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Dear Students,
Welcome to today’s quiz! This is your opportunity to demonstrate what you’ve learned so far, so do your best. Please keep in mind that you have a maximum of 30 minutes to complete all the questions. Make sure to manage your time wisely and answer each question thoughtfully.
Good luck!
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Pertanyaan 1 dari 8
1. Pertanyaan
1 pointsPerhatikan gambar berikut.
Diketahui:
QR // STSQ // TR
PQ = 5 cm
PR = 6 cm
RU = 16 cm
TU = 12 cm
Keliling jajar genjang QSTR adalah…
Benar
\(\color{blue}\triangle \text{PQR} \sim \triangle \text{PSU}\)
\(\dfrac{\text{PQ}}{\text{PS}} = \dfrac{\text{PR}}{\text{PU}}\)
\(\dfrac{5}{5 + \text{QS}} = \dfrac{6}{6 + 16}\)
\(\dfrac{5}{5 + \text{QS}} = \dfrac{6}{22}\)
\(\dfrac{5}{5 + \text{QS}} = \dfrac{3}{11}\)
Kali silang
\(5 \times 11 = 3(5 + \text{QS})\)
\(55 = 15 + 3\text{QS}\)
\(3\text{QS} = 55\:-\:15\)
\(\text{QS} = \dfrac{40}{3}\text{ cm}\)
\(\color{blue}\triangle \text{TUR} \sim \triangle \text{SUP}\)
\(\dfrac{\text{UT}}{\text{US}} = \dfrac{\text{UR}}{\text{UP}}\)
\(\dfrac{12}{12 + \text{TS}} = \dfrac{16}{16 + 6}\)
\(\dfrac{12}{12 + \text{TS}} = \dfrac{16}{22}\)
\(\dfrac{12}{12 + \text{TS}} = \dfrac{8}{11}\)
Kali silang
\(12 \times 11 = 8(12 + \text{TS})\)
\(132 = 96 + 8\text{TS}\)
\(8\text{TS} = 36\)
\(\text{TS} = \dfrac{36}{8} = \dfrac{9}{2}\text{ cm}\)
Keliling jajar genjang QSTR = 2(QS + TS)
Keliling jajar genjang QSTR = \(2\left( \dfrac{40}{3} +Â \dfrac{9}{2}\right)\)
Keliling jajar genjang QSTR = \(2 \cdot \dfrac{107}{6} = \dfrac{107}{3} \text{ cm}\)
Salah
\(\color{blue}\triangle \text{PQR} \sim \triangle \text{PSU}\)
\(\dfrac{\text{PQ}}{\text{PS}} = \dfrac{\text{PR}}{\text{PU}}\)
\(\dfrac{5}{5 + \text{QS}} = \dfrac{6}{6 + 16}\)
\(\dfrac{5}{5 + \text{QS}} = \dfrac{6}{22}\)
\(\dfrac{5}{5 + \text{QS}} = \dfrac{3}{11}\)
Kali silang
\(5 \times 11 = 3(5 + \text{QS})\)
\(55 = 15 + 3\text{QS}\)
\(3\text{QS} = 55\:-\:15\)
\(\text{QS} = \dfrac{40}{3}\text{ cm}\)
\(\color{blue}\triangle \text{TUR} \sim \triangle \text{SUP}\)
\(\dfrac{\text{UT}}{\text{US}} = \dfrac{\text{UR}}{\text{UP}}\)
\(\dfrac{12}{12 + \text{TS}} = \dfrac{16}{16 + 6}\)
\(\dfrac{12}{12 + \text{TS}} = \dfrac{16}{22}\)
\(\dfrac{12}{12 + \text{TS}} = \dfrac{8}{11}\)
Kali silang
\(12 \times 11 = 8(12 + \text{TS})\)
\(132 = 96 + 8\text{TS}\)
\(8\text{TS} = 36\)
\(\text{TS} = \dfrac{36}{8} = \dfrac{9}{2}\text{ cm}\)
Keliling jajar genjang QSTR = 2(QS + TS)
Keliling jajar genjang QSTR = \(2\left( \dfrac{40}{3} +Â \dfrac{9}{2}\right)\)
Keliling jajar genjang QSTR = \(2 \cdot \dfrac{107}{6} = \dfrac{107}{3} \text{ cm}\)
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Pertanyaan 2 dari 8
2. Pertanyaan
1 pointsPerhatikan gambar berikut.
