Area Under Curves
Area bounded by semicircle and line
MJAT_TS1_P1
Grade 12
Question:
If the area bounded by $x = \sqrt{r^2 - y^2}$ and $y = \tan\theta\,(x + r)$ is $\dfrac{1}{2}\pi r^2$, where $r \in \mathbb{R}^+$ and $0 < \theta < \dfrac{\pi}{2}$, then:
A) $2\theta - \sin 2\theta = \dfrac{\pi}{2}$
B) $\sin\theta = \theta$
C) $2\theta + \sin 2\theta = \dfrac{\pi}{2}$
D) $\sin 2\theta + 2\theta = \pi$
Step-by-Step Solution
Key Concept: $x = \sqrt{r^2 - y^2}$ is the right semicircle of radius $r$. The line $y = \tan\theta(x+r)$ passes through $(-r, 0)$ at angle $\theta$. Compute the area of the circular segment cut by this line and set it equal to $\frac{1}{2}\pi r^2$ (half the semicircle area).
The area of the circular segment is $r^2(\theta - \frac{1}{2}\sin 2\theta)$ plus the triangular part. Setting total area $= \frac{\pi r^2}{2}$ gives $2\theta + \sin 2\theta = \frac{\pi}{2}$.
Correct Answer: C