Question 7
A circle with radius $r$ and internal angle $\theta$ has a shaded segment as shown.
If $\theta$ is in radians, the area of the shaded segment is
Not quite. Multiplying by $\tfrac{\pi}{180}$ converts degrees to radians, but $\theta$ is already in radians. Converting again shrinks the sector term so much that this is negative for almost every angle up to $\pi$. In 2022, 56.3% of Queensland students chose (A).
That’s it. Step through the hints below to see why. In 2022, 29.9% of Queensland students chose (B).
Not quite. $\tfrac{r^2}{4}\cdot\tfrac{\theta\pi}{90}=\tfrac{\theta}{360}\times\pi r^2$, which is the sector area for $\theta$ in degrees. And $\tfrac{r^2}{4}$ is the triangle’s area only when $\sin\theta=\tfrac12$. With $\theta$ in radians, this is negative for every angle in the circle. In 2022, 9.8% of Queensland students chose (C).
Close. $\tfrac12r^2\theta$ is the sector, but $\tfrac12r^2$ is the triangle’s area only for a right angle, $\tfrac12\times r\times r$. The triangle’s height is $r\sin\theta$, so its area is $\tfrac12r^2\sin\theta$. In 2022, 3.1% of Queensland students chose (D).
(B) and (D) give the same value here, only because $\sin\tfrac{\pi}{2}=1$. Move $\theta$ and (D) breaks away. Only (B) equals the segment at every angle.
(D) is very close to (B) here, but not equal. It is short by $\tfrac12r^2(1-\sin\theta)$, which is zero only at $\theta=\tfrac{\pi}{2}$.
At $\theta=\pi$ the triangle is flat, so the segment is a semicircle, $\tfrac12\pi r^2$. (B) gives exactly that, because $\sin\pi=0$.
Only (B) equals the shaded segment at this angle.
The two radii and the arc make a sector. The two radii and the chord make a triangle. The shaded segment is what is left of the sector when you take the triangle away.
$$\text{segment}=\text{sector}-\text{triangle}$$
With $\theta$ in radians, the sector is a fraction $\tfrac{\theta}{2\pi}$ of the circle.
$\text{sector}=\dfrac{\theta}{2\pi}\times\pi r^2$$=\dfrac12r^2\theta$
The triangle has two sides of length $r$ with $\theta$ between them, so use $\tfrac12ab\sin C$.
$$\text{triangle}=\frac12r^2\sin\theta$$
Subtract and factorise.
$\dfrac12r^2\theta-\dfrac12r^2\sin\theta$$=\dfrac{r^2}{2}\bigl(\theta-\sin\theta\bigr)$
No $\tfrac{\pi}{180}$ is needed, because $\theta$ is already in radians. Use the slider to check: only (B) matches the segment at every angle.
$$\begin{aligned}\text{segment}&=\text{sector}-\text{triangle}\\&=\frac12r^2\theta-\frac12r^2\sin\theta\\&=\frac{r^2}{2}\bigl(\theta-\sin(\theta)\bigr)\end{aligned}$$
The sector formula $\tfrac12r^2\theta$ only works with $\theta$ in radians, which is what the question gives you.
How Queensland students answered in 2022
More students chose (A) than the answer. From the QCAA Mathematical Methods subject report for 2022. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2022 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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