Pasangan bangun datar yang saling sebangun adalah…
Benar
Segitiga siku-siku E dan F sebangun karena sisi-sisi yang bersesuaian memiliki perbandingan yang sama.
Salah
Segitiga siku-siku E dan F sebangun karena sisi-sisi yang bersesuaian memiliki perbandingan yang sama.
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Pertanyaan 3 dari 8
3. Pertanyaan
1 pointsSebuah foto berukuran 12 cm × 15 cm ditempelkan pada sebuah bingkai karton sehingga masih menyisakan karton pada bagian kiri, kanan, dan atas sepanjang \(x\) cm, serta pada bagian bawah menyisakan karton 3 cm.
Jika foto dan bingkai sebangun, nilai \(x = \dotso\)
Benar
Ukuran bingkai:
Lebar = (12 + 2x) cm
Tinggi = (15 + 3 + x) = (18 + x) cm
Ukuran Foto:
Lebar = 12 cm
Tinggi = 15 cm
Karena foto dan bingkai sebangun, maka sisi-sisi yang bersesuaian memiliki perbandingan yang sama.
\(\dfrac{\text{lebar foto}}{\text{lebar karton}} = \dfrac{\text{tinggi foto}}{\text{tinggi karton}}\)
\(\dfrac{12}{12 + 2x} = \dfrac{15}{18 + x}\)
Kali silang
\(12(18 + x) = 15(12 + 2x)\)
Bagi kedua ruas dengan 3
\(4(18 + x) = 5(12 + 2x)\)
\(72 + 4x = 60 + 10x\)
\(72\:-\:60 = 10x\:-\:4x\)
\(12 = 6x\)
\(x = 2\text{ cm}\)
Salah
Ukuran bingkai:
Lebar = (12 + 2x) cm
Tinggi = (15 + 3 + x) = (18 + x) cm
Ukuran Foto:
Lebar = 12 cm
Tinggi = 15 cm
Karena foto dan bingkai sebangun, maka sisi-sisi yang bersesuaian memiliki perbandingan yang sama.
\(\dfrac{\text{lebar foto}}{\text{lebar karton}} = \dfrac{\text{tinggi foto}}{\text{tinggi karton}}\)
\(\dfrac{12}{12 + 2x} = \dfrac{15}{18 + x}\)
Kali silang
\(12(18 + x) = 15(12 + 2x)\)
Bagi kedua ruas dengan 3
\(4(18 + x) = 5(12 + 2x)\)
\(72 + 4x = 60 + 10x\)
\(72\:-\:60 = 10x\:-\:4x\)
\(12 = 6x\)
\(x = 2\text{ cm}\)
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Pertanyaan 4 dari 8
4. Pertanyaan
1 pointsPerhatikan gambar berikut.
Diketahui panjang AE = 8 cm, BC = 30 cm, BD = 8 cm, dan AD = 12 cm. Sudut ADE besarnya sama dengan sudut ACB.
Panjang \(\textbf{DE} = \dotso\)
Benar
Segitiga DAE sebangun dengan segitiga CAB karena:
- \(m \angle \textbf{ADE} = m \angle \textbf{ACB}\:\:\:\text{diketahui}\)
- \(m \angle \textbf{DAE} = m \angle \textbf{CAB}\:\:\:\text{sudut berimpit}\)
Karena \(\triangle \textbf{DAE} \sim \triangle \textbf{CAB}\) maka sisi-sisi yang bersesuaian memiliki perbandingan yang sama.
\(\dfrac{\textbf{AE}}{\textbf{AB}} = \dfrac{\textbf{DE}}{\textbf{BC}}\)
\(\dfrac{8}{8 + 12} = \dfrac{\textbf{DE}}{30}\)
\(\dfrac{8}{20} = \dfrac{\textbf{DE}}{30}\)
\(\dfrac{2}{5} = \dfrac{\textbf{DE}}{30}\)
Kali silang
\(2 \times 30 = 5\cdot \textbf{DE}\)
\(60 = 5\cdot \textbf{DE}\)
\(\textbf{DE} = \dfrac{60}{5} = 12 \text{ cm}\)
Salah
Segitiga DAE sebangun dengan segitiga CAB karena:
- \(m \angle \textbf{ADE} = m \angle \textbf{ACB}\:\:\:\text{diketahui}\)
- \(m \angle \textbf{DAE} = m \angle \textbf{CAB}\:\:\:\text{sudut berimpit}\)
Karena \(\triangle \textbf{DAE} \sim \triangle \textbf{CAB}\) maka sisi-sisi yang bersesuaian memiliki perbandingan yang sama.
