = 17\,\mathrm{cm}\) exactly [2]

\(G\) is vertically above \(C\), so \(CG\) is perpendicular to the whole base plane and in particular to \(AC\). That makes angle \(ACG = 90^{\circ}\), and \(CG = AE = 8\,\mathrm{cm}\) since opposite vertical edges of a cuboid are equal.

\[AG^{2} = AC^{2} + CG^{2} = 15^{2} + 8^{2} = 225 + 64 = 289 \quad \textbf{[M1]}\] \[AG = \sqrt{289} = 17\,\mathrm{cm} \quad \textbf{[A1]}\]

Both targets are printed, so these four marks are for the working. Note how (a) feeds (b): the space diagonal is built from the base diagonal and the height, never from two edges directly.

(c) Angle between \(AG\) and the plane \(ABCD\), to \(0.1^{\circ}\) [3]

The angle between a line and a plane is the angle between the line and its projection onto that plane. Dropping a perpendicular from \(G\) to the base lands at \(C\), so the projection of \(AG\) is \(AC\) and the required angle is \(\angle GAC\). [M1]

Triangle \(GAC\) is right-angled at \(C\), with \(CG = 8\) opposite the angle and \(AC = 15\) adjacent to it, so use the tangent ratio:

\[\tan(\angle GAC) = \frac{CG}{AC} = \frac{8}{15} \quad \textbf{[M1]}\] \[\angle GAC = 28.072\ldots^{\circ} = 28.1^{\circ} \quad \textbf{[A1]}\]

Identifying the correct angle carries its own method mark, which is the real content of this part; using \(\sin^{-1}\dfrac{8}{17}\) is an equally valid route and gives the same value.

(d) Show \(BG = \sqrt{145}\,\mathrm{cm}\) and find angle \(AGB\), to \(0.1^{\circ}\) [3]

In the rectangular face \(BCGF\), angle \(BCG = 90^{\circ}\) because \(CG\) is vertical, so

\[BG^{2} = BC^{2} + CG^{2} = 9^{2} + 8^{2} = 81 + 64 = 145, \qquad BG = \sqrt{145}\,\mathrm{cm} \quad \textbf{[M1][A1]}\]

For the angle, note that \(AB\) is perpendicular to the plane \(BCGF\), so \(AB\) is perpendicular to \(BG\) and triangle \(ABG\) is right-angled at \(B\). With the angle at \(G\), \(AB = 12\) is opposite and \(BG = \sqrt{145}\) is adjacent:

\[\tan(\angle AGB) = \frac{12}{\sqrt{145}} \quad \Rightarrow \quad \angle AGB = 44.900\ldots^{\circ} = 44.9^{\circ} \quad \textbf{[A1]}\]

(e) Area of triangle \(ACG\) [3]

Part (b) established angle \(ACG = 90^{\circ}\). [B1] The two perpendicular sides therefore serve as base and height, so no trigonometry is needed:

\[\text{Area} = \frac{1}{2} \times AC \times CG = \frac{1}{2} \times 15 \times 8 \quad \textbf{[M1]}\] \[= 60\,\mathrm{cm^{2}} \quad \textbf{[A1]}\]

Using \(\tfrac{1}{2} \times AC \times AG\) is the error to avoid: \(AG\) is the hypotenuse of this triangle, not a height.

(f) Length of \(AM\), where \(M\) is the midpoint of \(GH\), to 3 s.f. [2]

Coordinates make this straightforward. Take \(A\) as the origin with \(AB\), \(AD\) and \(AE\) along the axes, so \(A(0,\,0,\,0)\), \(G(12,\,9,\,8)\) and \(H(0,\,9,\,8)\). The midpoint of \(GH\) is then

\[M\left(\frac{12 + 0}{2},\ 9,\ 8\right) = M(6,\, 9,\, 8).\] \[AM^{2} = 6^{2} + 9^{2} + 8^{2} = 36 + 81 + 64 = 181 \quad \textbf{[M1]}\] \[AM = \sqrt{181} = 13.4536\ldots = 13.5\,\mathrm{cm} \ (3\ \mathrm{s.f.}) \quad \textbf{[A1]}\]

(g) Angle between \(AM\) and the plane \(ABCD\), to \(0.1^{\circ}\) [2]

Apply the same projection principle as in part (c). The foot of the perpendicular from \(M\) to the base is \(N(6,\, 9,\, 0)\), so the required angle is \(\angle MAN\), and

\[AN = \sqrt{6^{2} + 9^{2}} = \sqrt{117}, \qquad MN = 8,\] \[\tan(\angle MAN) = \frac{8}{\sqrt{117}} \quad \textbf{[M1]}\] \[\angle MAN = 36.486\ldots^{\circ} = 36.5^{\circ} \quad \textbf{[A1]}\]

Common wrong turns on this question. Measuring the angle in part (c) or (g) between the line and a vertical edge instead of the base, which gives the complement, \(61.9^{\circ}\) and \(53.5^{\circ}\) respectively; a quick sanity check is that these space diagonals lie closer to the base than to the vertical, so their angles with the base must be under \(45^{\circ}\). Working in radian mode, which produces nonsense values. Rounding \(AC\) or \(AN\) before the final trigonometric step. Assuming \(M\) lies above the centre of the base, which it does not: \(N(6,\,9,\,0)\) is the midpoint of \(DC\), not the centre.

Check. \(AM = \sqrt{181} \approx 13.45\) must be shorter than the space diagonal \(AG = 17\) and longer than \(AH = \sqrt{81 + 64} = \sqrt{145} \approx 12.04\), since \(M\) sits between \(H\) and \(G\); it is. In part (d), \(AG^{2} = AB^{2} + BG^{2} = 144 + 145 = 289\), agreeing with part (b), and \(\angle GAB + \angle AGB\) should be \(90^{\circ}\): \(45.1^{\circ} + 44.9^{\circ} = 90^{\circ}\).