PAS MATEMATIKA BLOG

LINGKARAN-SEGI EMPAT TALI BUSUR-LANJUTAN

E. Segi Empat Tali Busur

Segi Empat Tali Busur adalah segi empat yang keempat titik sudutnya terletak pada satu lingkaran dan keempat sisinya merupakan tali busur.

Perhatikan ilustrasi berikut

LINGKARAN-SUDUT PUSAT DAN SUDUT KELILING-LANJUTAN

D. Sudut Pusat dan Sudut Keliling

Perhatikan gambar berikut

$\begin{aligned}&\textrm{Sudut pusat}=2\times \textrm{sudut keliling}\\ &\angle BOC=2\times\angle BAC\\\\ &\textbf{Sebagai pengingat}:\\ &\bullet \quad \textrm{Semua sudut keliling yang menghadap busur} \\ &\qquad\textrm{sama, maka besar sudutnya sama besar}\\ &\bullet \quad\textrm{Sudut keliling besarnya akan}\:\:\:90^{\displaystyle 0}\:\:\: \textrm{jika}\\ &\qquad\textrm{menghadap diameter} \end{aligned}$.

$\LARGE\fbox{CONTOH SOAL}$.

$\begin{aligned}1.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Jika besar}\:\: \angle ABD=46^{\displaystyle 0},\: \textrm{tentukan besar}\\ &\text{a}.\quad \angle ACD\\ &\text{b}.\quad \angle AOD\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{a}.\quad \angle ACD=\angle ABD=46^{\displaystyle 0}\\ &\qquad\textrm{karena sama-sama sudut keliling yang menghadap}\\ &\qquad\textrm{busur yang sama}\\ &\textrm{b}.\quad \angle AOD=2\times\angle ABD=2\times46^{\displaystyle 0}=92^{\displaystyle 0}\\ &\qquad\textrm{karena merupakan sudut pusat dari sudut}\\ &\qquad\textrm{keliling}\:\:\: \angle ABD\:\: \textrm{yang sama-sama menghadap}\:\: \widehat{AD}\end{aligned}\end{aligned}$.

$\begin{aligned}2.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Jika besar}\:\: \angle BAC=(2x+8)^{\displaystyle 0}\:\: \textrm{dan}\:\: \angle ACB=(4x-2)^{\displaystyle 0}\\ &\textrm{tentukan besar}\\ &\text{a}.\quad \angle ABC\\ &\text{b}.\quad \angle ACB\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{a}.\quad \angle ABC=\displaystyle \frac{1}{2}\angle AOC=\displaystyle \frac{1}{2}.180^{\displaystyle 0}=90^{\displaystyle 0}\\ &\qquad\textrm{karena sudut keliling yang menghadap}\\ &\qquad\textrm{diameter lingkaran}\\ &\textrm{b}.\quad \angle ACB+\angle CAB=90^{\displaystyle 0}\Leftrightarrow (2x+8)^{\displaystyle 0}+(4x-2)^{\displaystyle 0}=90^{\displaystyle 0}\\ &\qquad 6x+6=90^{\displaystyle 0}\Leftrightarrow x+1^{\displaystyle 0}=15^{\displaystyle 0}\Leftrightarrow x=14^{\displaystyle 0}\\ &\qquad\textrm{Sehingga besar}\:\: \angle ACB=(4x-2)^{\displaystyle 0}=(4.14-2)^{\displaystyle 0}=54^{\displaystyle 0} \end{aligned}\end{aligned}$.

SELAMAT MEMPERINGATI MAULID NABI MUHAMMAD SAW TAHUN 2026

CONTOH SOAL 5 LINGKARAN-BUSUR-JURING-TEMBERENG

$\begin{aligned}21.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\quad\qquad&\textrm{Luas daerah yang diarsir adalah}=\: .... \\ &\text{a}.\quad 154\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 77\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 44\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 38,5\:\:\textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}L_{\displaystyle \text{arsiran}}&=\displaystyle \frac{1}{2}\:\textrm{luas lingkaran}_{\text{besar}}-\textrm{lingkaran}_{\text{kecil}}\\ &=\displaystyle \frac{1}{2}(\displaystyle \frac{1}{4}\pi d^{\displaystyle 2}_{\textrm{besar}})-(\displaystyle \frac{1}{4}\pi d^{\displaystyle 2}_{\textrm{kecil}})\\ &=\displaystyle \frac{1}{2}.\displaystyle \frac{22}{4.7}.14^{\displaystyle 2}-\displaystyle \frac{1}{4}.\displaystyle \frac{22}{7}.7^{\displaystyle 2}\\ &=\displaystyle \frac{22}{28}.(7.14-49)\\&=\displaystyle \frac{22}{28}.49\\ &=38,5\:\: \textrm{cm}^{\displaystyle 2}\end{aligned}\end{aligned}$.

$\begin{aligned}22.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Luas daerah yang diarsir adalah}=\: .... \\ &\text{a}.\quad 57\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 62,8\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 107\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 114\:\:\textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}L_{\displaystyle \text{arsiran}}&=\textrm{Luas tembereng doubel}\\ &=2(\textrm{luas juring - luas segitiga siku-siku})\\ &=2\left( \displaystyle \frac{90^{\displaystyle 0}}{360^{\displaystyle 0}}\times\pi.r^{2}-\displaystyle \frac{1}{2}.r^{\displaystyle 2} \right)\\ &=2\left( \displaystyle \frac{1}{4}\pi.10^{\displaystyle 2}-\displaystyle \frac{1}{2}.10^{\displaystyle 2} \right)\\ &=2(78,5-50)=57\:\: \textrm{cm}^{\displaystyle 2}\end{aligned}\end{aligned}$.

$\begin{aligned}23.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Luas daerah yang diarsir adalah}=\: .... \\ &\text{a}.\quad 157\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 80\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 40\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 39,25\:\:\textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}L_{\displaystyle \text{arsiran}}&=(\textrm{luas juring besar - luas juring kecil})\\ &=\left( \displaystyle \frac{90^{\displaystyle 0}}{360^{\displaystyle 0}}\times\pi.r_{\text{besar}}^{2}-\displaystyle \frac{90^{\displaystyle 0}}{360^{\displaystyle 0}}\times\pi.r_{\text{kecil}}^{2} \right)\\ &=\left( \displaystyle \frac{1}{4}\pi.10^{\displaystyle 2}-\displaystyle \frac{1}{4}\pi.7^{\displaystyle 2} \right)\\ &=\displaystyle \frac{1}{4}(3,14).(100-49)\approx 40\:\: \textrm{cm}^{\displaystyle 2}\end{aligned}\end{aligned}$.

$\begin{aligned}24.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Luas daerah yang diarsir adalah}=\: .... \\ &\text{a}.\quad 112\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 70\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 42\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 28\:\:\textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}L_{\displaystyle \text{arsiran}}&=(\textrm{luas tembereng besar - 2 luas tembereng kecil})\\ &=\left( \displaystyle \frac{90^{\displaystyle 0}}{360^{\displaystyle 0}}\times\pi.r_{\text{besar}}^{2}-\displaystyle \frac{1}{2}\times r_{\text{besar}}^{2} \right)\\ &\quad-2\left( \displaystyle \frac{90^{\displaystyle 0}}{360^{\displaystyle 0}}\times \pi.r_{ \textrm{kecil}}^{\displaystyle 2}-\displaystyle \frac{1}{2}\times r_{\text{kecil}}^{2} \right)\\ &=\displaystyle \frac{1}{4}.\displaystyle \frac{22}{7}.14^{\displaystyle 2}-\displaystyle \frac{1}{2}.14^{\displaystyle 2}-2\left( \displaystyle \frac{1}{4}.\frac{22}{7}.7^{\displaystyle 2}-\displaystyle \frac{1}{2}.7^{\displaystyle 2} \right)\\ &=154-98-(77-49)=28\:\: \textrm{cm}^{\displaystyle 2}\end{aligned}\end{aligned}$.

