PAS MATEMATIKA BLOG

CONTOH 4-LIMIT FUNGSI

$\begin{array}{ll}\\ 16.&\textrm{Nilai}\: \: \: \underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2020}{2x^{2}-2021x}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 8\\ &\textrm{b}.\quad \displaystyle 4 \\ &\textrm{c}.\quad 2\\ &\textrm{d}.\quad 1\\ &\textrm{e}.\quad \displaystyle \frac{1}{2}\\\\ &\textbf{Jawab}:\qquad \textbf{b}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2020}{2x^{2}-2021x}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x^{2}+x-2020}{2x^{2}-2021x}\times \displaystyle \frac{\displaystyle \frac{1}{x^{2}}}{\displaystyle \frac{1}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{\displaystyle \frac{8x^{2}}{x^{2}}+\frac{x}{x^{2}}-\frac{2020}{x^{2}}}{\displaystyle \frac{2x^{2}}{x^{2}}-\frac{2021x}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8+\displaystyle \frac{1}{x}-\frac{2020}{x^{2}}}{2-\displaystyle \frac{2021}{x}}\\ &=\displaystyle \frac{8+\displaystyle \frac{1}{\infty }-\frac{2020}{\infty ^{2}}}{2-\displaystyle \frac{2021}{\infty }}\\ &=\displaystyle \frac{8+0-0}{2-0}=\frac{8}{2}\\ &=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textrm{Nilai}\: \: \: \underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2019}{\sqrt{9x^{2}-2020x}}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 3 \\ &\textrm{b}.\quad \displaystyle 1 \\ &\textrm{c}.\quad \displaystyle \frac{1}{3}\\ &\textrm{d}.\quad -\displaystyle \frac{1}{3}\\ &\textrm{e}.\quad \displaystyle -3\\\\ &\textbf{Jawab}:\qquad \textbf{d}\\ &\begin{aligned}&\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2021}{\sqrt{9x^{2}-2020x}}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{x+2021}{\sqrt{9x^{2}-2020x}}\times \displaystyle \frac{\left ( \displaystyle \frac{1}{x} \right )}{\left (-\sqrt{\displaystyle \frac{1}{x^{2}}} \right )}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{\displaystyle \frac{x}{x}+\frac{2021}{x}}{-\sqrt{\displaystyle \frac{9x^{2}}{x^{2}}-\frac{2020x}{x^{2}}}}\\ &=\underset{x\rightarrow -\infty }{\textrm{lim}}\: \displaystyle \frac{1+\displaystyle \frac{2021}{x}}{-\sqrt{9-\displaystyle \frac{2020}{x}}}\\ &=\displaystyle \frac{1+\displaystyle \frac{2021}{\infty }}{-\sqrt{9-\displaystyle \frac{2020}{\infty }}}=\displaystyle \frac{1}{-\sqrt{9}}\\ &=-\displaystyle \frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Nilai}\: \: \: \underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}=\: ....\\\\ &\textrm{a}.\quad \displaystyle 1\qquad\qquad\quad\quad\qquad \\ &\textrm{b}.\quad \displaystyle 4 \qquad\qquad\qquad\qquad \\ &\textrm{c}.\quad 9\\ &\textrm{d}.\quad 16\\ &\textrm{e}.\quad 25\\\\ &\textbf{Jawab}:\qquad \textbf{e}\\ &\begin{aligned}&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2^{x+1}+3^{x+1}+4^{x+1}+5^{x+1}}{2^{x-1}+3^{x-1}+4^{x-1}+5^{x-1}}\times \displaystyle \frac{\displaystyle \frac{1}{5^{x}}}{\displaystyle \frac{1}{5^{x}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2\left ( \displaystyle \frac{2}{5} \right )^{x}+3\left ( \displaystyle \frac{3}{5} \right )^{x}+4\left ( \displaystyle \frac{4}{5} \right )^{x}+5\left ( \displaystyle \frac{5}{5} \right )^{x}}{\displaystyle \frac{1}{2}\left ( \displaystyle \frac{2}{5} \right )^{x}+\frac{1}{3}\left ( \displaystyle \frac{3}{5} \right )^{x}+\frac{1}{4}\left ( \displaystyle \frac{4}{5} \right )^{x}+\frac{1}{5}\left ( \displaystyle \frac{5}{5} \right )^{x}}\\ &=\displaystyle \frac{0+0+0+5\left ( \displaystyle \frac{5}{5} \right )^{x}}{0+0+0+\displaystyle \frac{1}{5}\left ( \displaystyle \frac{5}{5} \right )^{x}}\\ &=\displaystyle \frac{5.1}{\displaystyle \frac{1}{5}.1}\\ &=25 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 19.&\textbf{(USM UGM Mat IPA)}\\ &\textrm{Nilai}\: \: \underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-x-1 \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{5}{3}\\\\ \textrm{b}.\quad \displaystyle \frac{2}{3}\\\\ \textrm{c}.\quad -\displaystyle \frac{1}{3}\\\\ \textrm{d}.\quad -\displaystyle \frac{2}{3}\\\\ \textrm{e}.\quad -\displaystyle \frac{5}{3}\\ \end{array}\\\\ &\textrm{Jawab}:\: \textbf{e} \end{array}$

