{"id":2356,"date":"2023-05-31T13:30:18","date_gmt":"2023-05-31T13:30:18","guid":{"rendered":"https:\/\/pikiranaceh.com\/index.php\/2023\/05\/31\/kunci-jawaban-matematika-kelas-8-halaman-319-320-321-soal-uji-kompetensi-semester-2\/"},"modified":"2023-05-31T13:30:18","modified_gmt":"2023-05-31T13:30:18","slug":"kunci-jawaban-matematika-kelas-8-halaman-319-320-321-soal-uji-kompetensi-semester-2","status":"publish","type":"post","link":"https:\/\/pikiranaceh.com\/index.php\/2023\/05\/31\/kunci-jawaban-matematika-kelas-8-halaman-319-320-321-soal-uji-kompetensi-semester-2\/","title":{"rendered":"Kunci Jawaban Matematika Kelas 8 Halaman 319 320 321 Soal Uji Kompetensi Semester 2"},"content":{"rendered":"

<\/strong><\/p>\n

PIKIRANACEH.COM<\/strong> – Hallo siswa! Yuk pelajari kunci jawaban Matematika kelas 8 halaman 319 320 321 pada Semester 2 di buku Kurikulum 2013.<\/p>\n

Pembahasan kunci jawaban Matematika kelas 8 halaman 319 320 321 ini dibuat sebagai bahan acuan siswa untuk dapat belajar secara mandiri.<\/p>\n

Baca Juga: Soal dan Kunci Jawaban PAI Kelas 11 Halaman 116-117<\/a><\/strong><\/p>\n

Sebelum menyimak kunci jawaban Matematika kelas 8 halaman 319 320 321 ini, diharapkan siswa untuk berusaha mengerjakannya terlebih dahulu.<\/p>\n

Berikut ini, simak baik-baik pembahasan kunci jawaban Matematika kelas 8 halaman 319 320 321 Uji Kompetensi Semester 2 soal bagian esai.<\/p>\n

1. Keliling Segitiga<\/p>\n

Berdasarkan gambar, segitiga ABC adalah segitiga siku-siku, dengan sudut siku-siku di B. Panjang BD = 6 m dan panjang AB = 8m.<\/p>\n

Jika luas segitiga ADC 50% lebih besar dari luas segitiga ABD, tentukan keliling segitiga ADC.<\/p>\n

Jawaban:
AD = (AB² + BD²) = √8² + 6²) = √100 = 10cm<\/p>\n

L\u2206ABD = 1\/2 x a x t = 1\/2 x 6 x 8 = 24m²<\/p>\n

L\u2206ADC = 150% x L\u2206ABD = 150\/100 x 24 = 36m²<\/p>\n

L.ABC = L.ABD + L.ADC = 24 + 36 = 60m²<\/p>\n

L\u2206ABC = 1\/2 x a x t
60 = 1\/2 x BC x 8
BC = (60 x 2)\/8 = 15m<\/p>\n

CD = BC – BD = 15 – 6 = 9m<\/p>\n

AC = (AB² + BC²) = (8² + 15²) = (289) = 17m<\/p>\n

Jadi, K = AD + CD + AC = 10 + 9 + 17 = 36m<\/p>\n

2. Amati gambar di samping. Tentukan keliling dan luas daerah yang diarsir. (pi = 3, 14).<\/p>\n

Jawaban:
L daerah = L.persegi + L.1\/4O + L.1\/4O
L daerah = L.persegi + L.1\/20
L daerah = (sxs) + 1\/2 x π x r²
L daerah = (10 x 10)+ 1\/2 x (3,14) x 5²
L daerah = 100 + 39.25
L daerah = 139.25 cm²<\/p>\n

K daerah = r + r + ¼k.o + 1½ s + s + s + ½s+ ¼ k.o
K daerah = 2r + ½.k + 3s
K daerah = 2r + 1\/2 x 2πr + 3s
K daerah = 2r + ½ x 2πr + 3s
K daerah = (2 x 5) + (3,14 x 5) + (3 x 10)
K daerah = 10 + 15,7 + 30
K daerah = 55,7 cm<\/p>\n