$\begin{aligned}25.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Jika}\:\: AB=5\:\: cm,\:\: AC=12\:\: cm,\:\: \textrm{dan}\:\: AB,\: AC\\ &\textrm{serta}\:\: BC\:\: \textrm{masing-masing adalah diamter lingkaran},\\ &\textrm{maka luas daerah terarsir adalah}=\: .... \\ &\text{a}.\quad 27\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 30\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 36\:\:\textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 37,5\:\:\textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}L_{\displaystyle \text{arsiran}}&=(2\:\textrm{luas setengah lingkaran - luas lingkaran besar})\\&\quad + \textrm{luas sebuah segitiga}\\ &=\displaystyle \frac{1}{2}.\frac{1}{4}\pi d_{2}^{\displaystyle 2}+\displaystyle \frac{1}{2}.\frac{1}{4}\pi d_{3}^{\displaystyle 2}-\displaystyle \frac{1}{2}.\frac{1}{4}\pi d_{1}^{\displaystyle 2}+\displaystyle \frac{1}{2}.d_{2} d_{3}\\ &=\displaystyle \frac{1}{8}(3,14) (d_{2}^{\displaystyle 2}+ d_{3}^{\displaystyle 2}- d_{1}^{\displaystyle 2})+\displaystyle \frac{1}{2}.d_{2} d_{3}\\ &=\displaystyle \frac{1}{8}(3,14) (12^{\displaystyle 2}+ 5^{\displaystyle 2}- 13^{\displaystyle 2})+\displaystyle \frac{1}{2}.12. 5\\ &=30\:\: \textrm{cm}^{\displaystyle 2} \end{aligned}\end{aligned}$.

CONTOH SOAL 4 LINGKARAN-BUSUR-JURING-TEMBERENG

$ \begin{aligned}16.\quad&\textrm{Diberikan pernyataan-pernyataan berikut}\\ &(i)\quad \pi(2r)\qquad\qquad\qquad (iii)\quad \displaystyle \frac{1}{2}\pi d\\ &(ii)\quad \pi.r^{\displaystyle 2}\qquad\qquad\qquad (iv)\quad \displaystyle \frac{1}{4}\pi d^{\displaystyle 2}\\&\textrm{jika r jari-jari lingkaran dan d adalah diameternya}\\&\textrm{Pernyataan di atas yang merupakan formula luas }\\ &\textrm{lingkaran adalah}\:....\\ &\text{a}.\quad (i)\quad \textrm{dan}\quad (ii)\\ &\text{b}.\quad (i)\quad \textrm{dan}\quad (iv)\\ &\text{c}.\quad (ii)\quad \textrm{dan}\quad (iii)\\ &\text{d}.\quad (ii)\quad \textrm{dan}\quad (iv)\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Cukup jelas}\end{aligned}$.

$ \begin{aligned}17.\quad&\textrm{Luas lingkaran yang berjari-jari}\quad 7\:\: \textrm{cm adalah}=\:....\\ &\text{a}.\quad 44\quad \textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 88\quad \textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 154\quad \textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 308\quad \textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &L=\pi r^{\displaystyle 2}=\displaystyle \frac{22}{7}.7^{\displaystyle 2}=154\:\: \textrm{cm}^{\displaystyle 2}\end{aligned}$.

$ \begin{aligned}18.\quad&\textrm{Luas lingkaran yang bediameter}\quad 20\:\: \textrm{cm adalah}=\:....\\ &\text{a}.\quad 12,56\quad \textrm{cm}^{\displaystyle 2}\\ &\text{b}.\quad 314\quad \textrm{cm}^{\displaystyle 2}\\ &\text{c}.\quad 62,8\quad \textrm{cm}^{\displaystyle 2}\\ &\text{d}.\quad 31,4\quad \textrm{cm}^{\displaystyle 2}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &L=\displaystyle \frac{1}{4}\pi d^{\displaystyle 2}=\displaystyle \frac{1}{4}.(3,14).20^{\displaystyle 2}=314\:\: \textrm{cm}^{\displaystyle 2}\end{aligned}$.

$ \begin{aligned}19.\quad&\textrm{Jari-jari lingkaran dengan luasnya}\quad 36\pi\:\: cm^{\displaystyle 2}\:\textrm{adalah}=\:....\\ &\text{a}.\quad 6\quad \textrm{cm}\\ &\text{b}.\quad 9\quad \textrm{cm}\\ &\text{c}.\quad 12\quad \textrm{cm}\\ &\text{d}.\quad 6\pi\quad \textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &L=\pi r^{\displaystyle 2}\Leftrightarrow 36\pi=\pi r^{\displaystyle 2}\Leftrightarrow r=\sqrt{36}=6\:\: \textrm{cm}\end{aligned}$.

$ \begin{aligned}20.\quad&\textrm{Dua lingkaran dengan jari-jari masing-masing}\\ &8\:\:\textrm{cm dan 10 cm. Perbandingan luas untuk}\\ &\textrm{kedua lingkaran tersebut adalah}=\:....\\ &\text{a}.\quad 4:5\\ &\text{b}.\quad 8:25\\ &\text{c}.\quad 16:25\\ &\text{d}.\quad 16:125\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\displaystyle \frac{L_{\displaystyle 8}}{L_{\displaystyle 10}}=\displaystyle \frac{\pi.8^{\displaystyle 2}}{\pi.10^{\displaystyle 2}}=\frac{16}{25}\end{aligned}$.

CONTOH SOAL 3 LINGKARAN-BUSUR-JURING-TEMBERENG

$\begin{aligned}11.\quad&\textrm{Sebuah lintasan lari berbentuk lingkaran yang}\\ &\textrm{berdiameter 56 meter. Banyak putaran yang harus}\\ &\textrm{dilakukan pelari jika ia ingin menempuh jarak}\\ &\textrm{902 meter adalah}\:....\\ &\text{a}.\quad 5\displaystyle \frac{3}{4}\quad \textrm{putaran}\\ &\text{b}.\quad 5\displaystyle \frac{1}{2}\quad \textrm{putaran}\\ &\text{c}.\quad 5\displaystyle \frac{1}{4}\quad \textrm{putaran}\\ &\text{d}.\quad 5\displaystyle \frac{1}{8}\quad \textrm{putaran}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Diketahui bahwa lingkaran dengan}\:\: K=\pi d\:\: \textrm{meter} \\ &\textrm{maka, K}=\displaystyle \frac{22}{7}.56\:\: \textrm{meter}=176\:\: \textrm{meter}\\ &\textrm{Sehingga banyak putaran yang perlu dilakukan adalah}\\ &=\displaystyle \frac{902}{176}=5\displaystyle \frac{1}{8}\:\: \textrm{putaran}\end{aligned}$.

