Tag Archives: HSC Advanced 2024

HSC Advanced 2024 Question 30

HSC Advanced 2024

Two circles have the same centre O. The smaller circle has a radius of 1 cm, while the larger has a radius of (x+1) cm. The circles enclose a region QRST, which is subtended by angle of {\theta} at O, as shaded.

The area of QRST is A cm2, where A is a constant and A>0

Let P cm be the perimeter of QRST

(a) By finding expressions for the area and perimeter of QRST show that P(x)=2x+\frac{2A}{x}

(b) Show that if the perimeter is minimised, then {\theta} must be less than 2.

(a) A=\frac{1}{2}\theta((x+1)^2-1^2)
A=\frac{1}{2}\theta(x^2+2x)
2A=\theta x^2 +2x \theta
\frac{2A}{x}=\theta x +2\theta

P=\theta(1)+\theta(1+x)+2x
P=2\theta +\theta x +2x
P=\frac{2A}{x}+2x

I like it when the first part requires the student to show something and the second part has them use it (that way they can still do the second part even if they couldn’t do the first part).

(b) \frac{dP}{dx}=2-\frac{2A}{x^2}
0=2-\frac{2A}{x^2}
x=\sqrt{A}

\frac{d^2P}{dx^2}=\frac{4A}{x^3}
Both x and A are greater than zero, therefore \frac{d^2P}{dx^2}>0 and x=\sqrt{A} is a minimum.

Substitute x=\sqrt{A} into the Area formula
2A=A\theta+2\sqrt{A}\theta
\theta=\frac{2A}{A+2\sqrt{A}}
\theta=\frac{2A}{\sqrt{A}(\sqrt{A}+2)}
\theta=\frac{2\sqrt{A}}{\sqrt{A}+2}

Now \frac{2\sqrt{A}}{\sqrt{A}+2}<\frac{2\sqrt{A}}{\sqrt{A}}
Hence \theta<\frac{2\sqrt{A}}{\sqrt{A}}
and \theta<2

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