$\begin{aligned}.\qquad \: \, \underset{x\rightarrow \infty }{\textrm{Lim}}\: &\: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-x-1 \right )\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \sqrt[3]{x^{3}-2x^{2}}-\sqrt[3]{\left ( x+1 \right )^{3}} \right )\\ &\: \: \: \textrm{ingat bentuk}\: \: a-b=\left ( \sqrt[3]{a}-\sqrt[3]{b} \right )\left ( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right )\\ &\: \: \: \textrm{dan untuk}\: \: \begin{cases} a & =\left ( x^{3}-2x^{2} \right ) \\ & \\ b & = \left ( x+1 \right )^{3} \end{cases}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( \sqrt[3]{a}-\sqrt[3]{b} \right )\left ( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right )}{\sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{a-b}{\sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( x^{3}-2x^{2} \right )-\left ( x+1 \right )^{3}}{\sqrt[3]{\left ( x^{3}-2x^{2} \right )^{2}}+\sqrt[3]{\left ( x^{3}-2x^{2} \right )\left ( x+1 \right )^{3}}+\sqrt[3]{\left ( \left ( x+1 \right )^{3} \right )^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{\left ( x^{3}-2x^{2} \right )-\left ( x^{3}+3x^{2}+3x+1 \right )}{\left ( x^{3}-2x^{2} \right )^{\frac{2}{3}}+\left ( x^{6}+... \right )^{\frac{1}{3}}+\left ( x+1 \right )^{\frac{6}{3}}}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \frac{-5x^{2}+...}{\left ( x^{3}-2x^{2} \right )^{\frac{2}{3}}+\left ( x^{6}+... \right )^{\frac{1}{3}}+\left ( x+1 \right )^{\frac{6}{3}}}\\ &=\displaystyle \frac{-5}{1+1+1}\\ &=-\displaystyle \frac{5}{3} \end{aligned}$

$\begin{array}{ll}\\ 20.&\textrm{Nilai}\: \: \underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 1\\\\ \textrm{b}.\quad \displaystyle \frac{3}{2}\\\\ \textrm{c}.\quad \displaystyle 2\\\\ \textrm{d}.\quad \displaystyle \frac{5}{2}\\\\ \textrm{e}.\quad \displaystyle \infty\\ \end{array}\\\\ &\textrm{Jawab}:\: \textbf{a}\\ &\begin{aligned}&\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{k\times (k+1)} \right )\\ &=\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( \left (1-\frac{1}{2} \right )+\left (\frac{1}{2}-\frac{1}{3} \right )+\left (\frac{1}{3}-\frac{1}{4} \right )+\cdots +\left (\frac{1}{k}-\frac{1}{k+1} \right )\right )\\ &=\underset{k\rightarrow \infty }{\textrm{Lim}}\: \: \displaystyle \left ( 1-\frac{1}{k+1} \right )\\ &=\displaystyle \left ( 1-\frac{1}{\infty +1} \right )\\ &=\displaystyle 1-\frac{1}{\infty }\\ &=1-0\\ &=1 \end{aligned} \end{array}$