$\begin{aligned}12.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Berdasarkan gambar di atas, keliling daerah arsiran}=\:....\\ &\text{a}.\quad 44\:\:\textrm{cm}\\ &\text{b}.\quad 66\:\:\textrm{cm}\\ &\text{c}.\quad 88\:\:\textrm{cm}\\ &\text{d}.\quad 132\:\:\textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}K&=\displaystyle \frac{1}{2}\textrm{lingkaran besar}+\textrm{1 lingkaran kecil penuh}\\ &=\displaystyle \frac{1}{2}\pi d_{\textrm{besar}}+\pi d_{\textrm{kecil}}\\ &=\displaystyle \frac{1}{2}\frac{22}{7}.28+\displaystyle \frac{22}{7}.14\\ &=44+44\\ &=88\:\: \textrm{cm}\end{aligned}\end{aligned}$.

$\begin{aligned}13.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Berdasarkan gambar di atas, keliling daerah arsiran}=\:....\\ &\text{a}.\quad 31,4\:\:\textrm{cm}\\ &\text{b}.\quad 62,8\:\:\textrm{cm}\\ &\text{c}.\quad 94,2\:\:\textrm{cm}\\ &\text{d}.\quad 125,6\:\:\textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}K&=2\:\textrm{lingkaran besar}\\ &=2\pi d\\ &=2(3,14).20\\ &=125,6\:\: \textrm{cm}\end{aligned}\end{aligned}$.

$\begin{aligned}14.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Berdasarkan gambar di atas, besar keliling daerah}\\ &\textrm{jika diketahui jari-jari lingkarannyanya 10 cm }\\ &\textrm{adalah}\: .... \\ &\text{a}.\quad 82,8\:\:\textrm{cm}\\ &\text{b}.\quad 67,1\:\:\textrm{cm}\\ &\text{c}.\quad 61,1\:\:\textrm{cm}\\ &\text{d}.\quad 47,1\:\:\textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}K&=\displaystyle \frac{3}{4}\:\textrm{lingkaran}+2r\\ &=\displaystyle \frac{3}{4}(2\pi r)+2r\\ &=\displaystyle \frac{3}{4}.2.(3,14).10+20\\ &=47,1+20\\ &=67,1\:\: \textrm{cm}\end{aligned}\end{aligned}$.

$\begin{aligned}15.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Berdasarkan gambar di atas, keliling daerah arsiran}=\: .... \\ &\text{a}.\quad 55\:\:\textrm{cm}\\ &\text{b}.\quad 62\:\:\textrm{cm}\\ &\text{c}.\quad 69\:\:\textrm{cm}\\ &\text{d}.\quad 83\:\:\textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\begin{aligned}K&=\displaystyle \frac{3}{4}\:\textrm{lingkaran}_{\text{kecil}}+\displaystyle \frac{1}{4}\:\textrm{lingkaran}_{\text{besar}}+7+7\\ &=\displaystyle \frac{3}{4}(2\pi r_{\textrm{kecil}})+\displaystyle \frac{1}{4}(2\pi r_{\textrm{besar}})+14\\ &=\displaystyle \frac{3}{4}.2.\displaystyle \frac{22}{7}.7+\displaystyle \frac{1}{4}.2.\displaystyle \frac{22}{7}.14\\ &=\displaystyle \frac{5}{4}.2.\displaystyle \frac{22}{7}.7\\ &=55\:\: \textrm{cm}\end{aligned}\end{aligned}$.

CONTOH SOAL 2 LINGKARAN-BUSUR-JURING-TEMBERENG

$\begin{aligned}6.\quad&\textrm{Keliling lingkaran yang berdiameter}\:\: d\:\: \textrm{adalah}\:....\\ &\text{a}.\quad 2\pi d\quad \textrm{satuan panjang}\\ &\text{b}.\quad \pi d\quad \textrm{satuan panjang}\\ &\text{c}.\quad \displaystyle \frac{1}{2}\pi d\quad \textrm{satuan panjang}\\ &\text{d}.\quad \displaystyle \frac{1}{4}\pi d\quad \textrm{satuan panjang}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Cukup jelas}\end{aligned}$.

$\begin{aligned}7.\quad&\textrm{Keliling lingkaran yang berdiameter}\:\: 10,5\:\: \textrm{cm}=\:....\\ &\text{a}.\quad 66\quad \textrm{cm}\\ &\text{b}.\quad 44\quad \textrm{cm}\\ &\text{c}.\quad 33\quad \textrm{cm}\\ &\text{d}.\quad 22\quad \textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Diketahui bahwa lingkaran dengan}\:\: d=10,5\:\: \textrm{cm} \\ &\textrm{maka, K}=\pi d=\displaystyle \frac{22}{7}(10,5\quad \textrm{cm})=\displaystyle \frac{22}{7}.\frac{21}{2}=33\:\: \textrm{cm}\end{aligned}$.

$\begin{aligned}8.\quad&\textrm{Keliling lingkaran yang berjari-jari}\:\: 14\:\: \textrm{cm}=\:....\\ &\text{a}.\quad 88\quad \textrm{cm}\\ &\text{b}.\quad 132\quad \textrm{cm}\\ &\text{c}.\quad 154\quad \textrm{cm}\\ &\text{d}.\quad 308\quad \textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Diketahui bahwa lingkaran dengan}\:\: r=14\:\: \textrm{cm} \\ &\textrm{maka, K}=2\pi r=2.\displaystyle \frac{22}{7}(14\quad \textrm{cm})=88\:\: \textrm{cm}\end{aligned}$.

$\begin{aligned}9.\quad&\textrm{Jari-jari lingkaran yang mempunyai keliling}\\ &132\:\: \textrm{cm adalah}\:....\\ &\text{a}.\quad 7\quad \textrm{cm}\\ &\text{b}.\quad 10,5\quad \textrm{cm}\\ &\text{c}.\quad 14\quad \textrm{cm}\\ &\text{d}.\quad 21\quad \textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Diketahui bahwa lingkaran dengan}\:\: K=132\:\: \textrm{cm} \\ &\textrm{maka, r}=\displaystyle \frac{K}{2\pi}=\displaystyle \frac{132}{2.\displaystyle \frac{22}{7}}=21\:\: \textrm{cm}\end{aligned}$.

$\begin{aligned}10.\quad&\textrm{Diameter lingkaran yang mempunyai keliling}\\ &154\:\: \textrm{cm adalah}\:....\\ &\text{a}.\quad 14\quad \textrm{cm}\\ &\text{b}.\quad 21\quad \textrm{cm}\\ &\text{c}.\quad 24,5\quad \textrm{cm}\\ &\text{d}.\quad 49\quad \textrm{cm}\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Diketahui bahwa lingkaran dengan}\:\: K=154\:\: \textrm{cm} \\ &\textrm{maka, d}=\displaystyle \frac{K}{\pi}=\displaystyle \frac{154}{\displaystyle \frac{22}{7}}=49\:\: \textrm{cm}\end{aligned}$.

CONTOH SOAL 1 LINGKARAN-BUSUR-JURING-TEMBERENG

$\begin{aligned}1.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Daerah yang diarsir pada gambar di atas adalah}\:....\\ &\text{a}.\quad \textrm{juring}\\ &\text{b}.\quad \textrm{tembereng}\\ &\text{c}.\quad \textrm{apotema}\\ &\text{d}.\quad \textrm{tali busur}\\\\ &\textrm{Jawab}:\quad \textbf{b}\\ &\textrm{Cukup jelas}\end{aligned}$.

