Shaded-Region Problems

The shaded region is triangle OAT minus sector OAB.

1. OT is the hypotenuse. Pythagoras gives $AT=\sqrt{10^2-6^2}=8$ cm.
2. For angle $\theta=\angle AOT$, cosine is adjacent/hypotenuse: $\cos\theta=6/10$. The inverse cosine finds the angle, so $\theta=\cos^{-1}(0.6)$ in RAD mode.
3. Triangle area: $\frac12(6)(8)=24$ cm². Sector area: $\frac12(6^2)\theta=18\theta$.
4. Subtract using the stored angle: $24-18\cos^{-1}(0.6)$.

CP is perpendicular to AB. The arc is centred at B and runs through C to Q on AB. Find angle ABC and the shaded area. (9709/12 Oct/Nov 2021 Q7)

1. Use right triangle APC: $AP=\sqrt{15^2-9^2}=12$, hence $BP=15-12=3$ cm.
2. In right triangle BPC, tangent is opposite/adjacent: $\tan(\angle ABC)=9/3=3$. Thus $\angle ABC=\tan^{-1}3=1.2490\ldots$ rad, which rounds to the requested 1.25 rad.
3. The sector radius is BC, not AC: $BC^2=3^2+9^2=90$. Sector area is $\frac12(90)\tan^{-1}3$.
4. This time the shaded region is sector − triangle. Subtract $\frac12(3)(9)=13.5$ cm².

A numerical “show that” can require a rounded target. Show the more accurate value first, then the stated rounding; keep the unrounded angle for the area.