CONTOH 3-LIMIT FUNGSI

$\begin{aligned}11.\quad&\textrm{Perhatikan gambar grafik fungsi}\:\: y=f(x)\: \: \textrm{berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Jika}\:\: a\in \left\{ 0,2,4 \right\}\:\: \textrm{dengan}\:\: \underset{x\rightarrow a }{\textrm{Lim}}\: f(x)\:\: \textrm{ada, maka}\\ &\textrm{untuk nilai semua}\:\: a\:\: \textrm{yang memenuhi adalah}\: ....\\ &\textrm{a}.\quad a=0\\ &\textrm{b}.\quad a=2\\ &\textrm{c}.\quad a=0\:\: \textrm{dan}\:\: a=2\\ &\textrm{d}.\quad a=0\:\: \textrm{dan}\:\: a=4\\ &\textrm{e}.\quad a=0\:,\: a=2\:,\: \textrm{dan}\:\: a=4\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Cukup jelas sesuai ilustrasi gambar, bahwa}\\ &\bullet \quad\underset{\displaystyle x\rightarrow 0^{\displaystyle -} }{\textrm{Lim}\qquad}\: f(x)=\underset{\displaystyle x\rightarrow 0^{\displaystyle +} }{\textrm{Lim}\qquad}\: f(x)=6\\ &\textrm{dan}\\ &\bullet \quad \underset{\displaystyle x\rightarrow 2^{\displaystyle -} }{\textrm{Lim}\qquad}\: f(x)=\underset{\displaystyle x\rightarrow 2^{\displaystyle +} }{\textrm{Lim}\qquad}\: f(x)=2\\ &\textrm{serta nilai}\\ &\bullet \quad \underset{\displaystyle x\rightarrow 4^{\displaystyle -} }{\textrm{Lim}\qquad}\: f(x)\neq \underset{\displaystyle x\rightarrow 4^{\displaystyle +} }{\textrm{Lim}\qquad}\: f(x) \end{aligned}$.

$\begin{array}{l}\\ 12.&\textrm{Diketahui}\\ &f(x)=\begin{cases} 3x^{\displaystyle 2}+8x+1 & ,&\textrm{untuk}\quad x<2 \\ ax+3&,&\textrm{untuk}\quad x\ge 2 \end{cases}\\ &\textrm{Besar}\quad a\quad \textrm{yang memenuhi agar nilai}\\ &\underset{x\rightarrow 2 }{\textrm{Lim}}\: f(x)\:\: \textrm{ada}\quad\textrm{adalah}\: ....\\ &\begin{array}{llllllll} \textrm{a}.&-13\qquad\qquad\qquad\qquad\qquad \textrm{d}.\quad 13\\ \textrm{b}.&-2\qquad\qquad \textrm{c}.\quad 2\:\quad\quad\qquad \textrm{e}.\quad 26\\ \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\textrm{Suatu fungsi}\quad f(x)\quad \textrm{memiliki nilai limit}\:\: \underset{x\rightarrow 2}{\textrm{Lim}}\: f(x)\\&\textrm{jika dan hanya jika nilai limit kiri sama dengan}\\ &\textrm{limit kanan pada}\quad x=2,\quad \textrm{yaitu}: \\ &\begin{aligned}&\underset{.\quad\displaystyle x\rightarrow 2^{\displaystyle -}}{\textrm{Lim}} f(x)\quad=\underset{.\quad\displaystyle x\rightarrow 2^{\displaystyle +}}{\textrm{Lim}}\: f(x)\\ &\Leftrightarrow 3(2)^{\displaystyle 2}+8(2)+1=a(2)+3\Leftrightarrow 29=2a+3\\ &\Leftrightarrow 2a=29-3\Leftrightarrow 2a=26\Leftrightarrow a=13\\ \end{aligned} \end{array}$.