Gunakan gambar berikut untuk menjawab soal nomor 2 sampai dengan nomor 5

$\begin{aligned}2.\quad&\textrm{Garis OD disebut}\:....\\ &\text{a}.\quad \textrm{apotema}\\ &\text{b}.\quad \textrm{tali busur}\\ &\text{c}.\quad \textrm{busur}\\ &\text{d}.\quad \textrm{jari-jari}\\\\ &\textrm{Jawab}:\quad \textbf{a}\\ &\textrm{Cukup jelas}\end{aligned}$.

$\begin{aligned}3.\quad&\textrm{Daerah yang diarsir adalah}\:....\\ &\text{a}.\quad \textrm{busur}\\ &\text{b}.\quad \textrm{tali busur}\\ &\text{c}.\quad \textrm{juring}\\ &\text{d}.\quad \textrm{tembereng}\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Cukup jelas}\end{aligned}$.

$\begin{aligned}4.\quad&\textrm{Tali busur ditunjukkan oleh}\:....\\ &\text{a}.\quad AC\\ &\text{b}.\quad OA\\ &\text{c}.\quad OD\\ &\text{d}.\quad AB\\\\ &\textrm{Jawab}:\quad \textbf{d}\\ &\textrm{Cukup jelas}\end{aligned}$.

$\begin{aligned}5.\quad&\textrm{Yang merupakan diameter adalah}\:....\\ &\text{a}.\quad AB\\ &\text{b}.\quad AC\\ &\text{c}.\quad BC\\ &\text{d}.\quad OB\\\\ &\textrm{Jawab}:\quad \textbf{c}\\ &\textrm{Cukup jelas}\end{aligned}$.

LINGKARAN-BUSUR, TALI BUSUR, JURING, DAN TEMBERENG-LANJUTAN

C. Keliling Lingkaran

Perhatikan kembali gambar lingkaran berikut

Keliling lingkaran dapat dirumuskan sebagai berikut:

$\begin{aligned}&\bullet \: K=\pi d=2\pi r\\ &\textrm{Keterangan}:\\ &\qquad K:\textrm{keliling lingkarang}\\ &\qquad d:\textrm{diameter lingkarang}\\ &\qquad r:\textrm{jar-jari lingkarang}\\ &\qquad \pi:\textrm{pi}=\displaystyle \frac{22}{7}=3,14\\ \end{aligned}$.

D. Panjang Busur Lingkaran

$\begin{aligned}&\bullet \:\displaystyle \frac{\textrm{Besar sudut pusat}}{\textrm{Besar sudut satu putaran}}=\frac{\textrm{Panjang busur}}{\textrm{Keliling lingkaran}}\\ &\bullet \:\textrm{Panjang busur}=\displaystyle \frac{\textrm{Besar sudut pusat}}{360^{\displaystyle 0}}\times 2\pi r\end{aligned}$.

E. Luas Lingkaran

$\begin{aligned}&\bullet \:L=\pi r^{\displaystyle 2}=\displaystyle \frac{1}{4}\pi d^{\displaystyle 2}\\ &\textrm{Keterangan}:\\ &\qquad L:\textrm{luas lingkarang}\\ &\qquad d:\textrm{diameter lingkarang}\\ &\qquad r:\textrm{jar-jari lingkarang}\\ &\qquad \pi:\textrm{pi}=\displaystyle \frac{22}{7}=3,14\\\end{aligned}$.

F. Luas Juring Lingkaran

$\begin{aligned}&\bullet \:\displaystyle \frac{\textrm{Besar sudut pusat}}{\textrm{Besar sudut satu putaran}}=\frac{\textrm{Luas juring}}{\textrm{Luas lingkaran}}\\ &\bullet \:\textrm{Luas juring}=\displaystyle \frac{\textrm{Besar sudut pusat}}{360^{\displaystyle 0}}\times \pi r^{\displaystyle 2}\end{aligned}$.

$\LARGE\fbox{CONTOH SOAL}$.

$\begin{aligned}1.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Tentukan panjang busur}\:\: \widehat{PQ}\\\\ &\textrm{Jawab}:\\ &\textrm{Panjang busur}\:\: \widehat{PQ}\\ &=\displaystyle \frac{\angle POQ}{360^{\displaystyle 0}}\text{Keliling lingkaran}_{r=OP}\\ &=\displaystyle \frac{144^{\displaystyle 0}}{360^{0}}.2\pi.r\\&=\displaystyle \frac{144^{\displaystyle 0}}{360^{0}}.2\pi.OP\\ &=\displaystyle \frac{2}{5}.2.\displaystyle \frac{22}{7}.7\\ &=\displaystyle \frac{2}{5}.44\\ &=\frac{88}{5}\\ &=17\displaystyle \frac{3}{5}\:\: cm\end{aligned}$.

$\begin{aligned}2.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Tentukan luas daerah yang diarsir}\\ &\textrm{jika}\quad OR=2\: cm\quad \textrm{dan}\:\:\quad PR=1\: cm\\\\ &\textrm{Jawab}:\\ &\textrm{Luas daerah arsiran}\\ &=\textrm{Luas juring POQ}-\textrm{Luas juring ROS}\\ &=\displaystyle \frac{\angle POQ}{360^{\displaystyle 0}}\text{Luas lingkaran}_{r=OP}-\displaystyle \frac{\angle ROS}{360^{\displaystyle 0}}\text{Luas lingkaran}_{r=OR}\\ &=\displaystyle \frac{72^{\displaystyle 0}}{360^{0}}.\pi.OP^{\displaystyle 2}-\displaystyle \frac{72^{\displaystyle 0}}{360^{0}}.\pi.OR^{\displaystyle 2}\\ &=\displaystyle \frac{1}{5}\pi.3^{\displaystyle 2}-\frac{1}{5}.\pi.2^{\displaystyle 2}\\ &=\displaystyle \frac{1}{5}\pi.9-\frac{1}{5}\pi.4\\ &=\frac{1}{5}\pi.(9-4)\\ &=\frac{1}{5}\pi.5\\ &=\pi\:\: cm^{\displaystyle 2}\qquad \textrm{atau}\\ &=\displaystyle \frac{22}{7}\:\: cm^{\displaystyle 2}\end{aligned}$.

$\begin{aligned}3.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Tentukan luas daerah yang diarsir (tembereng)}\\\\ &\textrm{Jawab}:\\ &\textrm{Luas daerah arsiran}\\ &=\textrm{Luas juring POQ}-\textrm{Luas segitiga POQ}\\ &=\displaystyle \frac{\angle POQ}{360^{\displaystyle 0}}\text{Luas lingkaran}_{r=OP}-\displaystyle \frac{1}{2}.\textrm{alas}\times \textrm{tinggi}\\ &=\displaystyle \frac{90^{\displaystyle 0}}{360^{0}}.\pi.OP^{\displaystyle 2}-\displaystyle \frac{1}{2}.r^{\displaystyle 2}\\ &=\displaystyle \frac{1}{4}(3,14).10^{\displaystyle 2}-\frac{1}{2}.10^{\displaystyle 2}\\ &=\frac{(3,14-2)}{4}.100\\ &=1,14\times 25\\ &=28,5\\ &=\displaystyle \frac{22}{7}\:\: cm^{\displaystyle 2}\end{aligned}$.