\(\dfrac{\textbf{AE}}{\textbf{AB}} = \dfrac{\textbf{DE}}{\textbf{BC}}\)
\(\dfrac{8}{8 + 12} = \dfrac{\textbf{DE}}{30}\)
\(\dfrac{8}{20} = \dfrac{\textbf{DE}}{30}\)
\(\dfrac{2}{5} = \dfrac{\textbf{DE}}{30}\)
Kali silang
\(2 \times 30 = 5\cdot \textbf{DE}\)
\(60 = 5\cdot \textbf{DE}\)
\(\textbf{DE} = \dfrac{60}{5} = 12 \text{ cm}\)
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Pertanyaan 5 dari 8
5. Pertanyaan
1 pointsPerhatikan gambar berikut.
Diketahui panjang AB = BC = CD = 18 cm dan ED = 12 cm. Panjang \(\textbf{BF} = \dotso\)
Benar
Misal panjang BF = x cm
\(\color{blue} \triangle \textbf{ABF} \sim \triangle \textbf{GCF}\)
\(\dfrac{\text{AB}}{\text{GC}} = \dfrac{\text{BF}}{\text{CF}}\)
\(\dfrac{18}{\text{GC}} = \dfrac{x}{18\:-\:x}\)
\(\textbf{CG} = \color{red} \dfrac{18(18\:-\:x)}{x}\)
\(\color{blue} \triangle \textbf{ABF} \sim \triangle \textbf{GDE}\)
\(\dfrac{\text{AB}}{\text{GD}} = \dfrac{\text{BF}}{\text{ED}}\)
\(\dfrac{18}{18\:-\:\text{CG}} = \dfrac{x}{12}\)
Kali silang
\(18 \times 12 = x(18 \:-\:\text{CG})\)
\(216 = 18x \:-\:\cancel{x} \cdot \color{red} \dfrac{18(18\:-\:x)}{\cancel{x}}\)
\(216 = 18x\:-\:324 + 18x\)
\(540 = 36x\)
\(x = \dfrac{540}{36}\)
\(x = 15\)
Jadi, panjang BF adalah 15 cm.
Salah
Misal panjang BF = x cm
\(\color{blue} \triangle \textbf{ABF} \sim \triangle \textbf{GCF}\)
\(\dfrac{\text{AB}}{\text{GC}} = \dfrac{\text{BF}}{\text{CF}}\)
\(\dfrac{18}{\text{GC}} = \dfrac{x}{18\:-\:x}\)
\(\textbf{CG} = \color{red} \dfrac{18(18\:-\:x)}{x}\)
\(\color{blue} \triangle \textbf{ABF} \sim \triangle \textbf{GDE}\)
\(\dfrac{\text{AB}}{\text{GD}} = \dfrac{\text{BF}}{\text{ED}}\)
\(\dfrac{18}{18\:-\:\text{CG}} = \dfrac{x}{12}\)
Kali silang
\(18 \times 12 = x(18 \:-\:\text{CG})\)
\(216 = 18x \:-\:\cancel{x} \cdot \color{red} \dfrac{18(18\:-\:x)}{\cancel{x}}\)
\(216 = 18x\:-\:324 + 18x\)
\(540 = 36x\)
\(x = \dfrac{540}{36}\)
\(x = 15\)
Jadi, panjang BF adalah 15 cm.
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Pertanyaan 6 dari 8
6. Pertanyaan
1 pointsPerhatikan gambar berikut.
