Exercise Zone : Segitiga
Table of Contents

Tipe:
No.
Tentukan panjang FG.ALTERNATIF PENYELESAIAN
Misal CG = x
OG=\dfrac32\sqrt{26}-x
\begin{aligned}
OF^2-OG^2&=CF^2-CG^2\\
\left(\dfrac92\sqrt2\right)^2-\left(\dfrac32\sqrt{26}-x\right)^2&=\left(6\sqrt2\right)^2-x^2\\
\dfrac{81}2-\left(\dfrac{117}2-3x\sqrt{26}+x^2\right)&=72-x^2\\
\dfrac{81}2-\dfrac{117}2+3x\sqrt{26}-\cancel{\color{red}x^2}&=72-\cancel{\color{red}x^2}\\
-18+3x\sqrt{26}&=72\\
3x\sqrt{26}&=90\\
x&=\dfrac{90}{3\sqrt{26}}\\
&=\dfrac{30}{\sqrt{26}}
\end{aligned}
\begin{aligned}
FG&=\sqrt{CF^2-CG^2}\\
&=\sqrt{\left(6\sqrt2\right)^2-\left(\dfrac{30}{\sqrt{26}}\right)^2}\\
&=\sqrt{72-\dfrac{900}{13}}\\
&=\sqrt{\dfrac{36}{13}}\\
&=\dfrac6{13}\sqrt{13}
\end{aligned}
Jadi, FG=\dfrac6{13}\sqrt{13} cm.
No.
Diketahui segitiga ABC siku-siku di A. Jika panjangALTERNATIF PENYELESAIAN
No.
Segitiga ABC siku siku di C jika panjangALTERNATIF PENYELESAIAN
\begin{aligned}\tan60\degree&=\dfrac{BC}{AC}\\\sqrt3&=\dfrac{BC}{20}\\BC&=\boxed{\boxed{20\sqrt3}}\end{aligned}
Jadi, BC=20\sqrt3 .
No.
Diketahui △ABC dengan panjang sisi a = 5 cm, b = 7 cm, dan c = 10 cm. Luas segitiga tersebut adalah ....ALTERNATIF PENYELESAIAN
\begin{aligned}
s&=\frac{a+b+c}2\\[3.7pt]
&=\frac{5+7+10}2\\[3.7pt]
&=\frac{22}2\\[3.7pt]
&=11
\end{aligned}
\begin{aligned}
L&=\sqrt{s(s-a)(s-b)(s-c)}\\
&=\sqrt{11(11-5)(11-7)(11-10)}\\
&=\sqrt{11(6)(4)(1)}\\
&=\boxed{\boxed{2\sqrt{66}}}
\end{aligned}
Jadi, luas segitiga tersebut adalah 2\sqrt{66} cm2.
Post a Comment