CONTOH 2-LIMIT FUNGSI

 $\begin{array}{ll}\\ 6.&\textrm{Diketahui bahwa}\: \: f(x)=x^{2}-2,\\ & \textrm{maka nilai}\: \: \: \underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle x^{2}-2&&\textrm{d}.\quad \displaystyle x\\\\ \textrm{b}.\quad \displaystyle x^{2}\quad &\textrm{c}.\quad \displaystyle 2x\quad &\textrm{e}.\quad \displaystyle 2x-2 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa}\: f(x)=x^{2}-2,\\ \textrm{maka nila}&\textrm{i untuk}\\ \underset{h\rightarrow 0 }{\textrm{Lim}}\: &\: \displaystyle \frac{f(x+h)-f(x)}{h}\, \, \, \, \\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\left ((x+h)^{2}-2 \right )-\left ( x^{2}-2 \right )}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{x^{2}+2xh+h^{2}-2-x^{2}+2}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{2xh+h^{2}}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \left (2x+h \right )\\ &= 2x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Diketahui}\: \: f(x)=\sqrt{x-1},\\ & \textrm{maka nilai}\: \: \: \underset{h\rightarrow 0 }{\textrm{Lim}}\: \:\displaystyle \frac{f(2+h)-f(2)}{h}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2}&&\textrm{d}.\quad \displaystyle 1\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{2}\quad &\textrm{c}.\quad \displaystyle 0\quad &\textrm{e}.\quad \displaystyle -1\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketa}&\textrm{hui bahwa}\: f(x)=\sqrt{x-1},\\ \textrm{maka}\: \: & \textrm{nilai untuk}\\ \underset{h\rightarrow 0 }{\textrm{Lim}}\: &\: \displaystyle \frac{f(2+h)-f(2)}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{(2+h)-1}-\sqrt{2-1}}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{\sqrt{h+1}-1}{h}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \displaystyle \frac{\sqrt{h+1}-1}{h}\times \displaystyle \frac{\sqrt{h+1}+1}{\sqrt{h+1}+1}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{(h+1)-1}{h\times \left ( \sqrt{h+1}+1 \right )}\\ &=\underset{h\rightarrow 0 }{\textrm{Lim}}\: \: \displaystyle \frac{1}{\left ( \sqrt{h+1}+1 \right )}\\ &=\displaystyle \frac{1}{\sqrt{0+1}+1}\\ &= \displaystyle \frac{1}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textbf{(Mat Das SIMAK UI 2013)}\\ &\textrm{Nilai}\: \: \underset{x\rightarrow 5}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \sqrt{3}+\sqrt{2}&&\textrm{d}.\quad 5\\ \textrm{b}.\quad 5-2\sqrt{6}\quad &\textrm{c}.\quad 2\sqrt{6}\quad &\textrm{e}.\quad  5+2\sqrt{6}\end{array}\\\\ &\textrm{Jawab}:\\ & \begin{aligned}\underset{x\rightarrow 5}{\textrm{Lim}}\: &\: \displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}\\ &=\displaystyle \frac{\sqrt{5+2\sqrt{5+1}}}{\sqrt{5-2\sqrt{5+1}}}\\ &=\displaystyle \frac{\sqrt{5+2\sqrt{6}}}{\sqrt{5-2\sqrt{6}}}\\ &=\displaystyle \frac{\sqrt{3+2+2\sqrt{3.2}}}{\sqrt{3+2-2\sqrt{3.2}}}\\ &=\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\ &=\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\ &=\displaystyle \frac{3+2+2\sqrt{6}}{3-2}\\ &= 5+2\sqrt{6} \end{aligned}\end{array}$

$\begin{array}{ll}\\ 9.&\textbf{(Mat IPA SBMPTN 2014)}\\\\ &\textrm{Jika}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)+\frac{1}{g(x)} \right )=4\\ & \textrm{dan}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)-\frac{1}{g(x)} \right )=-3,\\\\ &\textrm{maka nilai}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle f(x).g(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{14}&&\textrm{d}.\quad \displaystyle \frac{4}{14}\\\\ \textrm{b}.\quad \displaystyle \frac{2}{14}\quad &\textrm{c}.\quad \displaystyle \frac{3}{14}\quad &\textrm{e}.\quad \displaystyle \frac{5}{14}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\: ,\\ &\begin{array}{lll}\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)+\frac{1}{g(x)} \right )=4&&\\ &&\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \left ( f(x)-\frac{1}{g(x)} \right )=-3&+&\\ &&\\\hline &&\\ 2\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle f(x)\qquad\qquad\: \: =1&&\\ &&\\ \qquad\qquad\: \: \: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: f(x)=\displaystyle \frac{1}{2},&&\\\end{array}\\ &\textrm{sehingga}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: f(g)=\displaystyle \frac{2}{7}\\ &&\\ &\textrm{maka},\\ &\underset{x\rightarrow a}{\textrm{Lim}}\: \: f(x).g(x)=\displaystyle \frac{1}{2}\times \frac{2}{7}= \displaystyle \frac{2}{14} \end{aligned} \end{array}$

$\begin{array}{l}\\ 10.&\textrm{Nilai}\quad  \underset{x\rightarrow 10}{\textrm{Lim}}\: \displaystyle \frac{x^{\displaystyle 2}-14x+40}{\sqrt{2x}-\sqrt{x+10}}\quad \textrm{adalah}\: ....\\  &\begin{array}{llllllll} \textrm{a}.&6\sqrt{5}\qquad\qquad\qquad\qquad\qquad \textrm{d}.\quad 36\sqrt{5}\\ \textrm{b}.&12\sqrt{5}\qquad \textrm{c}.\quad 24\sqrt{5}\:\quad\qquad \textrm{e}.\quad 48\sqrt{5}\\  \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\underset{x\rightarrow 10}{\textrm{Lim}}\: \displaystyle \frac{x^{\displaystyle 2}-14x+40}{\sqrt{2x}-\sqrt{x+10}}=\displaystyle \frac{10^{\displaystyle 2}-14\times 10+40}{\sqrt{20}-\sqrt{10+10}}=\displaystyle \frac{0}{0}\\ &\textrm{bentuk di atas harus dihindari.  Selanjutnya}\\ &\textrm{gunakan opsi mengalikan penyebut dengan sekawan},\\ &\textrm{yaitu:}\\ &\underset{x\rightarrow 10}{\textrm{Lim}}\: \displaystyle \frac{x^{\displaystyle 2}-14x+40}{\sqrt{2x}-\sqrt{x+10}}\times \frac{\left( \sqrt{2x}+\sqrt{x+10} \right)}{\left( \sqrt{2x}+\sqrt{x+10} \right)}\\ &= \underset{x\rightarrow 10}{\textrm{Lim}}\: \displaystyle \frac{\left( x^{\displaystyle 2}-14x+40 \right)\left( \sqrt{2x}+\sqrt{x+10} \right)}{2x-(x+10)}\\ &=\underset{x\rightarrow 10}{\textrm{Lim}}\: \displaystyle \frac{(x-10)(x-4)\left( \sqrt{2x}+\sqrt{x+10} \right)}{x-10}\\ &=\underset{x\rightarrow 10}{\textrm{Lim}}\: (x-4)\left( \sqrt{2x}+\sqrt{x+10} \right)\\ &=(10-4)\left( \sqrt{20}+\sqrt{20} \right)\\ &=(6)\left( 2\sqrt{20} \right)=(6)(2)\left( 2\sqrt{5} \right)=24\sqrt{5}\end{aligned} \end{array}$.

