Question

Kuis ¹²¹ : Dari segibanyak tak beraturan ABCDE, tentukan: (i) Luas (ii) Keliling

Answer 1

[tex]\begin{aligned}\bullet\ &\textsf{Luas segibanyak ABCDE}\\&=\,\boxed{\vphantom{\Bigg|}\bf\left(3+\frac{11}{4}\sqrt{3}\right)\ cm^2\,}\\&\approx\,\boxed{\vphantom{\big|}\,\bf7{,}763\ cm^2\,}\\\bullet\ &\textsf{Keliling segibanyak ABCDE}\\&=\,\boxed{\vphantom{\bigg|}\,\bf19-\sqrt{6}\left(1+\sqrt{2}-\sqrt{3}\right)\ cm\,}\\&\approx\,\boxed{\vphantom{\big|}\,\bf17{,}329\ cm\,}\\\end{aligned}[/tex]
 

Penjelasan dengan langkah-langkah:

Untuk perhitungan luas dan keliling, kita persiapkan nilai sin(75°) terlebih dahulu, karena pelurus dari 105° adalah 75°.

sin(75°) = sin(30°) cos(45°) + cos(30°) sin(45°)
= ¼√2 + ¼√6
= ¼(√2 + √6)

Untuk menentukan besar beberapa sudut dan panjang ruas garis yang diperlukan, kita gunakan aturan sinus dan aturan cosinus.

Untuk besar ∠BDC:

[tex]\begin{aligned}\frac{BC}{\sin(\angle{BDC})}&=\frac{CD}{\sin(\angle{CBD})}\\\sin(\angle{BDC})&=\left(\frac{BC}{CD}\right)\sin(\angle{CBD})\\&=\left(\frac{3\sqrt{2}-\sqrt{6}}{2}\right)\sin(75^\circ)\\&=\left[\frac{\sqrt{3}\left(\sqrt{6}-\sqrt{2}\right)}{2}\right]\left(\frac{\sqrt{2}+\sqrt{6}}{4}\right)\\&=\frac{\sqrt{3}(6-2)}{8}\\&=\frac{\sqrt{3}}{2}\\\angle{BDC}&=\arcsin\left(\frac{\sqrt{3}}{2}\right),\ \angle{BDC} < 105^{\circ}\\\angle{BDC}&=\bf60^{\circ}\end{aligned}[/tex]

Oleh karena itu:

• ∠BCD = 105° – 60° = 45°.
• ∠BDE = 180° – 60° = 120°.

Untuk panjang BD:

[tex]\begin{aligned}\frac{BD}{\sin(\angle{BCD})}&=\frac{BC}{\sin(\angle{BDC})}\\BD&=\left[\frac{\sin(\angle{BCD})}{\sin(\angle{BDC})}\right]BC\\&=\left[\frac{\sin(45^{\circ})}{\sin(60^{\circ})}\right]\left(3\sqrt{2}-\sqrt{6}\right)\\&=\frac{\sqrt{2}}{\sqrt{3}}\left(3\sqrt{2}-\sqrt{6}\right)\\&=\sqrt{2}\left(\sqrt{6}-\sqrt{2}\right)\\BD&=\bf2\left(\sqrt{3}-1\right)\ cm\end{aligned}[/tex]

Oleh karena itu:

• AD = AB + BD
  ⇒ AD = 7 – 2√3 + 2(√3 – 1)
  ⇒ AD = 5 cm.

Untuk panjang DE:

[tex]\begin{aligned}AE^2&=AD^2+DE^2-2\cdot AD\cdot DE\cdot\cos(\angle{ADE})\\7^2&=5^2+DE^2-10DE\cos(120^\circ)\\49&=25+DE^2-10DE\left(-\frac{1}{2}\right)\\24&=DE^2+5DE\\0&=DE^2+5DE-24\\&=(DE-3)(DE+8)\\DE&=3\ \lor\ DE=-8\\DE&={\bf3\ cm}\quad\because DE > 0\end{aligned}[/tex]

Luas Segibanyak ABCDE (Segiempat ABCE)

[tex]\begin{aligned}L_{ABCDE}&=L_{\triangle{ADE}}+L_{\triangle{BCD}}\\&=\frac{1}{2}\left[\,\begin{matrix}AD\cdot DE\cdot\sin(\angle{ADE})\\+\\BD\cdot CD\cdot\sin(\angle{BDC})\end{matrix}\,\right]\\&\quad\rightarrow \sin(\angle{ADE})=\sin(\angle{BDC})\\&\qquad\because\angle{ADE}+\angle{BDC}=180^\circ\\&=\frac{1}{2}\sin(\angle{BDC})\left(AD\cdot DE+BD\cdot CD\right)\\&=\frac{1}{2}\sin(60^\circ)\left[5\cdot3+2\cdot2\left(\sqrt{3}-1\right)\right]\end{aligned}[/tex]
[tex]\begin{aligned}&=\frac{1}{4}\sqrt{3}\left(15+4\sqrt{3}-4\right)\\&=\frac{1}{4}\sqrt{3}\left(11+4\sqrt{3}\right)\\&=\frac{1}{4}\left(11\sqrt{3}+12\right)\\L_{ABCDE}&=\bf\left(3+\frac{11}{4}\sqrt{3}\right)\ cm^2\\L_{ABCDE}&\approx\bf7{,}763\ cm^2\\\end{aligned}[/tex]

Keliling Segibanyak ABCDE (Segiempat ABCE)

[tex]\begin{aligned}K_{ABCDE}&=AB+BC+CD+DE+AE\\&=7-2\sqrt{3}+3\sqrt{2}-\sqrt{6}+2+3+7\\&=19-2\sqrt{3}+3\sqrt{2}-\sqrt{6}\\&=19-\sqrt{6}\left(\sqrt{2}-\sqrt{3}+1\right)\\K_{ABCDE}&=\bf19-\sqrt{6}\left(1+\sqrt{2}-\sqrt{3}\right)\ cm\\K_{ABCDE}&\approx\bf17{,}329\ cm\end{aligned}[/tex]
[tex]\blacksquare[/tex]