Panjang \(\textbf{EF} = \dotso\)
Benar
\(\color{blue} \triangle \textbf{AEB} \sim \triangle \textbf{CED}\)
\(\dfrac{\text{AB}}{\text{CD}} = \dfrac{\text{BE}}{\text{ED}}\)
\(\dfrac{6}{10} = \dfrac{\text{BE}}{\text{ED}}\)
\(\dfrac{\text{BE}}{\text{ED}} = \dfrac{3}{5}\)
\(\textbf{BE} : \textbf{ED} = 3 : 5\)
\(\color{blue} \triangle \textbf{BFE} \sim \triangle \textbf{BCD}\)
\(\dfrac{\text{BE}}{\text{BD}} = \dfrac{\text{EF}}{\text{DC}}\)
\(\dfrac{3}{3 + 5} = \dfrac{\text{EF}}{10}\)
\(\dfrac{3}{8} = \dfrac{\text{EF}}{10}\)
Kali silang
\(3 \times 10 = 8\cdot \textbf{EF}\)
\(\textbf{EF} = \dfrac{30}{8} = 3,75 \text{ cm}\)
Salah
\(\color{blue} \triangle \textbf{AEB} \sim \triangle \textbf{CED}\)
\(\dfrac{\text{AB}}{\text{CD}} = \dfrac{\text{BE}}{\text{ED}}\)
\(\dfrac{6}{10} = \dfrac{\text{BE}}{\text{ED}}\)
\(\dfrac{\text{BE}}{\text{ED}} = \dfrac{3}{5}\)
\(\textbf{BE} : \textbf{ED} = 3 : 5\)
\(\color{blue} \triangle \textbf{BFE} \sim \triangle \textbf{BCD}\)
\(\dfrac{\text{BE}}{\text{BD}} = \dfrac{\text{EF}}{\text{DC}}\)
\(\dfrac{3}{3 + 5} = \dfrac{\text{EF}}{10}\)
\(\dfrac{3}{8} = \dfrac{\text{EF}}{10}\)
Kali silang
\(3 \times 10 = 8\cdot \textbf{EF}\)
\(\textbf{EF} = \dfrac{30}{8} = 3,75 \text{ cm}\)
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Pertanyaan 7 dari 8
7. Pertanyaan
1 pointsPerhatikan gambar berikut.
Diketahui titik E dan F berturut-turut adalah titik tengah diagonal AC dan BD. Panjang \(\textbf{EF} = \dotso\)
Benar
Perpanjang EF hingga memotong BC di titik G.
Perhatikan bahwa \(\color{blue} \triangle \textbf{CEG} \sim \triangle \textbf{CAB}\)
Karena E merupakan titik tengah AC, maka CE : EA = 1 : 1
\(\dfrac{\text{CE}}{\text{CA}} = \dfrac{\text{CG}}{\text{CB}}\)
\(\dfrac{1}{1 + 1} = \dfrac{\text{CG}}{\text{CB}}\)
\(\dfrac{\text{CG}}{\text{CB}} = \dfrac{1}{2}\)
\(\text{CG} : \text{CB} = 1 : 2\)
Artinya \(\text{CG} : \text{GB} = 1 : 1\)
Perhatikan bahwa \(\color{blue} \triangle \textbf{BGF} \sim \triangle \textbf{BCD}\)
\(\dfrac{\text{BG}}{\text{BC}} = \dfrac{\text{FG}}{\text{DC}}\)
\(\dfrac{\text{BG}}{\text{BC}} = \dfrac{\text{FG}}{\text{DC}}\)
\(\dfrac{1}{1 + 1} = \dfrac{\text{FG}}{12}\)
\(\dfrac{1}{2} = \dfrac{\text{FG}}{12}\)
\(\text{FG} = \dfrac{12}{2} = 6 \text{ cm}\)
Perhatikan bahwa \(\color{blue} \triangle \textbf{CEG} \sim \triangle \textbf{CAB}\)
\(\dfrac{\text{CE}}{\text{CA}} = \dfrac{\text{EG}}{\text{AB}}\)
\(\dfrac{1}{1 + 1} = \dfrac{\text{EG}}{18}\)
\(\dfrac{1}{2} = \dfrac{\text{EG}}{18}\)
\(\text{EG} = \dfrac{18}{2} = 9\text{ cm}\)
\(\textbf{EF} = \textbf{EG}\:-\:\textbf{FG}\)
\(\textbf{EF} = 9\:-\:6 = 3 \text{ cm}\)
Jadi, panjang EF = 3 cm.
Salah
Perpanjang EF hingga memotong BC di titik G.