$\begin{aligned}4.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Tentukan luas daerah yang diarsir (4 tembereng)}\\\\ &\textrm{Jawab}:\\ &\textrm{Diketahui bahwa lingkaran dengan}\:\: d=21\: cm\\&\textrm{Luas daerah arsiran (4 tembereng)}\\ &Alternatif\:1\\&=\textrm{Luas lingkaran}-\textrm{Luas persegi ABCD}\\ &=\displaystyle \frac{1}{4}\pi.d^{\displaystyle 2}-\displaystyle \frac{1}{2}d^{\displaystyle 2}=\displaystyle \frac{1}{4}.\frac{22}{7}.21^{\displaystyle 2}-\displaystyle \frac{1}{2}21^{\displaystyle 2}\\&=\displaystyle \frac{1}{2}.11.63-\frac{1}{2}.441\\&=126\:\: cm^{\displaystyle 2}\\&Alternatif\:2\\&\textrm{diserahkan ke pembaca yang budiman}\end{aligned}$.

$\begin{aligned}5.\quad&\textrm{Perhatikan gambar berikut}\end{aligned}$.

$\begin{aligned}\qquad&\textrm{Jika garis AB adalah diameter dan luas daerah}\\ &\textrm{arsiran}\:\: 22,5\:\: cm^{\displaystyle 2}\:\: \textrm{dan luas juring BOC}\:\: 10\:\: cm^{\displaystyle 2}\\ &\textrm{maka besar sudut}\:\: \angle \,\textrm{BOC}\:\: \textrm{adalah}\:....\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\textrm{Besar sudut pusat}}{\textrm{Besar sudut satu putaran}}=\frac{\textrm{Luas juring}}{\textrm{Luas lingkaran}}\end{aligned}\\ &\Leftrightarrow \displaystyle \frac{\angle BOC}{180^{\displaystyle 0}}=\frac{10\:\: cm^{\displaystyle 2}}{22,5\:\:cm^{\displaystyle 2}}\\ &\Leftrightarrow \angle BOC=\frac{10\:\: cm^{\displaystyle 2}}{22,5\:\:cm^{\displaystyle 2}}\times 180^{\displaystyle 0}\\ &\Leftrightarrow \angle BOC=80^{\displaystyle 0}\end{aligned}$.

DAFTAR PUSTAKA

Kurniawan. 2008. Mandiri Matematika Mengasah Kemampuan Diri SMP Kelas VIII Jilid 2 KTSP 2006. Jakarta: ERLANGGA.
Santoso, N.E., Sksin, N. 2024. Matematika untuk SMA/MA/SMK/MAK Kelas XII Kurikulum Merdeka. Yogyakarta: INTAN PARIWARA EDUKASI.

LINGKARAN-BUSUR, TALI BUSUR, JURING, DAN TEMBERENG

A. Pendahuluan

Lingkaran adalah bangun datar yang dibatasi oleh sekumpulan titik-titik pada bidang yang semuanya meiliki jarak yang sama dari satu titik tetap. Titik tetap tersebut disebut pusat lingkaran, sedangkan jarak dari pusat ke setiap titik pada lingkaran disebut jari-jari.

B. Beberapa Unsur Penting Lingkaran

Pusat: titik tengah lingkaran
Jari-jari (r): jarak dari pusat ke tepi lingkaran
Diameter (d): garis yang melalui pusat dan menghubungkan dua titik pada lingkaran. Diameter = 2 x jari-jari.
Busur: bagian lengkung pada keliling lingkaran
Tali busur: garis lurus yang menghubungkan dua titik pada lingkaran
Juring: daerah yang dibatasi oleh dua jari-jari dan sebuah busur
Temberang: daerah yang dibatasi oleh tali busur dan busur
Apotema: jarak tegak lurus dari pusat lingkaran ke sebuah tali busur
Sudut pusat: sudut yang titik sudutnya berada di pusat lingkaran
Sudut keliling: sudut yang titik sudutnya berada pada lingkaran
Garis singgung: garis yang menyentuh lingkaran tepat di satu titik
Titik-titik singgung: Titik tempat garis singgung menyentuh lingkaran
Garis sekan: garis yang memotong lingkaran di dua titik
Sudut antara tali busur: sudut yang terbentuk oleh dua tali busur
Busur setengah lingkaran: busur berukuran $180^{0}$, dibatasi oleh diameter.
Busur minor: busur yang ukurannya kurang dari $180^{0}$
Busur mayor: busur yang ukurannya lebih dari $180^{0}$.

$\LARGE\fbox{CONTOH SOAL}$.

Perhatikan gambar berikut

Yang tampak pada gambar di atas adalah:

Pusat: O
Jari-jari (r): OA, OB, OC, OD
Diameter (d): AC, BD
Busur: $\widehat{AF},\widehat{AB}, \widehat{AC},\widehat{AD}$, dan lain-lain
Tali busur: $\overline{\textrm{AC}},\overline{\textrm{BD}},\overline{\textrm{CF}},\overline{\textrm{CD}},\overline{\textrm{DF}}$
Juring: juring AOB, BOC, COD, AOD, AOC, dan lain-lain
Temberang: tembereng CD, tembereng CF, dan tembereng DF
Apotema: OE
Sudut pusat: $\angle \textrm{AOB}, \angle \textrm{BOC}, \angle \textrm{COD}, \angle \textrm{AOD},\angle \textrm{AOC}$, dan lain-lain
Sudut keliling: $\angle \textrm{FCD}, \angle \textrm{CDF}, \angle \textrm{CFD}$.
Garis singgung: tak tampak pada gambar
Titik-titik singgung: tak tampak pada gambar
Garis sekan: tak tampak pada gambar dan bedakan antara tali busur dengan garis sekan. Misalkan CD adalah tali busur, tetapi jika garis CD diperpanjang dapat menjadi garis sekan
Sudut antara tali busur: $\angle \textrm{AGF},\angle \textrm{CGD}$. Bedakan dengan sudut keliling. Jika sudut keliling melibatkan dua tai busur dan titik sudutnya pada lingkaran, tetapi untuk sudut antara tali busur adalah yang ada dalam lingkaran.
Busur setengah lingkaran: $\widehat{\textrm{AC}},\widehat{\textrm{BD}}$.
Busur minor: $\widehat{AF},\widehat{AB}, \widehat{AC},\widehat{AD}$
Busur mayor: $\widehat{ACF},\widehat{ACB},\widehat{ACD}$, dan lain-lain

CONTOH SOAL 10 MATRIKS DAN SPL

$\begin{array}{l}\\ 16.&\textrm{Jika diketahui}\quad x_{0}\quad \textrm{dan}\quad y_{0}\quad \textrm{adalah selesaian}\\ &\textrm{dari sistem persamaan}\quad \begin{cases} 2x+3y & =60 \\ 3x+2y & =60 \end{cases}\\ &\textrm{dengan}\quad x_{0}=\displaystyle \frac{a}{\begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix}}\quad \textrm{dan}\quad y_{0}=\displaystyle \frac{b}{\begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix}}\:, \: \textrm{maka}\\ &\textrm{nilai}\quad a+b\quad\textrm{adalah}\:....\\&\begin{array}{llllllll}\\ \textrm{a}.&-240\\ \textrm{b}.&-180\\ \textrm{c}.&-120\\ \textrm{d}.&-60\\ \textrm{e}.&0 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\textrm{Dengan aturan Cramer didapatkan}\\ &x_{0}=\displaystyle \frac{a}{\begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix}}=\displaystyle \frac{\begin{vmatrix} 60 & 3 \\ 60 & 2 \end{vmatrix}}{\begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix}}=\displaystyle \frac{120-180}{4-9}=\displaystyle \frac{-60}{-5}\\ &y_{0}=\displaystyle \frac{b}{\begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix}}=\displaystyle \frac{\begin{vmatrix} 2 & 60 \\ 3 & 60 \end{vmatrix}}{\begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix}}=\displaystyle \frac{120-180}{4-9}=\displaystyle \frac{-60}{-5}\\ &\textrm{Jadi, nilai}\quad a+b=(-60)+(-60)=-120 \end{array}$.