CONTOH 1-LIMIT FUNGSI

 $\begin{array}{ll}\\ 1&\textrm{Nilai}\: \: \underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{1}{x-2} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle \frac{1}{2}&&\textrm{d}.\quad \displaystyle \frac{1}{4}\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{4}\quad &\textrm{c}.\quad 0\quad &\textrm{e}.\quad \displaystyle \frac{1}{2}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{1}{x-2} \right )&= \left ( \displaystyle \frac{6-2}{2^{2}-4}-\frac{1}{2-2} \right )\\ &=\left ( \displaystyle \frac{4}{0}-\frac{1}{0} \right )= \infty -\infty \\ &\: \: \textrm{hal ini tidak diperkenankan}\\ \textrm{Sehingga},\, \qquad\qquad\qquad &\\ \underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{1}{x-2} \right )&=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{(x+2)}{(x-2)(x+2)} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{6-x}{x^{2}-4}-\frac{x+2}{x^{2}-4} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{4-2x}{x^{2}-4} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{-2(x-2)}{(x+2)(x-2)} \right )\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \left ( \displaystyle \frac{-2}{x+2} \right )\\ &=\displaystyle -\frac{2}{(2+2)}\\ &= -\displaystyle \frac{1}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Nilai}\: \: \underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{x-4}{2\sqrt{x}-x}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle 2&&\textrm{d}.\quad \displaystyle \frac{1}{2}\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{2}\quad &\textrm{c}.\quad 0\quad &\textrm{e}.\quad \displaystyle 2\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{x-4}{2\sqrt{x}-x}&= \left ( \displaystyle \frac{4-4}{2^{4}-4} \right )\\ &= \displaystyle \frac{0}{0} \\ &\: \: \textrm{hal ini juga tidak diperkenankan}\\ \textrm{Sehingga},\: \: \: \: \quad&\\ \underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{x-4}{2\sqrt{x}-x}&=\underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \sqrt{x}+2 \right )\left ( \sqrt{x}-2 \right )}{\sqrt{x}\left ( 2-\sqrt{x} \right )} \\ &=\underset{x\rightarrow 4}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \sqrt{x}+2 \right )\left ( \sqrt{x}-2 \right )}{-\sqrt{x}\left ( \sqrt{x}-2 \right )}\\ &=\underset{x\rightarrow 4}{\textrm{Lim}}\: \: -\displaystyle \frac{\left ( \sqrt{x}+2 \right )}{\sqrt{x}}\\ &=-\displaystyle \frac{\sqrt{4}+2}{\sqrt{4}}\\ &=-\displaystyle \frac{2+2}{2}\\ &= -2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Nilai}\: \: \underset{x\rightarrow 1}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( 2x-3\sqrt{x}+1 \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )^{2}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{4}&&\textrm{d}.\quad \displaystyle 2\\\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\quad &\textrm{c}.\quad 1\quad &\textrm{e}.\quad \displaystyle 4\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 1}{\textrm{Lim}}\:& \: \displaystyle \frac{\left ( 2x-3\sqrt{x}+1 \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )^{2}}\\ &= \displaystyle \frac{\left ( 2-3+1 \right )\left ( 1-1 \right )}{\left ( 1-1 \right )^{2}} \\ &=\displaystyle \frac{0\times 0}{0^{2}}\\ &= \displaystyle \frac{0}{0} \\ &\: \: \textrm{hal ini juga tidak diperkenankan}\\ \textrm{Sehi}&\textrm{ngga},\\ \underset{x\rightarrow 1}{\textrm{Lim}}\: &\: \displaystyle \frac{\left ( 2x-3\sqrt{x}+1 \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )^{2}}\\ &=\underset{x\rightarrow 1}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \left ( 2\sqrt{x}-1 \right )\times \left ( \sqrt{x}-1 \right ) \right )\left ( \sqrt{x}-1 \right )}{\left ( \sqrt{x}-1 \right )\left ( \sqrt{x}-1 \right )}\\ &=\underset{x\rightarrow 1}{\textrm{Lim}}\: \: \left ( 2\sqrt{x}-1 \right )\\ &=2.1-1\\ &=2-1\\ &=1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Nilai}\: \: \underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle \frac{1}{14}\sqrt{7}&&\textrm{d}.\quad \displaystyle \frac{1}{7}\sqrt{7}\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{7}\sqrt{7}\quad &\textrm{c}.\quad 0\quad &\textrm{e}.\quad \displaystyle \frac{1}{14}\sqrt{7}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 3}{\textrm{Lim}}\: &\: \displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{\sqrt{x+4}-\sqrt{2x+1}}{x-3}\times \displaystyle \frac{\sqrt{x+4}+\sqrt{2x+1}}{\sqrt{x+4}+\sqrt{2x+1}}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{\left ( x+4 \right )-\left ( 2x+1 \right )}{\left ( x-3 \right )\left ( \sqrt{x+4}+\sqrt{2x+1} \right )}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: -\displaystyle \frac{-x+3}{\left ( x-3 \right )\left ( \sqrt{x+4}+\sqrt{2x+1} \right )}\\ &=\underset{x\rightarrow 3}{\textrm{Lim}}\: \: \displaystyle \frac{-1}{\left ( \sqrt{x+4}+\sqrt{2x+1} \right )}\\ &=-\displaystyle \frac{1}{\left ( \sqrt{7}+\sqrt{7} \right )}\\ &=-\displaystyle \frac{1}{2\sqrt{7}}\\ &=-\displaystyle \frac{1}{2\sqrt{7}}\times \displaystyle \frac{\sqrt{7}}{\sqrt{7}}\\ &= -\displaystyle \frac{1}{14}\sqrt{7} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Jika}\: \: \underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{ax-2a}{\sqrt{2x}-x}=6,\: \: \textrm{maka nilai}\: \: a\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 2&&\textrm{d}.\quad \displaystyle -2\\\\ \textrm{b}.\quad \displaystyle 1\quad &\textrm{c}.\quad -1\quad &\textrm{e}.\quad \displaystyle -3\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\underset{x\rightarrow 2}{\textrm{Lim}}\: &\: \displaystyle \frac{ax-2a}{\sqrt{2x}-x}=6\\ &\textrm{dengan bantuan limit kanan }\\ &\textrm{yaitu}\: \: x=2+h\: \Rightarrow \: h\rightarrow 0\\ \underset{h\rightarrow 0}{\textrm{Lim}}\: &\: \displaystyle \frac{a\left ( 2+h \right )-2a}{\sqrt{2\left ( 2+h \right )}-\left ( 2+h \right )}=6\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{2a+ah-2a}{\sqrt{4+2h}-\left ( 2+h \right )}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah}{\sqrt{4+2h}-\left ( 2+h \right )}\times \displaystyle \frac{\left ( \sqrt{4+2h}+\left ( 2+h \right ) \right )}{\left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah\times \left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}{4+2h-\left ( 4+4h+h^{2} \right )}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah\times \left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}{-2h-h^{2}}\\ 6&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a\times \left (\sqrt{4+2h} +\left ( 2+h \right ) \right )}{-2-h}\\ 6&=\displaystyle \frac{a\times \left (\sqrt{4+0} +\left ( 2+0 \right ) \right )}{-2-0}\\ 6&=\displaystyle \frac{a\left ( \sqrt{4}+2 \right )}{-2}\\ \displaystyle \frac{a(4)}{-2}&=6\\ a(-2)&=6\\ a&= -3 \end{aligned} \end{array}$

LIMIT FUNGSI

 Kumpulan materi yang sudah ada

• materi limit aljabar
• contoh soal 1, contoh 2, 

• limit fungsi trigonometri (materi lama-perkenalan dengan limit fungsi aljabar), lanjutan materi lama, lanjutan rumus dasar limit trigonometri, 
• materi pendahuluan limit, menentukan nilai limit fungsi trigonometri, pembuktian teorema apit (identitas limit fungsi trigonometri), limit fungsi trigonometri di tak hingga,
• contoh soal 1, contoh 2, contoh 3, contoh 4, contoh 5,
• materi limit diketakhinggan.
• contoh soal 1, contoh 2, contoh 3, contoh 4, contoh 5,

ALHAMDULILLAH-HASIL OSKANU-MA FUTUHIYAH JEKETRO GUBUG GROBOGAN-SEPTEMBER 2026

 

CONTOH SOAL 13 LINGKARAN (LINGKARAN DAN SEGITIGA)

 $\begin{aligned}61.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Titik O adalah pusat lingkaranyang berjari-jari}\:\: r\\ &\textrm{Jika panjang ruas garis}\: ED=r,\:\: \textrm{maka rasio}\\ &\angle CED\:\: \textrm{terhadap}\:\: \angle AOB\:\: \textrm{adalah}\: ....\\ &\textrm{a}.\quad 1:3\\ &\textrm{b}.\quad 1:2\\ &\textrm{c}.\quad 2:3\\ &\textrm{d}.\quad 2:5\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\textrm{Bukti ada pada soal nomor 62}\\  \end{aligned}$.

$\begin{aligned}62.\quad&\textrm{Buktikan jawaban pada soal nomor 61 di atas}\\\\ &\textrm{Bukti}:\\ &\textrm{Perhatikan ilustrasi berikut} \end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Misalkan besar}\:\: \angle CED=x^{\displaystyle 0}\:\: \textrm{dan panjang ruas garis}\\ &ED=OA=OB=OD=r,\:\: \textrm{akibatnya}\:\: \bigtriangleup ODE\\ &\textrm{dan}\:\: \bigtriangleup OBD\:\: \textrm{adalah sama-sama segitiga sama kaki}\\ &\textrm{tetapi beda besar sudutnya. Jika untuk}\:\: \angle CED=x^{\displaystyle 0},\\ &\text{maka}\:\: \angle ODB=2x^{\displaystyle 0},\:\:  \angle CED=\angle EOD=x^{\displaystyle 0}.\\ &\textrm{Akibat lanjutannya juga, yaitu}:\: \angle AOB=3x^{\displaystyle 0},\\ &\text{karena}\:\: \angle ODB=\angle OBD=2x^{\displaystyle 0}.\:\: \textrm{Besar }\: \angle AOB=3x^{\displaystyle 0},\\ &\textrm{karena akibat dari sudut luar}\:\: \bigtriangleup EOB\\ &\textrm{Jadi, rasio}\:\: \angle DEC:\angle AOB=x^{\displaystyle 0}:3x^{\displaystyle 0}=1:3\quad (\textbf{terbukti})   \end{aligned}$.

