tolong di jawab yah kakak nomor 1 sampai 3.nomor 3 bagian a dan b juga yah soalnya pr untuk besok di kumpul

Soal No. 1
Jika :
Titik Awal = A
Titik Belok = B
Titik AKhir = C

AB = 50
BC = 120
AC = ?

[tex]AC= \sqrt{AB^2+BC^2} \\ \\ AC= \sqrt{(5\times10)^2+(12\times10)^2} \\ \\ AC= \sqrt{(25\times10^2)+(144\times10^2)} \\ \\ AC= \sqrt{(25+144)\times10^2} \\ \\AC= \sqrt{169}\times \sqrt{10^2} \\ \\ AC=13\times10 \\ \\ AC=130\ m [/tex]

Soal No. 2
Segitiga Siku-siku PQR. PQ : QR = 3 : 4. PR = 50 cm
Jika P = 3 maka Q = 4
Berdasarkan terema phytagoras type 1, maka PQ : QR : PR = 3 : 4 : 5

Dengan panjang PR = 50 cm, maka PQ = 30 cm dan QR = 40 cm

Keliling = PQ + QR + PR
               = 30 + 40 + 50
               = 120 cm

[tex]Luas= \frac{1}{2}\times40\times30 \\ \\ Luas = 20\times30 \\ \\ Luas = 600\ cm^2 [/tex]

Soal No. 3
AB = 250 km
BC = 300 km
AC = ....km

[tex]AC= \sqrt{AB^2+BC^2} \\ \\ AC= \sqrt{(25\times10)^2+(30\times10)^2} \\ \\ AC= \sqrt{(625\times10^2)+(900\times10^2)} \\ \\ AC= \sqrt{(625+900)\times10^2} \\ \\AC= \sqrt{1.525}\times \sqrt{10^2} \\ \\ AC= (\sqrt{25\times61}) \times10 \\ \\ AC=\sqrt{25}\times\sqrt{61} \times10\\ \\ AC=5\times\sqrt{61} \times10 \\ \\ AC=50\sqrt{61} \\ \\ AC=50\times(7,81) \\ \\ AC=390,51\ km [/tex]

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