2023 Paper 2, Q18
A company makes windows using glass that has a mass of 5.6 kg per square metre. A customer orders an unusual window in a partial parabolic shape, as shown.
Determine the mass of the window.
Four numbers, no equation, no axes and no angle. The glass is the shaded part only. The triangle is a hole.
Cut out the glass, then take out the triangle
It is two areas and a multiplication. The arch is an integral, the bite is a triangle, and the difference is what the glazier has to lift.
the build runs on a fourteen second loop
A triangle with no angle, and a curve with no axes
There are three sides and no angle, so use the cosine rule backwards. $\cos C=\frac{9^2-7^2-8^2}{-2\times 7\times 8}$ gives $C=73.398^\circ$, and then $\frac12\times 7\times 8\sin C=26.8328$ m².
Nothing is on axes until you put it there. Intercepts at $(0,0)$ and $(8,0)$ with vertex $(4,12)$ give $y=a(x-4)^2+12$, and $(8,0)$ forces $a=-\frac34$.
You choose the axes, but the area is always 64 m²
Move the origin and the equation changes completely. The area never does. Sweep the shading to watch the integral fill up to 64.
the blue cross is your origin
$\displaystyle\int_0^{x}-\tfrac34(t-4)^2+12\,dt$
What is the 9 m for?
QCAA marking guide · 5 marks
$$C=\cos^{-1}\!\left(\frac{c^2-a^2-b^2}{-2ab}\right)=\cos^{-1}\!\left(\frac{81-49-64}{-112}\right)=73.39845^\circ$$
$$\text{area}=\tfrac12\times 7\times 8\times\sin(73.39845^\circ)=26.8328\ \text{m}^2$$
Marker: correctly determines the area of the ‘removed’ triangle. Equivalent decimal values accepted.
Consider an inverted parabola with $x$-intercepts $(0,0)$ and $(8,0)$ and vertex $(4,12)$. Vertex form, then substitute an intercept:
$$0=a(8-4)^2+12\ \Rightarrow\ a=-\frac{12}{16}=-\frac34\ \Rightarrow\ y=-\tfrac34(x-4)^2+12$$
Marker: correctly determines the parabola equation. Alternative methods accepted, including fitting it on a GDC.
$$\text{area}=\int_0^8 -\tfrac34(x-4)^2+12\,dx=64\ \text{m}^2$$
Marker: determines the area between the parabola and the $x$-axis. On Paper 2 this is a GDC evaluation, and it comes out exactly 64.
$$64-26.8328=37.1672\ \text{m}^2,\qquad 37.1672\times 5.6=208.1363\ \text{kg}$$
About 208 kg of glass
Marker: determines the mass of the window glass. The 5.6 kg per square metre is the last step, not the first. Nothing can be weighed until the hole has been taken out.
State the axes you chose before you use them, keep the two areas separate and labelled, and carry the units through to kilograms.
Marker: shows logical organisation, communicating key steps.
The answer is The answer is The answer is The answer is 208 kg, and two ways to check it
A parabolic segment is always $\frac23$ of the rectangle around it, so $\frac23\times 8\times 12=64$ m². If your integral is not 64, you have the wrong value of $a$.
Heron with $s=12$: $\sqrt{12\times 5\times 4\times 3}=\sqrt{720}=12\sqrt5=26.8328$ m², exact, and no rounded angle to carry.
$5.6\left(64-12\sqrt5\right)=358.4-67.2\sqrt5\approx 208.14$ kg. Rounding the angle to $73.4^\circ$ instead costs about 20 grams. Harmless here, but keep the full value in your calculator.
There are two separate things to build here, and the question asks for neither of them. The picture has no axes, so you have to set up your own before the word “parabola” means anything. Any choice of axes works, and I find that is exactly what makes students hesitate. The triangle then has three sides and no angle, so you need the cosine rule from Year 11 to find one. Only once you have both areas does the 5.6 kg per square metre do anything. Every number in the question has a different job, and the word “mass” only appears once, at the very end.
Question wording and marking-guide steps are from the 2023 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.