$\begin{aligned}63.\quad&(\textbf{LM UGM ke-26 Th 2015 Tk.SMA})\\ &\textrm{Perhatikan gambar berikut}\end{aligned}$

$\begin{aligned}\quad\qquad&\textrm{Diketahui lingkaran tersebut berpusat di O dengan}\\ &\textrm{dua buah garis singgung lingkaran masing-masing di}\\ &\textrm{titik A dan B dan dua garis singgung tersebut berpo-}\\ &\textrm{tongan di titik P. Garis QR juga menyinggung ling-}\\ &\textrm{karan di S. Jika diameter lingkaran = 14, serta pan-}\\ &\textrm{jang}\:\:OP=25,\:\: \textrm{keliling}\:\: \bigtriangleup PQR\:\: \:  \textrm{adalah}\: ....\\  &\textrm{a}.\quad 24\\ &\textrm{b}.\quad 48\\ &\textrm{c}.\quad 72\\ &\textrm{d}.\quad 96\\ &\textrm{e}.\quad 120 \\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan bahwa}:PA=PB=24\:\: \textrm{ingat untuk}\\ &\textrm{tripel Pythagoras}:(7,24,25),\: \textrm{karena},\: r=7.\\ &\textrm{Perhatikan pula bahwa}: QS=QA\: \textrm{dan}\: RB=RS\\ &\textrm{akibat dari garis singgung lingkaran. Sehingga}\\ &\textrm{keliling}\:\: \bigtriangleup PQR=PQ+QS+SR+RP\\ &\:\,\qquad\qquad\qquad\quad=PQ+QA+BR+RP\\ &\:\,\qquad\qquad\qquad\quad=PA+BP\\ &\:\,\qquad\qquad\qquad\quad=24+24\\ &\:\,\qquad\qquad\qquad\quad=48  \end{aligned}$.

CONTOH SOAL 12 LINGKARAN (LINGKARAN DALAM DAN LUAR SEGITIGA)

 $\begin{aligned}56.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika PQ}=6\:\: \textrm{cm dan PR}=8\:\: \textrm{cm, maka panjang}\\ &\textrm{jari-jari lingkaran dalam}\:\: \bigtriangleup PQR=\:....\\ &\textrm{a}.\quad 1\:\:\textrm{cm}\\ &\textrm{b}.\quad 2\:\:\textrm{cm}\\ &\textrm{c}.\quad 2,5\:\:\textrm{cm}\\ &\textrm{d}.\quad 3\:\:\textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Cukup jelas bahwa panjang RQ}=10\:\: \textrm{cm}\\ &\textrm{Sehingga panjang jari-jari lingkran dalamnya}:\\ &r_{dalam}=\displaystyle \frac{\left[ PQR \right]}{s}=\frac{\left[ PQR \right]}{\displaystyle \frac{p+q+r}{2}}=\frac{2\left[ PQR \right]}{p+q+r}\\ &\:\:\:\,\quad\quad=\displaystyle \frac{2\left( \displaystyle \frac{1}{2}.6.8 \right)}{6+8+10}=\frac{48}{24}=2\:\: \textrm{cm} \end{aligned}$.

$\begin{aligned}57.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Berdasar gambar di atas, luas}\:\: \bigtriangleup BOC=\:....\\ &\textrm{a}.\quad 10\:\:\textrm{cm}^{\displaystyle 2}\\ &\textrm{b}.\quad 12\:\:\textrm{cm}^{\displaystyle 2}\\ &\textrm{c}.\quad 13\:\:\textrm{cm}^{\displaystyle 2}\\ &\textrm{d}.\quad 14\:\:\textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Cukup jelas bahwa panjang BC}=13\:\: \textrm{cm}\\ &\textrm{Sehingga panjang jari-jari lingkran dalamnya}:\\ &r_{dalam}=\displaystyle \frac{\left[ ABC \right]}{s}=\frac{\left[ ABC \right]}{\displaystyle \frac{a+b+c}{2}}=\frac{2\left[ ABC \right]}{a+b+c}\\ &\:\:\:\,\quad\quad=\displaystyle \frac{2\left( \displaystyle \frac{1}{2}.5.12 \right)}{5+12+13}=\frac{60}{30}=2\:\: \textrm{cm}\\ &\textrm{Selanjutnya untuk luas}\: \bigtriangleup BOC\\ &=\displaystyle \frac{1}{2}\times \textrm{alas}\times \textrm{tinggi}=\displaystyle \frac{1}{2}.13.2=13\:\: \textrm{cm}^{\displaystyle 2} \end{aligned}$.

$\begin{aligned}58.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Berdasar gambar di atas,}\:\: \bigtriangleup ABC\:\: \textrm{siku-siku di A}\\ &\textrm{Luas daerah arsiran adalah}=\:....\\ &\textrm{a}.\quad 47,54\:\:\textrm{cm}^{\displaystyle 2}\\ &\textrm{b}.\quad 46,74\:\:\textrm{cm}^{\displaystyle 2}\\ &\textrm{c}.\quad 25,74\:\:\textrm{cm}^{\displaystyle 2}\\ &\textrm{d}.\quad 28,26\:\:\textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Cukup jelas bahwa panjang AC}=12\:\: \textrm{cm}\\ &\textrm{Sehingga panjang jari-jari lingkran dalamnya}:\\ &r_{dalam}=\displaystyle \frac{\left[ ABC \right]}{s}=\frac{\left[ ABC \right]}{\displaystyle \frac{a+b+c}{2}}=\frac{2\left[ ABC \right]}{a+b+c}\\ &\:\:\:\,\quad\quad=\displaystyle \frac{2\left( \displaystyle \frac{1}{2}.9.12 \right)}{9+12+15}=\frac{108}{36}=3\:\: \textrm{cm}\\ &\textrm{Selanjutnya untuk luas arsiran}:\: \bigtriangleup BOC-L_{\bigcirc }\\ &=\displaystyle \frac{1}{2}\times \textrm{alas}\times \textrm{tinggi}-\pi.r^{\displaystyle 2}\\ &=\displaystyle \frac{1}{2}\times 9\times 12-(3,14).3^{\displaystyle 2}=54-28,26\\ &=25,74\:\: \textrm{cm}^{\displaystyle 2} \end{aligned}$.

$\begin{aligned}59.\quad&\textrm{Sebuah segitiga siku-siku dengan sisi penyikunya}\\ &\textrm{12 cm dan 16 cm. Luas lingkaran terkecil yang dapat }\\ &\textrm{menutupi segitiga tersebut adalah}\:....\\ &\textrm{a}.\quad 314\:\:\textrm{cm}^{\displaystyle 2}\\ &\textrm{b}.\quad \textrm{154  cm}^{\displaystyle 2}\\ &\textrm{c}.\quad \textrm{72  cm}^{\displaystyle 2}\\ &\textrm{d}.\quad \textrm{38,5  cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\textrm{tripel Pythagoras yang terjadi adalah}:(12,16,20)\\ &\textrm{Lingkaran terkecil yang dapat menutupi segitiga}\\ &\textrm{tersebut haruslah}\: diameternya=\textrm{sisi miring segitiga}\\&\textrm{yaitu}=20\:\: \textrm{cm}\\ &\textrm{Sehingga luas lingkarannya}\\ &=\frac{1}{4}\pi.d^{2}=\displaystyle \frac{1}{4}.(3,14).20^{\displaystyle 2}=314\:\: \textrm{cm}^{\displaystyle 2}\\ &\textrm{Berikut sebagai ilustrasi gambarnya}\end{aligned}$.

$\begin{aligned}60.\quad&R_{D}\:\: \textrm{dan}\:\: R_{L}\:\: \textrm{adalah panjang jari-jari lingkaran dalam}\\ &\textrm{dan luar segitiga ABC. Jika}\:\: \bigtriangleup ABC\:\: \textrm{siku-siku di A}\\ &\textrm{dengan AB= 6 cm dan AC = 8 cm, maka}\:\: \displaystyle \frac{R_{D}}{R_{\displaystyle L}}=\:....\\ &\textrm{a}.\quad \displaystyle \frac{2}{3}\\ &\textrm{b}.\quad \displaystyle \frac{1}{2}\\ &\textrm{c}.\quad \displaystyle \frac{2}{5}\\ &\textrm{d}.\quad \displaystyle \frac{3}{5}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{tripel Pythagoras yang terjadi adalah}:(6,8,10)\\  &\textrm{Untuk lingkaran dalam segitiga}:\\ &r_{dalam}=\displaystyle \frac{\left[ ABC \right]}{s}=\frac{\left[ ABC \right]}{\displaystyle \frac{a+b+c}{2}}=\frac{2\left[ ABC \right]}{a+b+c}\\ &\:\:\:\,\quad\quad=\displaystyle \frac{2\left( \displaystyle \frac{1}{2}.6.8 \right)}{6+8+10}=\frac{48}{24}=2\:\: \textrm{cm}\\ &\textrm{Sedangkan untuk lingkaran luar segitiga }\\ &\textrm{akan berdiamter}= 10\:\: \textrm{cm dan jari-jarinya}\\ &\textrm{adalah 5 cm}\\ &\textrm{Sehingga perbandingan}\: \displaystyle \frac{R_{D}}{R_{\displaystyle L}}=\frac{2}{5} \end{aligned}$.