$\begin{array}{l}\\ 13.&\textrm{Diketahui}\quad f(x)\quad \text{adalah fungsi dari polinom berderajat}\\ &3\quad \textrm{yang memenuhi kondisi}\quad \underset{x\rightarrow -1 }{\textrm{Lim}}\: \displaystyle \frac{f(x)}{x+1}=6\quad \textrm{dan}\\ &\underset{x\rightarrow 2 }{\textrm{Lim}}\: \displaystyle \frac{f(x)}{x-2}=30\:,\: \textrm{maka fungsi}\quad f(x)=\: ....\\ &\begin{array}{llllllll} \textrm{a}.&4x^{\displaystyle 3}+2x^{\displaystyle 2}-10x-4\quad\quad\textrm{d}.\quad 4x^{\displaystyle 3}-2x^{\displaystyle 2}+10x-4\\ \textrm{b}.&4x^{\displaystyle 3}-2x^{\displaystyle 2}-10x+4\quad\quad \textrm{e}.\quad 4x^{\displaystyle 3}-2x^{\displaystyle 2}-10x-4\\ \textrm{c}.&4x^{\displaystyle 3}-2x^{\displaystyle 2}+10x+4\end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\textbf{Analisis polinom (suku banyak)}\\ &\circ \quad (x+1)\quad \textrm{dan}\quad (x-2)\quad \textrm{adalah faktor dari}\quad f(x)\\&\circ \quad f(x)=(x+1)(x-2)(ax+b)\quad \textrm{dikalikan dengan}\\ &\:\:\quad (ax+b)\quad \textrm{agar menjadi polinom berderjat 3}\\ &\textbf{Kondisi limit}\\&\circ \quad \textrm{limit pertama}\\ &\:\:\quad \underset{x\rightarrow -1 }{\textrm{Lim}}\: \displaystyle \frac{f(x)}{x+1}=6\Leftrightarrow \underset{x\rightarrow -1 }{\textrm{Lim}}\: \displaystyle \frac{(x+1)(x-2)(ax+b)}{x+1}=6\\ &\:\:\quad \Leftrightarrow \underset{x\rightarrow -1 }{\textrm{Lim}}\: (x-2)(ax+b)=6 \\ &\:\:\quad \textrm{selanjutnya kita substitusikan}\quad x=-1,\quad \textrm{yaitu}:\\ &\:\:\quad (-1-2)(a(-1)+b)=6\Leftrightarrow -a+b=-2\: .......(1)\\&\circ \quad \textrm{limit kedua}\\ &\:\:\quad \underset{x\rightarrow 2 }{\textrm{Lim}}\: \displaystyle \frac{f(x)}{x-2}=30\Leftrightarrow \underset{x\rightarrow 2 }{\textrm{Lim}}\: \displaystyle \frac{(x+1)(x-2)(ax+b)}{x-2}=30\\ &\:\:\quad \Leftrightarrow \underset{x\rightarrow 2 }{\textrm{Lim}}\: (x+1)(ax+b)=30 \\ &\:\:\quad \textrm{selanjutnya kita substitusikan}\quad x=2,\quad \textrm{yaitu}:\\ &\:\:\quad (2+1)(a(2)+b)=30\Leftrightarrow 2a+b=10\: .......(2)\\ &\textbf{Menentukan nilai a dan b}\\ &\textrm{dengan eliminasi akan kita peroleh}\\ &\begin{array}{rl} -a+b=-2&\\ 2a+b=10&-\\\hline -3a=-12&\Leftrightarrow a=4\: \Rightarrow b=2\end{array}\\ &\textbf{Menyusun}\quad f(x)\\&f(x)=(x+1)(x-2)(4x+2)=4x^{\displaystyle 3}-2x^{\displaystyle 2}-10x-4\\ \end{array}$.