$\begin{array}{l}\\ 17.&\textrm{Sebuah garis}\quad k\quad \textrm{dinayatkan dalam bentuk}\\ &\begin{vmatrix} 1 & x & y\\ a & 1 & 1\\ 1 & 2 & 3 \end{vmatrix}=0.\quad \textrm{Nilai}\quad a\quad \textrm{saat garis tersebut}\\ &\textrm{melalui titik}\quad (1,1)\quad \textrm{adalah}\:....\\&\begin{array}{llllllll}\\ \textrm{a}.&5\\ \textrm{b}.&4\\ \textrm{c}.&3\\ \textrm{d}.&2\\ \textrm{e}.&1 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\textrm{Dengan substitusi langsung ke garis, maka}\\ &\begin{vmatrix} 1 & x & y\\ a & 1 & 1\\ 1 & 2 & 3 \end{vmatrix}=\begin{vmatrix} 1 & 1 & 1\\ a & 1 & 1\\ 1 & 2 & 3 \end{vmatrix}=3+1+2a-1-3a-1=0\\ &\Leftrightarrow 4+2a-3a-2=0\Leftrightarrow 2-a=0\Leftrightarrow a=2 \end{array}$.

$\begin{array}{l}\\ 18.&\textrm{Jika garis yang dinyatakan dengan bentuk}\\ &\begin{vmatrix} x & y & 1\\ 2 & 1 & 1\\ 4 & 5 & 1 \end{vmatrix}=0\quad \textrm{sejajar dengan garis yang }\\ &\textrm{dinyatakan pula dalam bentuk}\quad \begin{vmatrix} x & y & 1\\ 4 & a & 1\\ 3 & 6 & 1 \end{vmatrix}=0\\ &\textrm{maka nilai}\quad a\quad \textrm{adalah}\:....\\&\begin{array}{llllllll}\\ \textrm{a}.&2\\ \textrm{b}.&4\\ \textrm{c}.&5\\ \textrm{d}.&6\\ \textrm{e}.&8 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\textrm{Perhatikan bahwa untuk kedua garis tersebut adalah}\\ &\bullet \quad\begin{vmatrix} x & y & 1\\ 2 & 1 & 1\\ 4 & 5 & 1 \end{vmatrix}=0\Leftrightarrow x+4y+10-4-2y-5x=0\\ &\Leftrightarrow -4x+2y+6=0\Leftrightarrow y=2x-3,\quad \textrm{maka},\:m_{1}=2\\ &\bullet \quad \begin{vmatrix} x & y & 1\\ 4 & a & 1\\ 3 & 6 & 1 \end{vmatrix}=0\Leftrightarrow ax+3y+24-3a-6x-4y=0\\ &\Leftrightarrow (a-6)x-y+24-3a=0\Leftrightarrow y=(a-6)x+24-3a\\ &\textrm{maka}\: m_{2}=a-6\\ &\textrm{Syarat dua garis lurus sejajar adalah}:m_{1}=m_{2}\\ &\Leftrightarrow 2=a-6 \Leftrightarrow a=8 \end{array}$.

$\begin{array}{l}\\ 19.&\textrm{Diketahui suatu sistem persamaan berupa}\\ &\begin{cases} x+y+2z & =4 \\ 2x-y-2z & =-1\\ 3x-2y-z&=3 \end{cases}\quad \textrm{memiliki selesaian}\\ &(x_{0},y_{0},z_{0}).\quad \textrm{Jika}\quad x_{0}=\displaystyle \frac{a}{\begin{vmatrix} 1 & 1 & 2\\ 2 & -1 & -2\\ 3 & -2 & -1 \end{vmatrix}}\\&\textrm{dan}\quad z_{0}=\displaystyle \frac{b}{\begin{vmatrix} 1 & 1 & 2\\ 2 & -1 & -2\\ 3 & -2 & -1 \end{vmatrix}}\\ &\textrm{maka nilai}\quad a+b\quad \textrm{adalah}\:....\\&\begin{array}{llllllll}\\ \textrm{a}.&27\\ \textrm{b}.&25\\ \textrm{c}.&0\\ \textrm{d}.&-25\\ \textrm{e}.&-27 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\textrm{Dengan aturan Cramer kita akan mendapatkan}\\ &x_{0}=\displaystyle \frac{a}{\begin{vmatrix} 1 & 1 & 2\\ 2 & -1 & -2\\ 3 & -2 & -1 \end{vmatrix}}=\displaystyle \frac{\begin{vmatrix} 4 & 1 & 2\\ -1 & -1 & -2\\ 3 & -2 & -1 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 2\\ 2 & -1 & -2\\ 3 & -2 & -1 \end{vmatrix}}\\ &\quad=\displaystyle \frac{4+(-6)+4-(-6)-16-1}{1+(-6)+(-8)-(-6)-4-(-2)}\\ &\quad =\displaystyle \frac{-9}{-9}\\ &z_{0}=\displaystyle \frac{b}{\begin{vmatrix} 1 & 1 & 2\\ 2 & -1 & -2\\ 3 & -2 & -1 \end{vmatrix}}=\displaystyle \frac{\begin{vmatrix} 1 & 1 & 4\\ 2 & -1 & -1\\ 3 & -2 & 3 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 2\\ 2 & -1 & -2\\ 3 & -2 & -1 \end{vmatrix}}\\ &\quad=\displaystyle \frac{(-3)+(-3)+(-16)-(-12)-2-6}{1+(-6)+(-8)-(-6)-4-(-2)}\\ &\quad =\displaystyle \frac{-18}{-9}\\ &\textrm{maka nilai}\quad a+b=(-9)+(-18)=-27 \end{array}$.

$\begin{array}{l}\\ 20.&\textrm{Diketahui matriks transformasi berikut}\\ &\begin{pmatrix}\displaystyle \frac{1}{2}\sqrt{2} & \displaystyle \frac{1}{2}\sqrt{2}\\-\displaystyle \frac{1}{2}\sqrt{2} & \displaystyle \frac{1}{2}\sqrt{2}\end{pmatrix}\\ &\textrm{Peta berupa}\quad (x',y')\quad \textrm{merupakan}\:...\textrm{terhadap}\\&\textrm{titik}\quad (x,y)\\&\begin{array}{llllllll}\\ \textrm{a}.&\textrm{rotasi sejauh}\quad 30^{0}\\ \textrm{b}.&\textrm{rotasi sejauh}\quad 45^{0}\\ \textrm{c}.&\textrm{rotasi sejauh}\quad -45^{0}\\ \textrm{d}.&\textrm{rotasi sejauh}\quad 60^{0}\\ \textrm{e}.&\textrm{rotasi sejauh}\quad -60^{0} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\textrm{Ingat matriks rotasi dengan bentuk}\\ &\begin{pmatrix}\cos\alpha & -\sin\alpha\\\sin\alpha & \cos\alpha\end{pmatrix}\\ &\textrm{dengan memilih}\quad \alpha=-45^{0},\quad \textrm{maka akan sama}\\ &\textrm{dengan yang diketahui pada soal di atas} \end{array}$.

