CONTOH SOAL 11 LINGKARAN (GARIS SINGGUNG PERSEKUTUAN DUA LINGKARAN)

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

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

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

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

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

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

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

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

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

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

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

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