$\begin{array}{l}\\ 14.&\textrm{Sebuah virus menyebar luas di suatu daerah dengan }\\ &\textrm{sangat cepat dan luas daerah sebarannya dirumuskan}\\ &A(t)=\displaystyle \frac{1}{36}\pi t^{\displaystyle 2}\quad \textrm{dengan}\quad t\:\: \textrm{adalah hari penyebaran}\\ &\textrm{virus. Laju penyebaran virus pada hari ke}-t\\ &\textrm{diformulasikan dengan rumus}\quad \underset{t\rightarrow t_{1} }{\textrm{Lim}}\: \displaystyle \frac{A(t)-A\left( t_{1} \right)}{t-t_{1}}.\:\\ &\textrm{Laju sebaran virus pada hari ke-7 sejak wabah}\\&\textrm{menyebar adalah}\quad(\textrm{luas dalam}\:\: m^{\displaystyle 2})\: ....\: \displaystyle \frac{m^{\displaystyle 2}}{\textrm{hari}}\\ &\begin{array}{llllllll} \textrm{a}.&0,8\:\:\qquad\qquad\qquad\qquad\quad\qquad \textrm{d}.\quad 2,4\\ \textrm{b}.&1,2\qquad\qquad \textrm{c}.\quad 1,6\:\quad\quad\qquad \textrm{e}.\quad 2,8\\ \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\textrm{Diketahui fungsi luas}\quad A(t)=\displaystyle \frac{1}{36}\pi t^{\displaystyle 2}\:\: \textrm{dan rumus laju}\\ & \textrm{penyebaran virus}\quad V=\underset{t\rightarrow t_{1} }{\textrm{Lim}}\: \displaystyle \frac{A(t)-A\left( t_{1} \right)}{t-t_{1}}\: ,\: \textrm{maka hari} \\&\textrm{ke-7 laju sebaran virusnya}\:\: \left( t_{1}=7 \right)\:\: \textrm{adalah}:\\ &V=\underset{t\rightarrow 7 }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{1}{36}\pi t^{\displaystyle 2}-\displaystyle \frac{1}{36}\pi (7)^{\displaystyle 2}}{t-7}=\underset{t\rightarrow 7 }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{1}{36}\pi(t^{\displaystyle 2}-49)}{t-7}\\&\quad =\underset{t\rightarrow 7 }{\textrm{Lim}}\: \displaystyle \frac{\displaystyle \frac{1}{36}\left( \displaystyle \frac{22}{7} \right)(t+7)(t-7)}{t-7}=\underset{t\rightarrow 7 }{\textrm{Lim}}\: \displaystyle \frac{1}{36}\left( \displaystyle \frac{22}{7} \right)(t+7)\\ &\quad =\displaystyle \frac{1}{36}\left( \displaystyle \frac{22}{7} \right)(7+7)=\displaystyle \frac{44}{36}=1,\overline{222}\approx 1,2 \end{array}$.

$\begin{array}{l}\\ 15.&\textrm{Dalam sebuah riset terkait peningkatan penggunaan}\\ &\textrm{data internet pada perangkat seluler, pertumbuhan}\\ &\textrm{penggunaan data (dalam gigabita) dimodelkanan}\\ &\textrm{oleh fungsi}\:\:f(x)=\displaystyle \frac{100x^{\displaystyle 2}+200x}{2x^{\displaystyle 2}+100}\:\: \textrm{dengan}\:\:x\:\: \textrm{adalah}\\ &\textrm{waktu dalam bulan. Jika kecenderungan peningkatan}\\ &\textrm{waktu yang dihasilkan oleh fungsi}\quad f(x)\quad \textrm{terus}\\&\textrm{berlanjut, perkiraan penggunaan data mendekati}\\ &\textrm{tak hingga adalah}\: ....\: \textrm{gigabita}\\ &\begin{array}{llllllll} \textrm{a}.&300\:\:\qquad\qquad\qquad\qquad\quad\qquad \textrm{d}.\quad 100\\ \textrm{b}.&200\qquad\qquad \textrm{c}.\quad 150\:\quad\quad\qquad \textrm{e}.\quad 50\\ \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\textrm{Diketahui fungsi }\quad f(x)=\displaystyle \frac{100x^{\displaystyle 2}+200x}{2x^{\displaystyle 2}+100}\:\: \textrm{untuk}:x\to \infty \\ & \textrm{maka}\quad \underset{x\rightarrow \infty }{\textrm{Lim}}\: f(x)=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{100x^{\displaystyle 2}+200x}{2x^{\displaystyle 2}+100}\\ &=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{100x^{\displaystyle 2}+200x}{2x^{\displaystyle 2}+100}\times \displaystyle \frac{\displaystyle \frac{1}{x^{\displaystyle 2}}}{\displaystyle \frac{1}{x^{\displaystyle 2}}}=\underset{x\rightarrow \infty }{\textrm{Lim}}\: \displaystyle \frac{100+\displaystyle \frac{200}{x}}{2+\displaystyle \frac{100}{x^{\displaystyle 2}}}\\&=\displaystyle \frac{100+\displaystyle \frac{200}{\infty }}{2+\displaystyle \frac{100}{\infty ^{\displaystyle 2}}}=\displaystyle \frac{100+0}{2+0}=50\\ \end{array}$.

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

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)

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

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

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

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

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

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

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

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

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

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

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

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