DAFTAR PUSTAKA

1. Kurniawan. 2008. Mandiri Matematika Mengasah Kemampuan Diri SMP Kelas VIII Jilid 2 KTSP 2006. Jakarta: ERLANGGA.

CONTOH SOAL 11 LINGKARAN (GARIS SINGGUNG PERSEKUTUAN DUA LINGKARAN)

 $\begin{aligned}51.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika panjang PQ = 20 cm, maka jarak AB}=\:....\\ &\textrm{a}.\quad \textrm{12 cm}\\ &\textrm{b}.\quad \textrm{21 cm}\\ &\textrm{c}.\quad \textrm{24 cm}\\ &\textrm{d}.\quad \textrm{25 cm}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Perhatikan ilustrasi berikut} \end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Arahkan jawaban kita ke bentuk tripel Pythagoras}.\\ &\textrm{yaitu}:(15,20,25)\Rightarrow AB=25\:\: \textrm{cm}\end{aligned}$.

$\begin{aligned}52.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika panjang PQ = 20 cm, maka jarak terdekat T ke A}=\:....\\ &\textrm{a}.\quad \textrm{25 cm}\\ &\textrm{b}.\quad \textrm{15 cm}\\ &\textrm{c}.\quad \textrm{10 cm}\\ &\textrm{d}.\quad \textrm{6 cm}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan ilustrasi berikut} \end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Arahkan jawaban kita ke bentuk tripel Pythagoras}.\\ &\textrm{yaitu}:(15,20,25)\Rightarrow AB=25\:\: \textrm{cm}\Rightarrow \bigtriangleup ABQ'\\ &\textrm{Pandang juga}\: \bigtriangleup BTQ,\: \textrm{karena sebangun dengan}\\ &\bigtriangleup ABQ',\: \textrm{maka tripelnya adalah}:(6,8,10)\\ &\textrm{Selanjutnya jarak TB}=10\:\: \textrm{cm, sehingga jarak}\\&TA=25-10=15\:\: \textrm{cm}\end{aligned}$.

$\begin{aligned}53.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika PS}=a\:\: \textrm{cm dan QR}=b\:\: \textrm{cm, maka SR}=\:....\\ &\textrm{a}.\quad ab\sqrt{2}\:\:\textrm{cm}\\ &\textrm{b}.\quad 2\sqrt{ab}\:\:\textrm{cm}\\ &\textrm{c}.\quad 2ab\:\:\textrm{cm}\\ &\textrm{d}.\quad 4ab\:\:\textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan ilustrasi berikut} \end{aligned}$.

$\begin{aligned}\quad\qquad&\text{dan perhatikan}\:\: \bigtriangleup PQP',\: \textrm{dengan}\:\: PQ=a+b\\ &\textrm{serta}\:\: PP'=a-b.\:\: \textrm{Sehingga panjang}\:\: SR=P'Q\\ &\textrm{yaitu}:(\textrm{gunakan dalil Pythagoras})\\ &SR^{\displaystyle 2}+\left( PP' \right)^{\displaystyle 2}=PQ^{\displaystyle 2}\\ &\Leftrightarrow SR^{\displaystyle 2}+(a-b)^{\displaystyle 2}=(a+b)^{\displaystyle 2}\\&\Leftrightarrow SR^{\displaystyle 2}=(a+b)^{\displaystyle 2}-(a-b)^{\displaystyle 2}\\ &\Leftrightarrow SR^{\displaystyle 2}=a^{\displaystyle 2}+b^{\displaystyle 2}+2ab-(a^{\displaystyle 2}+b^{\displaystyle 2}-2ab)=4ab\\ &\Leftrightarrow SR=\sqrt{4ab}=2\sqrt{ab}\:\: \textrm{cm}\end{aligned}$.

$\begin{aligned}54.\quad&\textrm{Perhatikan gambar berikut dan nomor soal 55 juga}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika}\:r_{ A}=10\:\: \textrm{cm},\: r_{ B}=6\:\: \textrm{cm dan}\: \: r_{ C}=3\:\: \textrm{cm,}\\ &\textrm{maka panjang KM}=\:....\\ &\textrm{a}.\quad \textrm{25 cm}\\ &\textrm{b}.\quad \textrm{24 cm}\\ &\textrm{c}.\quad \textrm{20 cm}\\ &\textrm{d}.\quad \textrm{19 cm}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan ilustrasi berikut}\\ &\textrm{dan arahkan jawaban kita kebentuk tripel Pythagoras}.\\ &\textrm{yaitu}:(7,24,25)\Rightarrow KM=24\:\: \textrm{cm dengan}\:\: AC=25\: \textrm{cm}\end{aligned}$.

$\begin{aligned}55.\quad&\textrm{Jika}\:r_{ A}=9\:\: \textrm{cm},\: r_{ c}=1\:\: \textrm{cm dan}\: \: KM=15\:\: \textrm{cm,}\\ &\textrm{maka}\:\: r_{C}=\:....\\ &\textrm{a}.\quad 3\displaystyle \frac{1}{2}\:\:\textrm{cm}\\ &\textrm{b}.\quad \textrm{4  cm}\\ &\textrm{c}.\quad \textrm{6  cm}\\ &\textrm{d}.\quad \textrm{7  cm}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\textrm{Perhatikan ilustrasi berikut}\\ &\textrm{dan arahkan jawaban kita kebentuk tripel Pythagoras}.\\ &\textrm{yaitu}:(8,15,17)\Rightarrow KM=15\:\: \textrm{cm dengan}\:\: AC=17\: \textrm{cm}\\ &\textrm{Sehingga}\:\: r_{C}=\displaystyle \frac{7}{2}\:\: \textrm{cm}=3\displaystyle \frac{1}{2}\:\: \textrm{cm}\end{aligned}$.

CONTOH SOAL 10 LINGKARAN (GARIS SINGGUNG LINGKARAN)

 $\begin{aligned}46.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika panjang jari-jari lingkaran 8 cm dan jarak O ke P}\\ &\textrm{adalah 17 cm, maka panjang PQ adalah}\:....\\ &\text{a}.\quad \textrm{12 cm}\\ &\text{b}.\quad \textrm{13 cm}\\ &\text{c}.\quad \textrm{14 cm}\\ &\text{d}.\quad \textrm{15 cm}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Perhatikan}\:\: \bigtriangleup OPQ\:\: \textrm{dengan siku-siku di Q akan}\\ &\textrm{berlaku dalil Pythagoras}: PQ^{\displaystyle 2}+QO^{\displaystyle 2}=PO^{\displaystyle 2}\\ &\Leftrightarrow PQ^{\displaystyle 2}+8^{\displaystyle 2}=17^{\displaystyle 2}\Leftrightarrow PQ^{\displaystyle 2}=15^{\displaystyle 2}\Leftrightarrow PQ=15\: \textrm{cm}\\\\ &\textrm{Pengingat untuk mempermudah penyelesaian}\\ &\textrm{terkait tripel Pythagoras, berikut untuk diingat}\\&\bullet\quad (3,4,5)\\ &\bullet\quad (5,12,13)\\ &\bullet\quad (7,24,25)\\ &\bullet\quad (8,15,17)\\ &\bullet\quad (20,21,29)\\\\ &\textrm{Termasuk juga pembesarannya sekian kali, misal}\\ &\bullet \quad (3,4,5)\Rightarrow \times 2\:\:\textrm{akan menjadi}=(6,8,10)\\ &\bullet \quad (3,4,5)\Rightarrow \times 3\:\:\textrm{akan menjadi}=(9,12,15)\\ &\bullet \quad (3,4,5)\Rightarrow \times 4\:\:\textrm{akan menjadi}=(12,16,20)\\ &\bullet \quad (5,12,13)\Rightarrow \times 2\:\:\textrm{akan menjadi}=(10,24,26)\\ &\bullet \quad (5,12,13)\Rightarrow \times 3\:\:\textrm{akan menjadi}=(15,36,39)\\&\textrm{Demikian juga tripel Pythagoras yang lainnya}\end{aligned}\end{aligned}$.

 $\begin{aligned}47.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika panjang OQ 5 cm dan OP 13 cm, maka PQ}=\:....\\ &\textrm{a}.\quad \textrm{8 cm}\\ &\textrm{b}.\quad \textrm{9 cm}\\ &\textrm{c}.\quad \textrm{10 cm}\\ &\textrm{d}.\quad \textrm{12 cm}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Cukup jelas, lihat uraian di atas pada nomor sebelumnya} \end{aligned}$. 

 $\begin{aligned}48.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika panjang PR = 16 cm dan QR = 20 cm, maka OP}=\:....\\ &\textrm{a}.\quad \textrm{4 cm}\\ &\textrm{b}.\quad \textrm{5 cm}\\ &\textrm{c}.\quad \textrm{6 cm}\\ &\textrm{d}.\quad \textrm{8 cm}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Cukup jelas, lihat uraian di atas pada nomor sebelumnya}\\ &\textrm{dan arahkan jawaban kita kebentuk tripel Pythagoras}.\\ &OP=\displaystyle \frac{1}{2}PQ\Leftrightarrow PQ=12\:\: \textrm{cm dan tersusunlah tripel}\\&\textrm{Pythagoras, yaitu}:(12,16,20) \end{aligned}$.

