tolong jawab dengan benar dan pakai cara dan rumusnya!!!!!!

Berikut ini adalah pertanyaan dari yoschristian66 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Pertama

Tolong jawab dengan benar dan pakai cara dan rumusnya!!!!!!

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

$\begin{aligned}\sf a.\ &TQ=\bf\frac{25\sqrt{3}}{2}\ cm\\\textsf{-----}&\textsf{-----------------------------------}\\\sf b.\ &PT=\bf37.5\ cm\\\textsf{-----}&\textsf{-----------------------------------}\\\sf c.\ &PR=\bf25\left(1+\sqrt{3}\right)\ cm\\\textsf{-----}&\textsf{-----------------------------------}\end{aligned}$
 

Explanation

Right-Angled Triangle (Right Triangle)

Based on the given picture, it’s known that:

QS = 25 cm
∠QSR = 45°
∠QST = 60°
∠QTP = ∠SQP = ∠RQS = 90° (right-angle)
QS is the hypothenuse of ΔQST
RS is the hypothenuse of ΔQRS
PQ is the hypothenuse of ΔPQT

Hence:

TQ = QS × sin(∠QST)
⇒ TQ = 25 × sin(60°)
⇒ TQ = 25 × ½√3
$\therefore\ \boxed{\ \bf TQ=\frac{25\sqrt{3}}{2}\ cm\ }$

PT = TQ / tan(∠QPT)
⇒ PT = TQ / tan(90°–(90°–∠SQT))
⇒ PT = TQ / tan(∠SQT)
⇒ PT = TQ / tan(90°–∠QST)
⇒ PT = TQ / tan(90°–60°)
⇒ PT = TQ / tan(30°) ∴ [ ∠QPT = 30° ]
⇒ PT = TQ × cot(30°)
⇒ PT = (25√3)/2 × cos(30°) / sin(30°)
⇒ PT = (25√3)/2 × ½√3 / ½
⇒ PT = (25√3)/2 × √3
⇒ PT = 75/2 cm
$\therefore\ \boxed{\ \bf PT=37.5\ cm\ }$

PR = PQ + QR
⇒ PR = PQ + QS [ because ∠QSR = ∠QRS = 45° ]
⇒ PR = [TQ / sin(∠QPT)] + QS
⇒ PR = [(25√3)/2 / sin(30°)] + 25
⇒ PR = [(25√3)/2 / ½)] + 25
⇒ PR = [(25√3)/2 × 2)] + 25
$\therefore\ \boxed{\ \begin{aligned}\bf PR&\bf=25+25\sqrt{3}\ cm\\&\bf=25\left(1+\sqrt{3}\right)\ cm\end{aligned}\ }$