Perhatikan bahwa \(\color{blue} \triangle \textbf{CEG} \sim \triangle \textbf{CAB}\)
Karena E merupakan titik tengah AC, maka CE : EA = 1 : 1
\(\dfrac{\text{CE}}{\text{CA}} = \dfrac{\text{CG}}{\text{CB}}\)
\(\dfrac{1}{1 + 1} = \dfrac{\text{CG}}{\text{CB}}\)
\(\dfrac{\text{CG}}{\text{CB}} = \dfrac{1}{2}\)
\(\text{CG} : \text{CB} = 1 : 2\)
Artinya \(\text{CG} : \text{GB} = 1 : 1\)
Perhatikan bahwa \(\color{blue} \triangle \textbf{BGF} \sim \triangle \textbf{BCD}\)
\(\dfrac{\text{BG}}{\text{BC}} = \dfrac{\text{FG}}{\text{DC}}\)
\(\dfrac{\text{BG}}{\text{BC}} = \dfrac{\text{FG}}{\text{DC}}\)
\(\dfrac{1}{1 + 1} = \dfrac{\text{FG}}{12}\)
\(\dfrac{1}{2} = \dfrac{\text{FG}}{12}\)
\(\text{FG} = \dfrac{12}{2} = 6 \text{ cm}\)
Perhatikan bahwa \(\color{blue} \triangle \textbf{CEG} \sim \triangle \textbf{CAB}\)
\(\dfrac{\text{CE}}{\text{CA}} = \dfrac{\text{EG}}{\text{AB}}\)
\(\dfrac{1}{1 + 1} = \dfrac{\text{EG}}{18}\)
\(\dfrac{1}{2} = \dfrac{\text{EG}}{18}\)
\(\text{EG} = \dfrac{18}{2} = 9\text{ cm}\)
\(\textbf{EF} = \textbf{EG}\:-\:\textbf{FG}\)
\(\textbf{EF} = 9\:-\:6 = 3 \text{ cm}\)
Jadi, panjang EF = 3 cm.
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Pertanyaan 8 dari 8
8. Pertanyaan
1 pointsPerhatikan gambar di bawah ini:
Panjang CF adalah… cm
Benar
Tarik garis sejajar dengan AD melalui titik C, namakan CA’
Tarik juga garis sejajar EF melalui titik B, namakan BF’
Perhatikan bahwa segitiga CK’L dan CA’B sebangun.
\(\dfrac{\text{CL}}{\text{CB}} = \dfrac{\text{K’L}}{\text{A’B}}\)
\(\dfrac{\text{CL}}{\text{CB}} = \dfrac{2}{4}\)
\(\dfrac{\text{CL}}{\text{CB}} = \dfrac{1}{2}\)
Artinya:
\(\text{CL} : \text{LB} = 1 : 1\)
\(\color{blue} \text{BL} : \text{BC} = 1 : 2\)
Perhatikan juga bahwa segitiga BLM’ dan BCF’ sebangun.
\(\dfrac{\text{BL}}{\text{BC}} = \dfrac{\text{LM’}}{\text{CF’}}\)
\(\dfrac{1}{2} = \dfrac{1}{x\:-\:6}\)
Kali silang
\(x\:-\:6 = 2\)
\(x = 2 + 6 = 8\)
Jadi, panjang CF = \(x\) = 8 cm
Salah
Tarik garis sejajar dengan AD melalui titik C, namakan CA’
Tarik juga garis sejajar EF melalui titik B, namakan BF’
Perhatikan bahwa segitiga CK’L dan CA’B sebangun.
\(\dfrac{\text{CL}}{\text{CB}} = \dfrac{\text{K’L}}{\text{A’B}}\)
\(\dfrac{\text{CL}}{\text{CB}} = \dfrac{2}{4}\)
\(\dfrac{\text{CL}}{\text{CB}} = \dfrac{1}{2}\)
Artinya:
\(\text{CL} : \text{LB} = 1 : 1\)
\(\color{blue} \text{BL} : \text{BC} = 1 : 2\)
Perhatikan juga bahwa segitiga BLM’ dan BCF’ sebangun.
\(\dfrac{\text{BL}}{\text{BC}} = \dfrac{\text{LM’}}{\text{CF’}}\)
\(\dfrac{1}{2} = \dfrac{1}{x\:-\:6}\)
Kali silang
\(x\:-\:6 = 2\)
\(x = 2 + 6 = 8\)
Jadi, panjang CF = \(x\) = 8 cm