 $\begin{aligned}49.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika panjang OS = 6 cm, maka panjang PQ}=\:....\\ &\textrm{a}.\quad \textrm{17 cm}\\ &\textrm{b}.\quad \textrm{20 cm}\\ &\textrm{c}.\quad \textrm{21 cm}\\ &\textrm{d}.\quad \textrm{24 cm}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Cukup jelas, lihat uraian di atas pada nomor sebelumnya}\\ &\textrm{dan arahkan jawaban kita kebentuk tripel Pythagoras}.\\ &OS=\displaystyle \frac{1}{2}RS\Leftrightarrow PQ=12\:\: \textrm{cm dan tersusunlah tripel}\\&\textrm{Pythagoras, yaitu}:(5,12,13)\:\: \textrm{dan}\:\:(12,16,20)\\ &\textrm{Sehingga panjang PQ}=5+16=21\:\: \textrm{cm} \end{aligned}$.

$\begin{aligned}50.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika panjang OB = 6 cm dan BC = 8 cm, maka panjang AB}=\:....\\ &\textrm{a}.\quad \textrm{10 cm}\\ &\textrm{b}.\quad \textrm{9,6 cm}\\ &\textrm{c}.\quad \textrm{8,4 cm}\\ &\textrm{d}.\quad \textrm{7 cm}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Cukup jelas, lihat uraian di atas pada nomor sebelumnya}\\ &\textrm{dan arahkan jawaban kita kebentuk tripel Pythagoras}.\\ &\textrm{yaitu}:(6,8,10)\Rightarrow OC=10\:\: \textrm{cm}\\&\textrm{Sekarang perhatikan untuk luas}\:\: \bigtriangleup OBC\:\:\textrm{atau}\:\:\left[ OBC \right]\\&\left[ OBC \right]=\left[ OBC \right]\Leftrightarrow \displaystyle \frac{1}{2}\textrm{alas}\times \textrm{tinggi}=\displaystyle \frac{1}{2}\textrm{alas}\times \textrm{tinggi}\\&\Leftrightarrow \displaystyle \frac{1}{2}\times OC\times \textrm{tinggi}_{\frac{1}{2}\textrm{AB}}=\displaystyle \frac{1}{2}\times OB\times BC\\ &\Leftrightarrow \displaystyle \frac{1}{2}\times 10\times \textrm{tinggi}_{\frac{1}{2}\textrm{AB}}=\displaystyle \frac{1}{2}\times 6\times 8\\ &\Leftrightarrow  \textrm{tinggi}_{\frac{1}{2}\textrm{AB}}=\displaystyle \frac{48}{10}\:\: \textrm{cm}.\:\: \textrm{Sehingga panjang tali busur AB}\\ &=2\times \textrm{tinggi}_{\frac{1}{2}\textrm{AB}}=2\left(\displaystyle  \frac{48}{10} \right)=9,6\:\: \textrm{cm}\\ \end{aligned}$.

CONTOH SOAL 9 LINGKARAN (GARIS SINGGUNG)

 $\begin{aligned}41.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Banyak garis singgung yang melalui titik A adalah}\:....\\ &\text{a}.\quad 0\\ &\text{b}.\quad 1\\ &\text{c}.\quad 2\\ &\text{d}.\quad \textrm{banyak sekali}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&\textrm{Cukup jelas}\end{aligned}\end{aligned}$.

$\begin{aligned}42.\quad&\textrm{Besar sudut yang dibentuk oleh garis singgung}\\ &\textrm{dan jari-jari lingkarannya adalah}\:....\\ &\text{a}.\quad 45^{\displaystyle 0}\\ &\text{b}.\quad 60^{\displaystyle 0}\\ &\text{c}.\quad 90^{\displaystyle 0}\\ &\text{d}.\quad 180^{\displaystyle 0}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Cukup jelas}\\ &\textrm{Nama lain dari garis terondisi ini adalah garis}\: tangen\end{aligned}$.

$\begin{aligned}43.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Banyak garis singgung yang dapat ditarik dari titik P adalah}\:....\\ &\text{a}.\quad 1\\ &\text{b}.\quad 2\\ &\text{c}.\quad 3\\ &\text{d}.\quad \textrm{banyak sekali}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&\textrm{hanya dua dan cukup jelas}\end{aligned}\end{aligned}$.

$\begin{aligned}44.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Yang merupakan titik singgung pada gambar di atas adalah}\:....\\ &\text{a}.\quad O\\ &\text{b}.\quad A\\ &\text{c}.\quad B\\ &\text{d}.\quad C\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Cukup jelas}\\\end{aligned}$.

$\begin{aligned}45.\quad&\textrm{Banyak garis singgung yang melalui seuah titik pada}\\ &\textrm{lingkaran adalah}\:....\\ &\text{a}.\quad \textrm{1 buah}\\ &\text{b}.\quad \textrm{2 buah}\\ &\text{c}.\quad \textrm{3 buah}\\ &\text{d}.\quad \textrm{banyak sekali}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\textrm{Cukup jelas}\\\end{aligned}$.

CONTOH SOAL 8 LINGKARAN (SUDUT ANTARA DUA TALI BUSUR)

 $\begin{aligned}36.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika}\:\: \angle AOC=151^{\displaystyle 0},\: \angle BOD=23^{\displaystyle 0},\:\: \textrm{maka}\:\:\: \angle AEC=\: .... \\ &\text{a}.\quad 87^{\displaystyle 0}\\ &\text{b}.\quad 78^{\displaystyle 0}\\ &\text{c}.\quad 69^{\displaystyle 0}\\ &\text{d}.\quad 64^{\displaystyle 0}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\textrm{Ingat sudut antara dua tali busur, yaitu:}\\ &\angle AEC=\displaystyle \frac{1}{2}\left( \angle AOC+\angle BOD \right)=\displaystyle \frac{1}{2}\left( 151^{\displaystyle 0}+23^{\displaystyle 0} \right)\\ &\:\quad\qquad=\displaystyle \frac{1}{2}\left( 174^{\displaystyle 0} \right)=87^{\displaystyle 0} \end{aligned}\end{aligned}$.

 $\begin{aligned}37.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika}\:\: \angle BOD=50^{\displaystyle 0},\: \angle AOC=110^{\displaystyle 0},\:\: \textrm{maka}\:\:\: \angle AEC=\: .... \\ &\text{a}.\quad 60^{\displaystyle 0}\\ &\text{b}.\quad 75^{\displaystyle 0}\\ &\text{c}.\quad 80^{\displaystyle 0}\\ &\text{d}.\quad 120^{\displaystyle 0}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\textrm{Ingat sudut antara dua tali busur, yaitu:}\\ &\angle AEC=\displaystyle \frac{1}{2}\left( \angle AOC+\angle BOD \right)=\displaystyle \frac{1}{2}\left( 110^{\displaystyle 0}+50^{\displaystyle 0} \right)\\ &\:\quad\qquad=\displaystyle \frac{1}{2}\left( 160^{\displaystyle 0} \right)=80^{\displaystyle 0} \end{aligned}\end{aligned}$.

$\begin{aligned}38.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika}\:\: \angle BCD=68^{\displaystyle 0},\: \angle AEC=105^{\displaystyle 0},\:\: \textrm{maka}\:\:\: \angle ADC=\: .... \\ &\text{a}.\quad 87^{\displaystyle 0}\\ &\text{b}.\quad 73^{\displaystyle 0}\\ &\text{c}.\quad 37^{\displaystyle 0}\\ &\text{d}.\quad 240^{\displaystyle 0}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\textrm{Ingat sudut keliling}\\ &\angle CEB=180^{\displaystyle 0}-105^{\displaystyle 0}=75^{\displaystyle 0}\:\: (\textrm{pelurus sudut})\\ &\textrm{Selanjutnya perhatikan bahwa pada}\: \bigtriangleup CEB\\ &\textrm{total sudutnya}=180^{\displaystyle 0}\\&\angle C+\angle E+\angle B=180^{\displaystyle 0}\Leftrightarrow 68^{\displaystyle 0}+75+\angle B^{\displaystyle 0}=180^{\displaystyle 0}\\ &\Leftrightarrow \angle B=\angle EBC=37^{\displaystyle 0}\\ &\textrm{Dan kita juga tahu bahwa}:\angle EBC=\angle ADC=37^{\displaystyle 0}\\ &(\textrm{Sama-sama sudut keliling menghdap busur yang sama}) \end{aligned}\end{aligned}$.

$\begin{aligned}39.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika}\:\: \angle BCD=41^{\displaystyle 0},\: \angle ABC=83^{\displaystyle 0},\:\: \textrm{maka}\:\:\: \angle AEC=\: .... \\ &\text{a}.\quad 42^{\displaystyle 0}\\ &\text{b}.\quad 41^{\displaystyle 0}\\ &\text{c}.\quad 40^{\displaystyle 0}\\ &\text{d}.\quad 21^{\displaystyle 0}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\textrm{Perhatikan bahwa pada}\:\: \bigtriangleup CEB\\ &\angle BCE+\angle BEC=\angle ABC\\ &\Leftrightarrow 41^{\displaystyle 0}+\angle BEC=83^{\displaystyle 0}\\ &\Leftrightarrow \angle BEC=\angle AEC=83^{\displaystyle 0}-41^{\displaystyle 0}=42^{\displaystyle 0}\end{aligned}\end{aligned}$.

