2022 Paper 1, Q19
Two triangles are said to be similar if their corresponding angles are congruent and the corresponding sides are in proportion, e.g. if $\triangle UVW$ is similar to $\triangle XYZ$ then $\angle U=\angle X$, $\angle V=\angle Y$ and $\angle W=\angle Z$ and $\frac{UV}{XY}=\frac{VW}{YZ}=\frac{UW}{XZ}$.
Two parallel walls $AB$ and $CD$, where the northern ends are $A$ and $C$ respectively, are joined by a fence from $B$ to $C$. The wall $AB$ is 20 metres long, the angle $ABC=30^\circ$ and the fence $BC$ is 10 metres long.
A new fence is being built from $A$ to a point $P$ somewhere along $CD$. The new fence $AP$ will cross the original fence $BC$ at $O$. Let $OB=x$ metres, where $0<x\le 10$.
Determine the value of $x$ that minimises the total area enclosed by $\triangle OBA$ and $\triangle OCP$. Verify that this total area is a minimum.
Seven marks, and the first one is for the drawing. QCAA required a labelled diagram showing 20, 30, $x$ and either 10 or $10-x$. Try it on paper before you look.
Drawn with north to the right, so the 5 m strip between the walls fits the page. QCAA accepts any layout: “diagrams do not need to be accurate in terms of the relative sizes of any drawn angles or lengths.”
Swing the new fence and the two paddocks trade area
As $O$ slides up the old fence, the near triangle grows and the far one shrinks, but not at the same rate, so the total has a minimum.
Two triangles that share an angle and a ratio
Use two sides and the angle between them, so $\frac12\times 20\times x\times\sin 30^\circ=5x$.
The walls are parallel, so $\angle ABO=\angle PCO=30^\circ$ (alternate angles) and the angles at $O$ are vertically opposite. The provided definition then gives $\frac{CP}{AB}=\frac{OC}{OB}$.
$\frac{CP}{20}=\frac{10-x}{x}$, so $CP=\frac{20(10-x)}{x}$, and the far triangle is $\frac12(10-x)\cdot CP\sin 30^\circ=\frac{5(10-x)^2}{x}$.
That simplification is worth a mark of its own, and it is the one that makes the calculus easy. Written as $10x+500x^{-1}-100$, the derivative is only two terms, and you do not need the quotient rule.
The two areas, drawn to the same scale
Both triangles keep the $30^\circ$ angle. Drag $x$: the near one grows steadily, the far one collapses like $\frac{1}{x}$, and the sum bottoms out at $x=\sqrt{50}$.
Why does the paper define similarity for you?
QCAA marking guide · 7 marks
A labelled diagram of the two parallel walls, the fence $BC$, the new fence $AP$ and the crossing point $O$.
Marker: correctly uses all of the given information to draw a labelled diagram. The diagram must include 20, 30, $x$ and either 10 or $(10-x)$. The $10-x$ label may be implied by later working. Relative sizes need not be accurate.
$$A_T=\tfrac12\times 20\times x\times\sin 30^\circ+\tfrac12\times(10-x)\times CP\times\sin 30^\circ$$
Marker: correctly establishes a formula for the total area.
$$\frac{CP}{AB}=\frac{OC}{OB}\ \Rightarrow\ \frac{CP}{20}=\frac{10-x}{x}\ \Rightarrow\ CP=\frac{20(10-x)}{x}$$
Marker: correctly determines an expression for $CP$ in terms of $x$.
$$A_T=5x+\frac{5}{x}(10-x)^2=\frac{10x^2-100x+500}{x}=10x+500x^{-1}-100$$
Marker: determines a simplified version of the formula for the total area. Other simplifications accepted, e.g. $\frac{5x^2-100x+500}{x}+5x$.
$$A_T'=10-500x^{-2},\qquad 0=10-\frac{500}{x^2}\ \Rightarrow\ \frac{500}{x^2}=10\ \Rightarrow\ x^2=50$$
Marker: determines an equation to solve for stationary points. Equivalent equations accepted, including the quotient-rule form set to zero.
$x=\pm\sqrt{50}$, but $x$ is a positive length. And the domain is $0<x\le 10$. So $x=\sqrt{50}=5\sqrt2\approx 7.07$ m.
Marker: evaluates the reasonableness of solutions. A whole mark for one sentence about the domain.
$$A_T''=\frac{1000}{x^3},\qquad A_T''\!\left(\sqrt{50}\right)>0\ \therefore\ \text{minimum}$$
$x=\sqrt{50}$ minimises the total area, at $100\sqrt2-100\approx 41.4$ m²
Marker: verifies the solution. Unlike the fence question in 2025, here the verification is explicitly demanded, and skipping it costs a seventh of the question.
Where the marks actually are
They are the diagram, the area formula, $CP$ and the simplification. Differentiating is only the fifth thing that happens. And a student who never gets past the picture still banks a mark.
$A_T\!\left(5\sqrt2\right)=50\sqrt2+50\sqrt2-100=100\sqrt2-100\approx 41.42$ m². The question only asks for $x$, but the value gives you a free check, because it must be less than $A_T(10)=50$.
At $x=10$ the point $P$ collides with $C$, the far triangle vanishes and the total is 50 m². As $x\to 0$ the total runs to infinity. That is why the floor sits inside the domain, not at an end of it.
There is no picture. Seven marks of geometry are described in two sentences of prose, and the QCAA gave a full mark for turning that prose into a labelled diagram. It is the only question in the set that does this. You then have to work out where the provided definition fits. Similar triangles are defined for you, but the question never tells you that these two triangles are similar. That comes from the walls being parallel. The last mark is for doing what the final sentence says, which is to verify. Two of the seven marks here are for habits, not mathematics, so I would always draw the diagram, even when you think you do not need it.
Question wording and marking-guide steps are from the 2022 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.