CONTOH SOAL 9 MATRIKS DAN SPL

$\begin{array}{l}\\ 11.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 4x+3y-11z & =-216 \\ 2x+5y+8z & =63\\ x+y+z & =0 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=-6,\quad y=-9\quad \textrm{dan}\quad z=15\\ \textrm{b}.&x=-6,\quad y=9\quad \textrm{dan}\quad z=15\\ \textrm{c}.&x=6,\quad y=-9\quad \textrm{dan}\quad z=15\\ \textrm{d}.&x=-6,\quad y=-9\quad \textrm{dan}\quad z=-15\\ \textrm{e}.&x=-6,\quad y=9\quad \textrm{dan}\quad z=-15 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{ccc|c} 4&3&-11&-216\\ 2&5&8&63\\ 1&1&1&0 \end{array} \right]\quad\begin{matrix}\bullet R_{\displaystyle 2}\longleftarrow 2R_{\displaystyle 2}-R_{\displaystyle 1}\\ \bullet R_{\displaystyle 3}\longleftarrow 4R_{\displaystyle 3}-R_{\displaystyle 1}\end{matrix}\\ &=2(2,5,8|63)-(4,3,-11|-216)=(0,7,27|342)\:\:\textrm{dan}\\ &=4(1,1,1|0)-(4,3,-11|-216)=(0,1,15|216)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{ccc|c} 4&3&-11&-216\\ 0&7&27&342\\ 0&1&15&216 \end{array} \right]\quad \bullet R_{\displaystyle 3}\longleftarrow 7R_{\displaystyle 3}-R_{\displaystyle 2}\\ &=7(0,1,15|216)-(0,7,27|342)=(0,0,78|1170)\\ &\textrm{sehingga}\quad z=\displaystyle \frac{1170}{78}=15,\quad \textrm{maka}\quad y=-9,\:\: x=-6 \end{array}$

$\begin{array}{l}\\ 12.&\textrm{Bayu membeli sebuah buku tulis dan dua buah pensil}\\ &\textrm{Rp}10.000,00.\:\: \textrm{Di toko yang sama, Giri membeli dua }\\ &\textrm{buku tulis dan tiga buah pensil seharga Rp17.500,00}\\ &\textrm{Bentuk perkalian matriks dari narasi di atas adalah}....\\ &\begin{array}{llllllll} \textrm{a}.&\begin{pmatrix} 2 & 1\\ 2 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 10.000\\ 17.500 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} 1 & 2\\ 3 & 2 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 10.000\\ 17.500 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} 1 & 2\\ 2 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 17.500\\ 10.000 \end{pmatrix}\\ \textrm{d}.&\begin{pmatrix} 1 & 2\\ 2 & 3 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 10.000\\ 17.500 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} 2 & 1\\ 3 & 2 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 17.500\\ 10.000 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &x=\textrm{buku},\:\: y=\textrm{pensil}\\ & \end{array}$.

$\begin{array}{l}\\ 13.&\textrm{Penyelesaian dari sistem persamaan berikut}\\ &\begin{cases} 4x+3y+9& =0 \\ -x+y & =4 \end{cases}\\ &\textrm{Model matriks yang menyatakan bentuk di atas}....\\ &\begin{array}{llllllll} \textrm{a}.&\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{7}\begin{pmatrix} 1 & -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ -4 \end{pmatrix}\\ \textrm{b}.&\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{7}\begin{pmatrix} 1 & 3\\ -1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ 4 \end{pmatrix}\\ \textrm{c}.&\begin{pmatrix} x\\ y \end{pmatrix}=-\displaystyle \frac{1}{7}\begin{pmatrix} 1 & -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ 4 \end{pmatrix}\\ \textrm{d}.&\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{7}\begin{pmatrix} 1 & -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ 4 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{7}\begin{pmatrix} 1 & -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} 9\\ 4 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{cases} 4x+3y & =-9 \\-x+y & =4 \end{cases}\Rightarrow \begin{pmatrix} 4 & 3\\ -1 & 1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -9\\ 4 \end{pmatrix}\\ &\Leftrightarrow \begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{\begin{vmatrix}4 & 3\\ -1 & 1\end{vmatrix}}\begin{pmatrix} 1 & -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ 4 \end{pmatrix}\\ &\Leftrightarrow \begin{pmatrix} x\\ y \end{pmatrix}=\displaystyle \frac{1}{7}\begin{pmatrix} 1 & -3\\ 1 & 4 \end{pmatrix}\begin{pmatrix} -9\\ 4 \end{pmatrix} \end{array}$.

$\begin{array}{l}\\ 14.&\textrm{Diberikan SPL berikut}\\ &\begin{cases} 4x+3y & =12 \\ 3x+2y & =7 \end{cases}\\ &\textrm{Nilai}\quad x+y\quad \textrm{adalah}\:....\\&\begin{array}{llllllll}\\ \textrm{a}.&-5\\ \textrm{b}.&5\\ \textrm{c}.&11\\ \textrm{d}.&-11\\ \textrm{e}.&10 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{cc|c} 4&3&12\\ 3&2&7 \end{array} \right]\quad\bullet R_{\displaystyle 1}\longleftarrow R_{\displaystyle 1}-R_{\displaystyle 2}\\ &=(4,3|12)-(3,2|7)=(1,1|5)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{cc|c} 1&1&5\\ 3&2&7 \end{array} \right]\\ &\textrm{sehingga}\quad x+y=5 \end{array}$.

$\begin{array}{l}\\ 15.&\textrm{Perpotongan dua garis yang tersaji sebagai}\\ &\textrm{persamaan matriks berikut}\\ &\begin{pmatrix} -1 & 3\\ 1 & 2 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 5\\ 5 \end{pmatrix}\\ & \textrm{adalah}\:....\\&\begin{array}{llllllll}\\ \textrm{a}.&(2,-1)\\ \textrm{b}.&(1,-2)\\ \textrm{c}.&(-1,2)\\ \textrm{d}.&(-1,-2)\\ \textrm{e}.&(1,2) \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{e}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{cc|c} -1&3&5\\ 1&2&5 \end{array} \right]\quad \begin{aligned}&\bullet R_{\displaystyle 2}\longleftarrow R_{\displaystyle 1}+R_{\displaystyle 2}\\ &\bullet R_{\displaystyle 1}\longleftarrow -R_{\displaystyle 1}\end{aligned}\\ &=(1,2|5)+(-1,3|5)=(0,5|10)\quad \textrm{dan}\\ &=(1,-3|-5)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{cc|c} 1&-3&-5\\ 0&5&10 \end{array} \right]\quad \begin{aligned}&\bullet R_{\displaystyle 1}\longleftarrow \displaystyle \frac{3}{5}R_{\displaystyle 1}+R_{\displaystyle 2}\\ &\bullet R_{\displaystyle 2}\longleftarrow \displaystyle \frac{1}{5}R_{\displaystyle 1}\end{aligned}\\ &\textrm{Selanjutnya diperoleh}\\&\left[ \begin{array}{cc|c} 1&0&1\\ 0&1&2 \end{array} \right]\\ &\textrm{Jadi, koordinat titik potongnya}: (1,2) \end{array}$.