$\begin{aligned}40.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika}\:\: \angle AFC=105^{\displaystyle 0},\: \angle BCD=28^{\displaystyle 0},\:\: \textrm{maka}\:\:\: \angle AEC=\: .... \\ &\text{a}.\quad 75^{\displaystyle 0}\\ &\text{b}.\quad 47^{\displaystyle 0}\\ &\text{c}.\quad 45^{\displaystyle 0}\\ &\text{d}.\quad 40^{\displaystyle 0}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&\textrm{Ingat segiempat tali busur}\:\: \angle AFC+\angle ABC=180^{\displaystyle 0}\\ &\Leftrightarrow 105^{\displaystyle 0}+\angle ABC=180^{\displaystyle 0}\Leftrightarrow \angle ABC=75^{\displaystyle 0}\\ &\textrm{Perhatikan segitiga}\quad \bigtriangleup BEC\\ &\angle AEC+\angle BEC=\angle ABC\\&\Leftrightarrow \angle AEC+28^{\displaystyle 0}=75^{\displaystyle 0}\\ &\Leftrightarrow \angle AEC=75^{\displaystyle 0}-28^{\displaystyle 0}=47^{\displaystyle 0}\end{aligned}\end{aligned}$.

CONTOH SOAL 7 LINGKARAN (SEGI EMPAT TALI BUSUR)

 $\begin{aligned}31.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Besar}\:\: \angle C+\angle D=\: .... \\ &\text{a}.\quad 145^{\displaystyle 0}\\ &\text{b}.\quad 180^{\displaystyle 0}\\ &\text{c}.\quad 215^{\displaystyle 0}\\ &\text{d}.\quad 290^{\displaystyle 0}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned} &\textrm{Perhatikan bahwa}\:\: \angle A+\angle B+\angle C+\angle D=360^{\displaystyle 0}\\ &\textrm{akibat dari segi empat tali busur. Sehingga besar}\\ &\angle A+\angle B+\angle C+\angle D=360^{\displaystyle 0}\\ &80^{\displaystyle 0}+65^{\displaystyle 0}+\angle C+\angle D=360^{\displaystyle 0}\\ &\angle C+\angle D=360^{\displaystyle 0}-(80^{\displaystyle 0}+65^{\displaystyle 0})=215^{\displaystyle 0}\end{aligned}\end{aligned}$.

$\begin{aligned}32.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika}\:\: \angle PSO=3x^{\displaystyle 0},\: \angle PQO=2x^{\displaystyle 0},\:\: \textrm{serta}\:\:\: \angle PQRS=75^{\displaystyle 0}\\ &\textrm{maka nilai}\:\:x=\: .... \\ &\text{a}.\quad 15^{\displaystyle 0}\\ &\text{b}.\quad 21^{\displaystyle 0}\\ &\text{c}.\quad 45^{\displaystyle 0}\\ &\text{d}.\quad 50^{\displaystyle 0}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&\textrm{Perhatikan ilustrasi berikut} \end{aligned}\end{aligned}$.

$\begin{aligned}\quad\qquad&\angle SPQ=3x^{\displaystyle 0}+2x^{\displaystyle 0}=105^{\displaystyle 0}\\ &\textrm{Ingat bahwa}\:\:  \angle P +\angle R=180^{\displaystyle 0}\:\: \textrm{serta segitiga}\\ &\textrm{SPO dan QPO adalah segita sama kaki, sehingga}\\ &\angle SPQ=3x^{\displaystyle 0}+2x^{\displaystyle 0}=5x^{\displaystyle 0}=105^{\displaystyle 0}\\&\Leftrightarrow x^{\displaystyle 0}=21^{\displaystyle 0}\end{aligned}$.

$\begin{aligned}33.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika}\:\: \angle QRT=108^{\displaystyle 0},\: \:\: \textrm{dan}\:\:\: PQ=PS\\ &\textrm{maka}\:\:\angle PQO=\: .... \\ &\text{a}.\quad 36^{\displaystyle 0}\\ &\text{b}.\quad 42^{\displaystyle 0}\\ &\text{c}.\quad 48^{\displaystyle 0}\\ &\text{d}.\quad 54^{\displaystyle 0}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\textrm{Perhatikan ilustrasi berikut} \end{aligned}\end{aligned}$.

$\begin{aligned}\quad\qquad &\angle QRS=180^{\displaystyle 0}-\angle QRT\quad (\textrm{sudut pelurus})\\ &\:\qquad\quad=180^{\displaystyle 0}-108^{\displaystyle 0}=72^{\displaystyle 0}\\ &\angle QPS+\angle QRS=180^{\displaystyle 0}\Leftrightarrow \angle QPS=180^{\displaystyle 0}-\angle QRS\\ &(\textrm{akibat segi empat tali busur})\\&\bullet \quad\textrm{Bagian Pertama}:\bigtriangleup PSQ\\&\angle QPS=180^{\displaystyle 0}-72^{\displaystyle 0}=108^{\displaystyle 0}.\: \textrm{Karena}\: \bigtriangleup PSQ\: \textrm{sama kaki},\\ &\textrm{maka}\:\: \angle PQS=\angle PSQ.\:\: \textrm{Selanjutnya untuk}\:\: \bigtriangleup PSQ\\ &\angle PQS+\angle PSQ+\angle QPS=180^{\displaystyle 0}\\ &\Leftrightarrow 2\angle PQS+108^{\displaystyle 0}=180^{\displaystyle 0}\Leftrightarrow \angle PQS=36^{\displaystyle 0}\\ &\bullet \quad\textrm{Bagian Kedua}:\bigtriangleup SQO\\ &\angle QPS\:\: \textrm{dan}\:\: \angle QOS\:\: \textrm{menghadap busur besar yang}\\&\textrm{sama, yaitu}\:\:  \widehat{QS},\:\: \textrm{akibatnya}:\angle QOS=2\angle QPS=216^{\displaystyle 0}.\\ &\textrm{Akibat lanjutannya adalah}\\ &\angle QOS_{\displaystyle \textrm{busur kecil}}=360^{\displaystyle 0}-216^{\displaystyle 0}=144^{\displaystyle 0}.\\ &\textrm{Karena}\:\: \bigtriangleup SQO\:\: \textrm{sama kaki juga, maka}\:\: \angle OQS=\angle OSQ\\ &\textrm{Selanjutnya}:\angle OQS+\angle OSQ+\angle QOS=180^{\displaystyle 0}\\  & \Leftrightarrow 2\angle OQS+144^{\displaystyle 0}=180^{\displaystyle 0}\Leftrightarrow \angle OQS=18^{\displaystyle 0}.\\ &\textrm{Sehingga besar}\:\: \angle PQO=\angle PQS+\angle OQS\\ &\qquad\qquad\qquad\qquad\qquad=36^{\displaystyle 0}+18^{\displaystyle 0}=54^{\displaystyle 0}\end{aligned}$.

$\begin{aligned}34.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Nilai}\:\:x-y=\: .... \\ &\text{a}.\quad 51^{\displaystyle 0}\\ &\text{b}.\quad 37^{\displaystyle 0}\\ &\text{c}.\quad 28^{\displaystyle 0}\\ &\text{d}.\quad 21^{\displaystyle 0}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&\textrm{Perhatikan bahwa}\: \begin{cases} 2x+3^{\displaystyle 0} &+&75^{\displaystyle 0} =180 \\ 3y-3^{\displaystyle 0} &+&140^{\displaystyle 0} =180 \end{cases}\\ &\textrm{Selanjutnya}\:\: \begin{cases} 2x & =180-75^{\displaystyle 0}-3^{\displaystyle 0}&\Leftrightarrow x=51^{\displaystyle 0} \\ 3y& =180-140^{\displaystyle 0}+3^{\displaystyle 0}&\Leftrightarrow y=14,\overline{33}^{\displaystyle 0} \end{cases}\\ &\textrm{Sehingga nilai}:x-y=51^{\displaystyle 0}-14,\overline{33}^{\displaystyle 0}=36,67^{\displaystyle 0}\simeq  37^{\displaystyle 0}\end{aligned}\end{aligned}$.

$\begin{aligned}35.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika}\:\: \angle P:\angle Q:\angle R=5:8:7,\: \:\: \textrm{maka}\:\:\angle R=\: .... \\ &\text{a}.\quad 112^{\displaystyle 0}\\ &\text{b}.\quad 105^{\displaystyle 0}\\ &\text{c}.\quad 75^{\displaystyle 0}\\ &\text{d}.\quad 56^{\displaystyle 0}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{misalkan}:\angle P=5n,\:\angle Q=8n,\: \textrm{dan}\:\: \angle R=7n\\ &\textrm{Selanjutnya}\\ &\angle P+\angle Q=180^{\displaystyle 0}\Leftrightarrow 5n+7n=180^{\displaystyle 0}\Leftrightarrow 12n=180^{\displaystyle 0}\\&\textrm{sehingga}\quad n=15^{\displaystyle 0}\:\: \textrm{dan besar}\:\: \angle R=7n=7\times 15^{\displaystyle 0}=105^{\displaystyle 0}\end{aligned}$.

