To move from the origin (0,0,0) to A, we only need to move 8 units in the positive \(x\)-direction.

We don't need to move in the \(y\)- or \(z\)-directions.

So A is (8,0,0).

To move from the origin O to D, we could go via C.

OC is 5 units in the positive \(y\)-direction and then CD is 6 units in the positive \(z\)-direction.

So D is (0,5,6).

To find the coordinates of E, we could start at A(8,0,0) and go via B.

AB is 5 units in the positive \(y\)-direction and then BE is 6 units in the positive \(z\)-direction.

So E is (8,5,6).

DE = OA = 8 units.

DM and ME are each 8 ÷ 2 = 4 units.

We know that E is (8,5,6). So one way of doing this is to start at E and move 4 units in the negative \(x\)-direction. So the \(x\)-coordinate of M is 8 – 4 = 4.

Moving from E to M does not involve any movement parallel to the \(y\)- or \(z\)-axes. So those coordinates don't change.

So M is (4,5,6).

N is the midpoint of the base diagonal OB.

B is (8,5,0). We just need to halve each of these values.

So N is (4,\( \frac{5}{2} \),0).

From the coordinates of A and C, we can see that the base has length 8 units and breadth 6 units.

The coordinates of the centre of the base are therefore (4,3,0).

As the height of the pyramid equals 8, the coordinates of the apex D are (4,3,8).

The \(y\)-coordinate of A is 4, so we know that the cube has edge length 4.

B is therefore 4 units above A, so we add 4 to the \(z\)-coordinate of A, giving us B(8,4,10).

One way to get to C from a known point is to start at B(8,4,10) and subtract 4 from each of the \(x\)- and \(y\)-coordinates. So C is (4,0,10).

The \(x\)-coordinate of A is 6, so we know that the cube has edge length 6.

From the origin to B is 6 in the positive \(y\)-direction and 6 in the positive \(z\)-direction. So B is (0,6,6).

The height of the pyramid is 6 ÷ 2 = 3 units. C is therefore 6 + 3 = 9 units above the centre of the base, which is (3,3,0). So C has coordinates (3,3,9).

The easiest way to get to B from a point whose coordinates we know is to start at A(4,0,5) and travel along the line AB.

AB is the same length as DC, which we are told is 8 units. AB is in the direction of the positive \(y\)-axis.

So we add 8 to the \(y\)-coordinate of A. This tells us that B has coordinates (4,8,5).

To work out the coordinates of C, it makes sense to start at E(2,0,0) and travel along ED, followed by DC.

The difference between the \(x\)-coordinates of A and E is 2, so ED is double that: 4.

As before, DC = 8 units. So we add 4 to the \(x\)-coordinate of E and we add 8 to the \(y\)-coordinate of E.

This tells us that C has coordinates (6,8,0).

The diameter of the base is 6 units, so its radius is 3 units. Point A is therefore (3,0,0).

The centre of the base is (3,3,0) and point B is 8 units directly above it, so B is (3,3,8).

Because A is A(6,0,0) we know that C is (3,0,0) and the radius of the hemisphere is 3 units.

B is therefore 3 units vertically below C. So we subtract 3 from the \(z\)-coordinate to get B(3,0,–3).