The diagram shows a triangular field with base \(46\) m and perpendicular height \(28\) m. (a) Work out the area of the field. [2] (b) Maize is planted on \...

Both parts test one idea: find a quantity first, then take a percentage of it. The area of any triangle is half the base times the perpendicular height, and the perpendicular height must be the one measured at right angles to the base you use. Here the stem states the base as \(46\) m and the perpendicular height as \(28\) m, so those two values go straight into the formula.

(a) Substituting into \(A=\frac{1}{2}bh\):

\[ A=\frac{1}{2}\times 46\times 28 \] [M1]

\(46\times 28=1288\), and half of that is \(644\), so the area is \(644\) m\(^2\). [A1]

Note the unit: two lengths in metres multiply to an area in square metres.

(b) "Maize is planted on \(62.5\%\) of the field" means you take \(62.5\%\) of the answer to part (a). Convert the percentage to a decimal multiplier by dividing by \(100\): \(62.5\div 100=0.625\).

\[ 644\times 0.625 \] [M1]

\[ =402.5 \]

The area planted with maize is \(402.5\) m\(^2\). [A1]

An equivalent route is \(\frac{644}{100}\times 62.5\), or noticing that \(62.5\%=\frac{5}{8}\) so \(644\times\frac{5}{8}=402.5\). All three give the same value and earn the same marks. The mark for the method here follows through from your own area, so an arithmetic slip in (a) does not cost you the method mark in (b).

A common error is to use a sloping side of the triangle instead of the perpendicular height. Only the height at right angles to the chosen base works in \(\frac{1}{2}bh\).