Exercise Zone : Segitiga

Berikut ini adalah kumpulan soal mengenai Segitiga. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

Standar SBMPTN HOTS

No.

Tentukan panjang FG.

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Misal CG=x
OG=\dfrac32\sqrt{26}-x

$$\begin{array}{rl}O{F}^2-O{G}^2& =C{F}^2-C{G}^2\\ {\left({\displaystyle \frac{9}{2}}\sqrt{2}\right)}^2-{({\displaystyle \frac{3}{2}}\sqrt{26}-x)}^2& ={\left(6\sqrt{2}\right)}^2-x^2\\ {\displaystyle \frac{81}{2}}-({\displaystyle \frac{117}{2}}-3x\sqrt{26}+x^2)& =72-x^2\\ {\displaystyle \frac{81}{2}}-{\displaystyle \frac{117}{2}}+3x\sqrt{26}-\cancel{x^2}& =72-\cancel{x^2}\\ -18+3x\sqrt{26}& =72\\ 3x\sqrt{26}& =90\\ x& ={\displaystyle \frac{90}{3\sqrt{26}}}\\ & ={\displaystyle \frac{30}{\sqrt{26}}}\end{array}$$

$$\begin{array}{rl}FG& =\sqrt{C{F}^2-C{G}^2}\\ & =\sqrt{{\left(6\sqrt{2}\right)}^2-{\left({\displaystyle \frac{30}{\sqrt{26}}}\right)}^2}\\ & =\sqrt{72-{\displaystyle \frac{900}{13}}}\\ & =\sqrt{{\displaystyle \frac{36}{13}}}\\ & ={\displaystyle \frac{6}{13}}\sqrt{13}\end{array}$$

Jadi, FG=\dfrac6{13}\sqrt{13} cm.

No.

Diketahui segitiga ABC siku-siku di A. Jika panjang AB=6, dan panjang BC=10, maka tinggi segitiga ABC pada sisi BC adalah

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$\begin{array}{rl}AC& =\sqrt{B{C}^2-A{B}^2}\\ & =\sqrt{{10}^2-6^2}\\ & =\sqrt{100-36}\\ & =\sqrt{64}\\ & =8\end{array}$

$\begin{array}{rl}t& ={\displaystyle \frac{AB\cdot AC}{BC}}\\ & ={\displaystyle \frac{6\cdot 8}{10}}\\ & ={\displaystyle \frac{48}{10}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{24}{5}}}}}}\end{array}$

Jadi, tinggi segitiga ABC pada sisi BC adalah \dfrac{24}5.

No.

Segitiga ABC siku siku di C jika panjang {AC = 20} cm dan besar sudut A = 60\degree tentukan panjang BC!​

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$$\begin{array}{rl}\tan 60{}^{\circ }& ={\displaystyle \frac{BC}{AC}}\\ \sqrt{3}& ={\displaystyle \frac{BC}{20}}\\ BC& =\boxed{{\displaystyle \boxed{{\displaystyle 20\sqrt{3}}}}}\end{array}$$

Jadi, BC=20\sqrt3.