CONTOH SOAL 6 LINGKARAN (SUDUT PUSAT SUDUT KELILING)

 $\begin{aligned}26.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika}\:\: \angle BOC=100^{\displaystyle 0},\: \textrm{maka}\:\:\angle OAC=\: .... \\ &\text{a}.\quad 25^{\displaystyle 0}\\ &\text{b}.\quad 50^{\displaystyle 0}\\ &\text{c}.\quad 75^{\displaystyle 0}\\ &\text{d}.\quad 80^{\displaystyle 0}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&\angle OAC+\angle OCA=\angle BOC,\quad \textrm{dengan}\:\:\angle OAC=\angle OCA\\ &\textrm{akibat segitiga sama kaki, sehingga} \\ &2\angle OAC=100^{\displaystyle 0}\Leftrightarrow \angle OAC=50^{\displaystyle 0}\end{aligned}\end{aligned}$.

 $\begin{aligned}27.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika}\:\: \angle OCB=65^{\displaystyle 0},\: \textrm{maka}\:\:\angle AOC=\: .... \\ &\text{a}.\quad 50^{\displaystyle 0}\\ &\text{b}.\quad 65^{\displaystyle 0}\\ &\text{c}.\quad 70^{\displaystyle 0}\\ &\text{d}.\quad 130^{\displaystyle 0}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&\angle OCB+\angle OBC=\angle AOC,\quad \textrm{dengan}\:\:\angle OCB=\angle OBC\\ &\textrm{akibat segitiga sama kaki, sehingga} \\ &2(65^{\displaystyle 0})=130^{\displaystyle 0}=\angle AOC\Leftrightarrow \angle AOC=130^{\displaystyle 0}\end{aligned}\end{aligned}$.

 $\begin{aligned}28.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika}\:\: \angle AOB=40^{\displaystyle 0},\: \textrm{maka}\:\:\angle ACD=\: .... \\ &\text{a}.\quad 70^{\displaystyle 0}\\ &\text{b}.\quad 72^{\displaystyle 0}\\ &\text{c}.\quad 80^{\displaystyle 0}\\ &\text{d}.\quad 83^{\displaystyle 0}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}&\angle AOB\quad \textrm{adalah pelurus}\:\:\angle AOD\\ &\textrm{selanjutnya}\\ &\angle AOD=180^{\displaystyle 0}-\angle AOB=180^{\displaystyle 0}-40^{\displaystyle 0}=140^{\displaystyle 0} \\ &\textrm{Dan}\:\: \angle ACD=\displaystyle \frac{1}{2}\angle AOD\\ &\textrm{karena sudut keliling yang menghadap busur}\:\: \widehat{AD}\\ &\textrm{Sehingga}\:\: \angle ACD=\displaystyle \frac{1}{2}\times 140^{\displaystyle 0}=70^{\displaystyle 0} \end{aligned}\end{aligned}$.

$\begin{aligned}29.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika}\:\: \angle AOC=150^{\displaystyle 0},\: \textrm{maka}\:\:\angle ABC=\: .... \\ &\text{a}.\quad 50^{\displaystyle 0}\\ &\text{b}.\quad 65^{\displaystyle 0}\\ &\text{c}.\quad 70^{\displaystyle 0}\\ &\text{d}.\quad 75^{\displaystyle 0}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned} &\textrm{Perhatikan bahwa}\:\: \angle ABC=\displaystyle \frac{1}{2}\angle AOC\\ &\textrm{karena sudut keliling yang menghadap busur}\:\: \widehat{AC}\\ &\textrm{Sehingga}\:\: \angle ABC=\displaystyle \frac{1}{2}\times 150^{\displaystyle 0}=75^{\displaystyle 0} \end{aligned}\end{aligned}$.

$\begin{aligned}30.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika}\:\: \angle AOC=150^{\displaystyle 0},\: \textrm{maka}\:\:\angle ABC=\: .... \\ &\text{a}.\quad 30^{\displaystyle 0}\\ &\text{b}.\quad 90^{\displaystyle 0}\\ &\text{c}.\quad 105^{\displaystyle 0}\\ &\text{d}.\quad 210^{\displaystyle 0}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned} &\textrm{Perhatikan bahwa}\:\: \angle ABC=\displaystyle \frac{1}{2}\angle AOC\\ &\textrm{karena sudut keliling yang menghadap busur}\:\: \widehat{AC}\\ &\textrm{Sehingga}\:\: \angle ABC=\displaystyle \frac{1}{2}\times \left(360^{\displaystyle 0}- 150^{\displaystyle 0} \right)=105^{\displaystyle 0}\\ &\textrm{Sebagai catatan}:\\ &\angle ABC\:\: \textrm{menghadap busur besar}\:\: \widehat{AC}\:\: \textrm{dan tidak}\\ & \textrm{menghadap busur kecil}\:\: \widehat{AC} \end{aligned}\end{aligned}$.

LINGKARAN-DALAM, LUAR DAN SINGGUNG SEGITIGA-LANJUTAN

 I. Lingkaran Dalam, Luar, dan Singgung Segitiga

Perhatikan ilustasi berikut

$\Large\begin{array}{|c|}\hline \displaystyle \frac{a}{\sin A}= \frac{b}{\sin B}=\frac{c}{\sin C}=2R\\\hline \end{array}$.

$\begin{aligned}1.\quad a&=b.\displaystyle \frac{\sin \angle A}{\sin \angle B}=c.\displaystyle \frac{\sin \angle A}{\sin \angle C}=2R\sin \angle A\\ 2.\quad b&=c.\displaystyle \frac{\sin \angle B}{\sin \angle C}=a.\displaystyle \frac{\sin \angle B}{\sin \angle A}=2R\sin \angle B\\ 3.\quad c&=a.\displaystyle \frac{\sin \angle C}{\sin \angle A}=b.\displaystyle \frac{\sin \angle C}{\sin \angle B}=2R\sin \angle C \end{aligned}$.

Sehingga luas segitiga dapat dituliskan sebagai berikut:

$\begin{aligned} 1.\quad L\bigtriangleup ABC&=\displaystyle \frac{1}{2}ab\sin \angle C\\ &=\displaystyle \frac{1}{2}a\left ( a.\displaystyle \frac{\sin \angle B}{\sin \angle A} \right )\sin \angle C\\ &=\displaystyle \frac{1}{2}a^{2}\displaystyle \frac{\sin \angle B\sin \angle C}{\sin \angle A}\\ 2.\quad L\bigtriangleup ABC&=\displaystyle \frac{1}{2}bc\sin \angle A\\ &=\displaystyle \frac{1}{2}b\left ( b.\displaystyle \frac{\sin \angle C}{\sin \angle B} \right )\sin \angle A\\ &=\displaystyle \frac{1}{2}b^{2}\displaystyle \frac{\sin \angle B\sin \angle A}{\sin \angle B}\\ 3.\quad L\bigtriangleup ABC&=\displaystyle \frac{1}{2}ac\sin \angle B\\ &=\displaystyle \frac{1}{2}\left ( c.\displaystyle \frac{\sin \angle A}{\sin \angle C} \right )c\sin \angle B\\ &=\displaystyle \frac{1}{2}c^{2}\displaystyle \frac{\sin \angle A\sin \angle B}{\sin \angle C} \end{aligned}$.

I.1 Luas segitiga berdasar tiga sisinya (Heron's formula)

Bukti Luas Segitiga dengan sisi a, b, dan c

$\begin{aligned}& \textrm{Bagaimana membuktikan luas suatu}\\ &\textrm{segitiga jika diketahui sisinya}\: a,b\: \textrm{dan}\: c\\ &\textrm{berupa rumus}\\ &L_{\bigtriangleup }=\left [ ABC \right ]=\sqrt{s(s-a)(s-b)(s-c)}\\ & \textrm{dengan}\\ &\qquad s=\displaystyle \frac{1}{2}(a+b+c) \end{aligned}$.

Berikut akan dipaparkan buktinya

$\begin{aligned}\displaystyle \textrm{L}{\bigtriangleup }\textrm{ABC}&=\frac{1}{2}bc\sin\angle A\\ &=\displaystyle \frac{1}{2}bc\sqrt{\sin ^{2}\angle A}\\ &=\displaystyle \frac{1}{2}\sqrt{b^{2}c^{2}\left ( \sin ^{2}\angle A \right )}\\ &=\displaystyle \frac{1}{2}\sqrt{b^{2}c^{2}\left ( 1-\cos ^{2}\angle A \right )},\\ &\textrm{ingat bahwa};\: \cos \angle A=\frac{b^{2}+c^{2}-a^{2}}{2bc}\\ &=\displaystyle \frac{1}{2}\sqrt{b^{2}c^{2}\left ( 1-\left ( \frac{b^{2}+c^{2}-a^{2}}{2bc} \right )^{2} \right )}\\ &=\displaystyle \frac{1}{2}\sqrt{b^{2}c^{2}-\left ( \frac{b^{2}+c^{2}-a^{2}}{2} \right )^{2}}\\ &=\displaystyle \frac{1}{2}\sqrt{\frac{4b^{2}c^{2}-\left ( b^{2}+c^{2}-a^{2} \right )^{2}}{4}}\\ &=\displaystyle \frac{1}{2}.\frac{1}{2}\sqrt{\left ( 2bc \right )^{2}-\left ( b^{2}+c^{2}-a^{2} \right )^{2}}\\ &=\displaystyle \frac{1}{4}\sqrt{\left ( 2bc+b^{2}+c^{2}-a^{2} \right )\left (2bc-b^{2}-c^{2}+a^{2} \right )}\\ &=\frac{1}{4}\sqrt{\left \{ \left ( b+c \right )^{2}-a^{2} \right \}\left \{ a^{2}-\left ( b-c \right )^{2} \right \}}\\ &=\displaystyle \frac{1}{4}\sqrt{\left ( b+c+a \right )\left ( b+c-a \right )\left ( a+b-c \right )\left ( a-b+c \right )},\\ &\textrm{dengan mengingat bahwa}\: 2s=a+b+c\\ &=\displaystyle \frac{1}{4}\sqrt{(2s)(2s-2a)(2s-2b)(2s-2c)}\\ &=\displaystyle \frac{1}{4}.\sqrt{16.s(s-a)(s-b)(s-c)}\\ &=\displaystyle \frac{1}{4}.4\sqrt{s(s-a)(s-b)(s-c)}\\ \textrm{L}\bigtriangleup \textrm{ABC}&=\sqrt{s(s-a)(s-b)(s-c)}\quad \blacksquare \end{aligned}$.

