Put the zero of the ruler on the corner at F — the zero mark, not the end of the plastic, which is where most of the lost marks in this topic go — and turn the ruler so its edge runs along the line to D.

Read the scale at D.

\(FD = 7\) cm.

Every side of this triangle is a whole number of centimetres, so if your reading lands between two centimetre marks the ruler has slipped off the corner. Line it up again and read it a second time.

Line the zero of the ruler up with the corner at \(A\) and read the scale at \(B\).

\(AB = 6\) cm.

Reading to the nearest half centimetre means deciding between a whole centimetre mark and the mark halfway between two of them — nothing finer than that.

There are \(10\) millimetres in \(1\) centimetre, so multiply by \(10\).

\(6 \times 10 = 60\)

\(AB = 60\) mm.

Millimetres are the smaller unit, so there are more of them and the number gets bigger. If your answer came out smaller than the number of centimetres, you divided when you should have multiplied.

Zero of the ruler on \(A\), read the scale at \(B\).

\(AB = 7\) cm.

Move the ruler up to the top side and do the same again: zero on \(D\), read at \(C\).

\(DC = 4\) cm.

Both sides are a whole number of centimetres. Measure it rather than guessing from how much shorter it looks — the whole point of the two parallel sides being drawn accurately is that they can be compared by measuring.

Subtract the shorter measurement from the longer one.

\(7 - 4 = 3\)

\(AB\) is \(3\) cm longer than \(DC\).

Three letters name an angle by its two arms and its corner, and the middle letter is the corner. So angle \(ACB\) is the angle at C, between the lines CA and CB.

Put the cross in the middle of the protractor's straight edge on C, and turn the protractor until one of its two zeros sits along CA. Then follow the arm CB out to the scale.

Two numbers meet the arm there, one on each scale, and they add up to \(180\): \(95\) and \(85\). The one you want is on the same scale as the zero you started from — and the picture settles which that is, because the angle is clearly bigger than a right angle.

Angle \(ACB = 95^\circ\).

The two scales are the whole difficulty of a protractor. Decide from the picture whether the angle is bigger or smaller than \(90^\circ\) before you read anything, then take the number that agrees with you.

Put the centre cross of the protractor on the corner, line one zero up with one of the two sides that meet there, and read where the other side crosses the scale.

\(x = 35^\circ\).

Check the reading against the picture before writing it down: the corner is smaller than a square corner, so a reading of \(35^\circ\) fits and the other number on the protractor does not.

An acute angle is less than \(90^\circ\), an obtuse angle is between \(90^\circ\) and \(180^\circ\), and a reflex angle is more than \(180^\circ\).

\(x = 35^\circ\), which is smaller than \(90^\circ\), so \(x\) is acute.

No angle inside a parallelogram can be reflex, so the only decision here is between acute and obtuse — but it is still a decision worth making from the measurement rather than from how the corner looks.

A protractor only goes up to \(180^\circ\), so a reflex angle cannot be read off it directly. Measure the other angle at that corner instead — the one between the same two arms, on the outside of the marked angle — and take it away from \(360^\circ\).

Measured that way, the angle between the two arms is \(140^\circ\).

\(x = 360 - 140 = 220\)

\(x = 220^\circ\).

Check it against the picture: a reflex angle is always more than \(180^\circ\), so if your answer came out smaller than that you have written down the angle you measured instead of the one you were asked for.

Zero of the ruler on \(P\), read the scale at \(Q\).

\(PQ = 9\) cm on the drawing.

Each centimetre on the drawing stands for \(10\) m, so multiply the measurement by \(10\).

\(9 \times 10 = 90\)

The real field is \(90\) m long.

This is why measuring is worth doing carefully rather than quickly: at this scale, being 1 cm out on the drawing puts the answer \(10\) m out on the ground.

Put the zero of the ruler on the corner at P — the zero mark, not the end of the plastic, which is where most of the lost marks in this topic go — and turn the ruler so its edge runs along the line to Q.

Read the scale at Q.

\(PQ = 4\) cm.

Every side of this triangle is a whole number of centimetres, so if your reading lands between two centimetre marks the ruler has slipped off the corner. Line it up again and read it a second time.

Line the zero of the ruler up with the corner at \(A\) and read the scale at \(B\).

\(AB = 7.5\) cm.

Reading to the nearest half centimetre means deciding between a whole centimetre mark and the mark halfway between two of them — nothing finer than that.

There are \(10\) millimetres in \(1\) centimetre, so multiply by \(10\).

\(7.5 \times 10 = 75\)

\(AB = 75\) mm.

Millimetres are the smaller unit, so there are more of them and the number gets bigger. If your answer came out smaller than the number of centimetres, you divided when you should have multiplied.

Zero of the ruler on \(A\), read the scale at \(B\).

\(AB = 9\) cm.

Move the ruler up to the top side and do the same again: zero on \(D\), read at \(C\).

\(DC = 5\) cm.

Both sides are a whole number of centimetres. Measure it rather than guessing from how much shorter it looks — the whole point of the two parallel sides being drawn accurately is that they can be compared by measuring.

Subtract the shorter measurement from the longer one.

\(9 - 5 = 4\)

\(AB\) is \(4\) cm longer than \(DC\).