Measurement, STD2 2026 HSC 34

The diagram shows a right-angled triangle $OAB$ and a circle with centre $O$ and radius $r$ cm, where $OA$ and $OB$ cut the circle at $X$ and $Y$ respectively and $AB=5$ cm, $OB=12$ cm and $AX=7$ cm. The part of the triangle outside the circle is shaded.

What percentage of the triangle is shaded, correct to 1 decimal place? (5 marks)

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$76.3\%$

Show Worked Solution

$\text{Find $r$ (by Pythagoras):}$

$OA=\sqrt{12^2+5^2}=13$

$r=13-7=6\ \text{cm}$

$\text{Find}\ \angle AOB:$

$\tan \angle AOB=\dfrac{5}{12}$

$\angle AOB = \tan^{-1}\left(\dfrac{5}{12}\right)=22.619\ldots^{\circ}$

$\text{Sector area ($OXY$)}=\dfrac{\theta} {360^{\circ}}\times\pi r^2=\dfrac{22.6\ldots} {360^{\circ}}\times\pi(6)^2=7.10624\ldots\ \text{cm}^2$

$\text{Triangle area}=\dfrac{1}{2}\times12\times5=30\ \text{cm}^2$

$\text{Shaded triangle area}=30-7.1\ldots = 22.89\dots \text{cm}^2$

$\text{Percentage shaded}=\dfrac{22.89\dots}{30}\times100\% = 76.3\%\ \text{(1 d.p.)}$