$\begin{aligned}21.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.
$\begin{aligned}\quad\qquad&\textrm{Luas daerah yang diarsir adalah}=\: .... \\ &\text{a}.\quad 154\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 77\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 44\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 38,5\:\:\textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}L_{\displaystyle \text{arsiran}}&=\displaystyle \frac{1}{2}\:\textrm{luas lingkaran}_{\text{besar}}-\textrm{lingkaran}_{\text{kecil}}\\ &=\displaystyle \frac{1}{2}(\displaystyle \frac{1}{4}\pi d^{\displaystyle 2}_{\textrm{besar}})-(\displaystyle \frac{1}{4}\pi d^{\displaystyle 2}_{\textrm{kecil}})\\ &=\displaystyle \frac{1}{2}.\displaystyle \frac{22}{4.7}.14^{\displaystyle 2}-\displaystyle \frac{1}{4}.\displaystyle \frac{22}{7}.7^{\displaystyle 2}\\ &=\displaystyle \frac{22}{28}.(7.14-49)\\&=\displaystyle \frac{22}{28}.49\\ &=38,5\:\: \textrm{cm}^{\displaystyle 2}\end{aligned}\end{aligned}$.
$\begin{aligned}22.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.
$\begin{aligned}\qquad&\textrm{Luas daerah yang diarsir adalah}=\: .... \\ &\text{a}.\quad 57\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 62,8\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 107\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 114\:\:\textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}L_{\displaystyle \text{arsiran}}&=\textrm{Luas tembereng doubel}\\ &=2(\textrm{luas juring - luas segitiga siku-siku})\\ &=2\left( \displaystyle \frac{90^{\displaystyle 0}}{360^{\displaystyle 0}}\times\pi.r^{2}-\displaystyle \frac{1}{2}.r^{\displaystyle 2} \right)\\ &=2\left( \displaystyle \frac{1}{4}\pi.10^{\displaystyle 2}-\displaystyle \frac{1}{2}.10^{\displaystyle 2} \right)\\ &=2(78,5-50)=57\:\: \textrm{cm}^{\displaystyle 2}\end{aligned}\end{aligned}$.
$\begin{aligned}23.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.
$\begin{aligned}\qquad&\textrm{Luas daerah yang diarsir adalah}=\: .... \\ &\text{a}.\quad 157\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 80\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 40\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 39,25\:\:\textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}L_{\displaystyle \text{arsiran}}&=(\textrm{luas juring besar - luas juring kecil})\\ &=\left( \displaystyle \frac{90^{\displaystyle 0}}{360^{\displaystyle 0}}\times\pi.r_{\text{besar}}^{2}-\displaystyle \frac{90^{\displaystyle 0}}{360^{\displaystyle 0}}\times\pi.r_{\text{kecil}}^{2} \right)\\ &=\left( \displaystyle \frac{1}{4}\pi.10^{\displaystyle 2}-\displaystyle \frac{1}{4}\pi.7^{\displaystyle 2} \right)\\ &=\displaystyle \frac{1}{4}(3,14).(100-49)\approx 40\:\: \textrm{cm}^{\displaystyle 2}\end{aligned}\end{aligned}$.
$\begin{aligned}24.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.
$\begin{aligned}\qquad&\textrm{Luas daerah yang diarsir adalah}=\: .... \\ &\text{a}.\quad 112\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 70\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 42\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 28\:\:\textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}L_{\displaystyle \text{arsiran}}&=(\textrm{luas tembereng besar - 2 luas tembereng kecil})\\ &=\left( \displaystyle \frac{90^{\displaystyle 0}}{360^{\displaystyle 0}}\times\pi.r_{\text{besar}}^{2}-\displaystyle \frac{1}{2}\times r_{\text{besar}}^{2} \right)\\ &\quad-2\left( \displaystyle \frac{90^{\displaystyle 0}}{360^{\displaystyle 0}}\times \pi.r_{ \textrm{kecil}}^{\displaystyle 2}-\displaystyle \frac{1}{2}\times r_{\text{kecil}}^{2} \right)\\ &=\displaystyle \frac{1}{4}.\displaystyle \frac{22}{7}.14^{\displaystyle 2}-\displaystyle \frac{1}{2}.14^{\displaystyle 2}-2\left( \displaystyle \frac{1}{4}.\frac{22}{7}.7^{\displaystyle 2}-\displaystyle \frac{1}{2}.7^{\displaystyle 2} \right)\\ &=154-98-(77-49)=28\:\: \textrm{cm}^{\displaystyle 2}\end{aligned}\end{aligned}$.
$\begin{aligned}25.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.
$\begin{aligned}\qquad&\textrm{Jika}\:\: AB=5\:\: cm,\:\: AC=12\:\: cm,\:\: \textrm{dan}\:\: AB,\: AC\\ &\textrm{serta}\:\: BC\:\: \textrm{masing-masing adalah diamter lingkaran},\\ &\textrm{maka luas daerah terarsir adalah}=\: .... \\ &\text{a}.\quad 27\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 30\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 36\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 37,5\:\:\textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}L_{\displaystyle \text{arsiran}}&=(2\:\textrm{luas setengah lingkaran - luas lingkaran besar})\\&\quad + \textrm{luas sebuah segitiga}\\ &=\displaystyle \frac{1}{2}.\frac{1}{4}\pi d_{2}^{\displaystyle 2}+\displaystyle \frac{1}{2}.\frac{1}{4}\pi d_{3}^{\displaystyle 2}-\displaystyle \frac{1}{2}.\frac{1}{4}\pi d_{1}^{\displaystyle 2}+\displaystyle \frac{1}{2}.d_{2} d_{3}\\ &=\displaystyle \frac{1}{8}(3,14) (d_{2}^{\displaystyle 2}+ d_{3}^{\displaystyle 2}- d_{1}^{\displaystyle 2})+\displaystyle \frac{1}{2}.d_{2} d_{3}\\ &=\displaystyle \frac{1}{8}(3,14) (12^{\displaystyle 2}+ 5^{\displaystyle 2}- 13^{\displaystyle 2})+\displaystyle \frac{1}{2}.12. 5\\ &=30\:\: \textrm{cm}^{\displaystyle 2}\\\\ &\text{Di atas adalah aplikasi dari}\:\: \textbf{Teorema Hippocrates} \end{aligned}\end{aligned}$.
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