CONTOH SOAL 8 MATRIKS DAN SPL

$\begin{array}{l}\\ 6.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 2x-7y & =-19 \\ 3x-y & =0 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=-3\quad \textrm{dan}\quad y=-1\\ \textrm{b}.&x=3\quad \textrm{dan}\quad y=1\\ \textrm{c}.&x=1\quad \textrm{dan}\quad y=3\\ \textrm{d}.&x=-3\quad \textrm{dan}\quad y=1\\ \textrm{e}.&x=1\quad \textrm{dan}\quad y=-3 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{cc|c} 2&-7&-19\\ 3&-1&0 \end{array} \right]\quad\bullet R_{\displaystyle 2}\longleftarrow 2R_{\displaystyle 2}-3R_{\displaystyle 1}\\ &=2(3,-1|0)-3(2,-7|-19)=(0,19|57)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{cc|c} 2&-7&-19\\ 0&19&57 \end{array} \right]\\ &\textrm{sehingga}\quad y=\displaystyle \frac{57}{19}=3,\quad \textrm{maka}\quad x=1 \end{array}$.

$\begin{array}{l}\\ 7.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 4x-3y & =14 \\ -x+2y & =-1 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=2\quad \textrm{dan}\quad y=5\\ \textrm{b}.&x=5\quad \textrm{dan}\quad y=-2\\ \textrm{c}.&x=5\quad \textrm{dan}\quad y=2\\ \textrm{d}.&x=-5\quad \textrm{dan}\quad y=-2\\ \textrm{e}.&x=-2\quad \textrm{dan}\quad y=-5 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{c}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{cc|c} 4&-3&14\\ -1&2&-1 \end{array} \right]\quad\bullet R_{\displaystyle 2}\longleftarrow R_{\displaystyle 2}+\displaystyle \frac{1}{4}R_{\displaystyle 1}\\ &=(-1,2|-1)+\displaystyle \frac{1}{4}(4,-3|14)=(0,\displaystyle \frac{5}{4}|\displaystyle \frac{10}{4})\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{cc|c} 4&-3&14\\ 0&\displaystyle \frac{5}{4}&\displaystyle \frac{10}{4} \end{array} \right]\\ &\textrm{sehingga}\quad y=\displaystyle \frac{\displaystyle \frac{10}{4}}{\displaystyle \frac{5}{4}}=2,\quad \textrm{maka}\quad x=1 \end{array}$.

$\begin{array}{l}\\ 8.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 8x+3y & =37 \\ 4x+y & =15 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=2\quad \textrm{dan}\quad y=7\\ \textrm{b}.&x=2\quad \textrm{dan}\quad y=-7\\ \textrm{c}.&x=7\quad \textrm{dan}\quad y=-2\\ \textrm{d}.&x=7\quad \textrm{dan}\quad y=2\\ \textrm{e}.&x=-2\quad \textrm{dan}\quad y=-7 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{cc|c} 8&3&37\\ 4&1&15 \end{array} \right]\quad\bullet R_{\displaystyle 2}\longleftarrow R_{\displaystyle 2}-\displaystyle \frac{1}{2}R_{\displaystyle 1}\\ &=(4,1|15)-\displaystyle \frac{1}{2}(8,3|37)=(0,-\displaystyle \frac{1}{2}|-\displaystyle \frac{7}{2})\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{cc|c} 8&3&37\\ 0&-\displaystyle \frac{1}{2}&-\displaystyle \frac{7}{2} \end{array} \right]\\ &\textrm{sehingga}\quad y=\displaystyle \frac{-\displaystyle \frac{7}{2}}{-\displaystyle \frac{2}{2}}=7,\quad \textrm{maka}\quad x=2 \end{array}$.

$\begin{array}{l}\\ 9.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 4x+2y+3z & =0 \\ 2x+3y+5z & =9\\ 3x+y+7z & =9 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=-2,\quad y=1\quad \textrm{dan}\quad z=2\\ \textrm{b}.&x=2,\quad y=1\quad \textrm{dan}\quad z=2\\ \textrm{c}.&x=-2,\quad y=-1\quad \textrm{dan}\quad z=-2\\ \textrm{d}.&x=-2,\quad y=1\quad \textrm{dan}\quad z=-2\\ \textrm{e}.&x=-2,\quad y=-1\quad \textrm{dan}\quad z=2 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{a}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{ccc|c} 4&2&3&0\\ 2&3&5&9\\ 3&1&7&9 \end{array} \right]\quad\begin{matrix} \bullet R_{\displaystyle 2}\longleftarrow 2R_{\displaystyle 2}-R_{\displaystyle 1}\: .\\ \bullet R_{\displaystyle 3}\longleftarrow 4R_{\displaystyle 3}-3R_{\displaystyle 1} \end{matrix}\\ &=2(2,3,5|9)-(4,2,3|0)=(0,4,7|18)\quad \textrm{dan}\\ &=4(2,3,5|9)-3(4,2,3|0)=(0,-2,19|36)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{ccc|c} 4&2&3&0\\ 0&4&7&18\\ 0&-2&19&36 \end{array} \right]\quad \bullet R_{\displaystyle 3}\longleftarrow 2R_{\displaystyle 3}+R_{\displaystyle 2}\\ &=2(0,-2,19|36)+(0,4,7|18)=(0,0,45|90)\\ &\textrm{sehingga}\quad z=\displaystyle \frac{90}{45}=2,\quad \textrm{maka}\quad y=1,\:\: x=-2 \end{array}$.

$\begin{array}{l}\\ 10.&\textrm{SPL berikut memiliki selesaian}....\\ &\begin{cases} 7x+2y+z & =-40 \\ 2x+7y+4z & =45\\ 5x-4y+6z & =8 \end{cases}\\ &\begin{array}{llllllll}\\ \textrm{a}.&x=-8,\quad y=-3\quad \textrm{dan}\quad z=10\\ \textrm{b}.&x=-8,\quad y=3\quad \textrm{dan}\quad z=10\\ \textrm{c}.&x=-8,\quad y=3\quad \textrm{dan}\quad z=-10\\ \textrm{d}.&x=8,\quad y=3\quad \textrm{dan}\quad z=10\\ \textrm{e}.&x=8,\quad y=-3\quad \textrm{dan}\quad z=10 \end{array}\\\\ &\textbf{Jawab}:\quad \textbf{b}\\ &\textrm{Dengan metode eliminasi Gauss didapatkan}\\ &\left[ \begin{array}{ccc|c} 7&2&1&-40\\ 2&7&4&45\\ 5&-4&6&8 \end{array} \right]\quad\begin{matrix}\bullet R_{\displaystyle 2}\longleftarrow 7R_{\displaystyle 2}-2R_{\displaystyle 1}\\ \bullet R_{\displaystyle 3}\longleftarrow 7R_{\displaystyle 3}-5R_{\displaystyle 1}\end{matrix}\\ &=7(2,7,4|45)-2(7,2,1|-40)=(0,45,26|395)\:\:\textrm{dan}\\ &=7(5,-4,6|8)-5(7,2,1|-40)=(0,-38,37|256)\\ &\textrm{Sehingga diperoleh}\\ &\left[ \begin{array}{ccc|c} 7&2&1&-40\\ 0&45&26&395\\ 0&-38&37&256 \end{array} \right]\quad \bullet R_{\displaystyle 3}\longleftarrow \displaystyle \frac{1}{45}R_{\displaystyle 3}+\displaystyle \frac{1}{38}R_{\displaystyle 2}\\ &=\displaystyle \frac{1}{45}(0,-38,37|256)+\displaystyle \frac{1}{38}(0,45,26|395)=(0,0,\displaystyle \frac{2653}{1710}|\displaystyle \frac{26530}{1710})\\ &\textrm{sehingga}\quad z=\displaystyle \frac{\displaystyle \frac{226530}{1710}}{\displaystyle \frac{2653}{1710}}=10,\quad \textrm{maka}\quad y=3,\:\: x=-8 \end{array}$.