Rumus di atas lebih dikenal dengan istilah rumus Heron lihat Heron's formula di sini.

Sumber tulisan lagi di antara silahkan kunjungi di sini.

I.2 Luas segitiga sama sisi

$\begin{aligned}L_{\bigtriangleup }ABC&=\displaystyle \frac{1}{2}ab\sin \angle C,\quad a=b=c\\ &\qquad\quad\quad \textrm{dan}\: \: \angle A=\angle B\angle C=60^{\circ}\\ &=\displaystyle \frac{1}{2}a.a\sin 60^{\circ}\\ &=\displaystyle \frac{1}{2}a^{2}\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\displaystyle \frac{1}{4}a^{2}\sqrt{3} \end{aligned}$.

I.3 Lingkaran Luar Segitiga

Perhatikan lagi lingkaran luar segitiga di atas, dari sana kita akan mendapatkan rumus luas segitiga yang dapat kita munculkan harga R nya, yaitu:

$\begin{aligned}1.\quad L\bigtriangleup ABC&=\displaystyle \frac{1}{2}ab\sin \angle C\\ &=\displaystyle \frac{1}{2}(2R\sin \angle A)(2R\sin \angle B)\sin \angle C\\ &=2R^{2}\sin \angle A\sin \angle B\sin \angle C\\ 2.\quad L\bigtriangleup ABC&=\displaystyle \frac{1}{2}ab\sin \angle C\\ &=\displaystyle \frac{1}{2}ab\left ( \displaystyle \frac{c}{2R} \right )\\ &=\displaystyle \frac{abc}{4R} \end{aligned}$.

I.4 Lingkaran dalam segitiga

Perhatikanlah gambar berikut

$\begin{aligned}\textrm{Diketahu}&\textrm{i}\\ L_{\bigtriangleup }AOB&=\displaystyle \frac{1}{2}(AB)(OD)=\displaystyle \frac{1}{2}cr\\ L_{\bigtriangleup }AOC&=\displaystyle \frac{1}{2}(AC)(OF)=\displaystyle \frac{1}{2}br\\ L_{\bigtriangleup }BOC&=\displaystyle \frac{1}{2}(BC)(OE)=\displaystyle \frac{1}{2}ar\\ \textrm{Sehingga}&\\ L_{\bigtriangleup }ABC&=\left [ ABC \right ]\\ &=\displaystyle \frac{1}{2}ar+\displaystyle \frac{1}{2}br+\displaystyle \frac{1}{2}cr\\ &=\displaystyle \frac{1}{2}r(a+b+c)\\ &=\displaystyle \frac{1}{2}r(2s)\\ &=rs \end{aligned}$.

I.5 Lingkaran singgung segitiga

Sebagai ilustrasinya adalah gambar berikut

$\begin{aligned}&\textrm{Diketahui}\\ &DO=EO=FO=r_{a}\\ &\textrm{maka}\\ &1.\quad L_{\bigtriangleup}ABO=\displaystyle \frac{1}{2}(AB)(OD)=\displaystyle \frac{1}{2}cr_{a}\\ &2.\quad L_{\bigtriangleup}ACO=\displaystyle \frac{1}{2}(AC)(OE)=\displaystyle \frac{1}{2}br_{a}\\ &3.\quad L_{\bigtriangleup}BCO=\displaystyle \frac{1}{2}(BC)(OF)=\displaystyle \frac{1}{2}ar_{a} \end{aligned}$.

$\begin{aligned} \textrm{Sehingga}&\\ L_{\bigtriangleup }ABC&=\left [ ABC \right ]\\ &=\left [ ACO \right ]+\left [ ABO \right ]-\left [ BCO \right ]\\ &=\displaystyle \frac{1}{2}br_{a}+\displaystyle \frac{1}{2}cr_{a}-\displaystyle \frac{1}{2}ar_{a}\\ &=\displaystyle \frac{1}{2}r_{a}(b+c-a)\\ &=\displaystyle \frac{1}{2}r_{a}(a+b+c-2a)\\ &=\displaystyle \frac{1}{2}r_{a}(2s-2a)\\ &=r_{a}(s-a) \end{aligned}$.

$\LARGE\fbox{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diberikan sembarang}\: \: \bigtriangleup ABC\: .\: \textrm{Jika}\: \: r\\ & \textrm{merupakan jari-jari lingkaran singgung }\\ &\textrm{dalam pada}\: \: \bigtriangleup ABC\: \: \textrm{dan}\: \: r_{a},\: r_{b},\: r_{c}\\ &\textrm{adalah jari-jari singgung luar pad}\: \: \bigtriangleup ABC\\ &\textrm{tunjukkan bahwa}:\: \displaystyle \frac{1}{r_{a}}+\frac{1}{r_{b}}+\frac{1}{r_{c}}=\frac{1}{r}\\\\ &\textbf{Bukti}:\\ &\begin{aligned} \textrm{Diketahu}&\textrm{i}\\ L_{\bigtriangleup }ABC&=r_{a}(s-a),\: \Rightarrow r_{a}=\displaystyle \frac{\left [ ABC \right ]}{s-a}\\ L_{\bigtriangleup }ABC&=r_{b}(s-b),\: \Rightarrow r_{b}=\displaystyle \frac{\left [ ABC \right ]}{s-b}\\ L_{\bigtriangleup }ABC&=r_{c}(s-c),\: \Rightarrow r_{c}=\displaystyle \frac{\left [ ABC \right ]}{s-c}\\ \textrm{maka}\: \quad&\\ \displaystyle \frac{1}{r_{a}}+\frac{1}{r_{b}}&+\frac{1}{r_{c}}\\ &=\displaystyle \frac{1}{\displaystyle \frac{\left [ ABC \right ]}{s-a}}+\displaystyle \frac{1}{\displaystyle \frac{\left [ ABC \right ]}{s-b}}+\displaystyle \frac{1}{\displaystyle \frac{\left [ ABC \right ]}{s-c}}\\ &=\displaystyle \frac{s-a}{\left [ ABC \right ]}+\displaystyle \frac{s-b}{\left [ ABC \right ]}+\displaystyle \frac{s-c}{\left [ ABC \right ]}\\ &=\displaystyle \frac{s-a+s-b+s-c}{\left [ ABC \right ]}\\ &=\displaystyle \frac{3s-(a+b+c)}{\left [ ABC \right ]}\\ &=\displaystyle \frac{3s-2s}{\left [ ABC \right ]}\\ &=\displaystyle \frac{s}{\left [ ABC \right ]}\\ &=\displaystyle \frac{1}{\displaystyle \frac{\left [ ABC \right ]}{s}}\\ &=\displaystyle \frac{1}{r}\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Pada}\: \: \bigtriangleup ABC\: ,\: \textrm{Jika}\: \: AB=20\: \textrm{cm},\: BC=12\: \textrm{cm}\\ &AC=16\:\textrm{cm}.\quad \textrm{Tentukan jari-jari lingkaran}\\ & \textrm{dalam}\quad \bigtriangleup ABC\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi berikut} \end{array}$.

$\begin{aligned}\qquad&\textrm{Karena}\:\: BC^{\displaystyle 2}+AC^{\displaystyle 2}=AB^{\displaystyle 2},\:\: \textrm{maka}\:\: \bigtriangleup ABC\\ &\textrm{adalah segitiga siku-siku di}\:\: C\\ & \textrm{Dan karena}\:\: r\:\: \textrm{jari-jari lingkaran dalam}\:\: \bigtriangleup ABC,\\ &\textrm{maka}\\ &BC=12-r+16-r\\ &\Leftrightarrow 20=28-2r\\ &\Leftrightarrow r=4 \end{aligned}$.

DAFTRA PUSTAKA

1. Isnaini, H.F., Santoso, N.E. 2023. Matematika untuk SMA/SMK/MAK Kelas 11A Kurikulum Merdeka. Yogyakarta: PENERBIT INTAN PARIWARA.
2. Maulan, S.F. 2010. Juara Olimpiade Matematika SMA. Jakarta: WAHYUMEDIA.
3. Sembiring, S., Sukino. 2020. Super Master KSN Matematika SMA/MA. Bandung: YRAMA WIDYA.