Baca Juga: Kunci Jawaban PAI Kelas 12 Halaman 57-58<\/a><\/strong><\/p>\n

3. Dalam model ini, lantai loteng ABCD berbentuk persegi. Tiang yang menopang atap merupakan rusuk balok EFGH.KLMN.<\/p>\n

Titik E terletak di tengah AT, titik F di tengah BT, titik G di tengah CT, dan titik H di tengah DT. Semua rusuk piramid pada model tersebut panjangnya 12 m.<\/p>\n

a. Hitunglah luas lantai loteng ABCD.<\/p>\n

Jawaban:
Lantai ABCD = s x s
Lantai ABCD = 12 x 12
Lantai ABCD = 144 m²<\/p>\n

b. Hitunglah panjang EF.<\/p>\n

Jawaban:
\u2206ABT – \u2206EFT
EF\/AB = ET\/AT
EF\/12 = 6\/12
EF = 12 x 6 \/ 12
EF = 6 m<\/p>\n

Baca Juga: Kunci Jawaban PKN Kelas 9 Halaman 147 Uji Kompetensi 6<\/a><\/strong><\/p>\n

4. Kecepatan Mobil Balap<\/p>\n

Grafik berikut menunjukkan kecepatan mobil balap sepanjang lintasan 3 km selama putaran kedua.<\/p>\n

a. Pada jarak berapakah dari garis start yang menunjukkan awal lintasan lurus terpanjang?<\/p>\n

A. 0,5 km
B. 1,5 km
C. 2,3 km
D. 2,6 km<\/p>\n

Jawaban: B. 1,5 km<\/p>\n

b. Dimanakah posisi kecepatan terendah yang terekam selama putaran (lap) kedua?<\/p>\n

A. Pada garis start
B. Sekitar 0.8 km
C. Sekitar 1.3 km
D. Pertengahan jarak sepanjang lintasan<\/p>\n

Jawaban: C. Sekitar 0.8 km<\/p>\n

c. Menurut kalian apa yang terjadi pada kecepatan mobil yang berada pada jarak diantara 2.6 km sampai 2.8 km?<\/p>\n

A. Kecepatan mobil konstan\/tetap
B. Kecepatan mobil bertambah
C. Kecepatan mobil berkurang
D. Kecepatan mobil tidak dapat ditentukan<\/p>\n

Jawaban: B. Kecepatan mobil bertambah<\/p>\n

5. Ketika berjalan-jalan di berjalan-jalan di sebuah mall, Gunarso mendapatkan keberuntungan sebagai pengunjung mall terpilih pada hari itu.<\/p>\n

Gunarso berkesempatan memilih 1 hadiah dari 3 kotak yang sudah disediakan panitia mall.<\/p>\n

Setiap kotak tersebut berisi bola berwarna merah mewakili mobil, kuning mewakili motor, dan hijau mewakili TV dengan komposisi sebagai berikut.<\/p>\n

Gunarso hanya diberi kesempatan untuk mengambil 1 hadiah dari salah satu kotak. Tentukan kotak mana yang memiliki peluang terbesar mendapatkan mobil. Jelaskan.<\/p>\n

Jawaban:
A. P(mobil kotak A) = n(mobil kotak A) \/ n(S)
P(mobil kotak A) = 7 \/ 7+9+10
P(mobil kotak A) = 7\/26
P(mobil kotak A) = 0,269<\/p>\n

B. P(mobil kotak B) = n(mobil kotak B) \/ n(S)
P(mobil kotak B) = 8 \/ 8+12+8
P(mobil kotak B) = 8\/28
P(mobil kotak B) = 0,285<\/p>\n

C. P(mobil kotak C) = n(mobil kotak C) \/ n(S)
P(mobil kotak C) = 9 \/ 9+12+9
P(mobil kotak C) = 9\/30
P(mobil kotak C) = 0,3<\/p>\n

Jadi, peluang terbesar adalah kotak C.<\/p>\n

Perlu dicatat, artikel ini tidak mutlak menjamin kebenaran seluruh jawaban. Oleh sebab itu, siswa dapat mengeksplorasi jawaban-jawaban yang tentu bisa lebih baik lagi.<\/p>\n

Baca Juga: Kunci Jawaban Matematika Kelas 8 Halaman 306 307 Uji Kompetensi 10<\/a><\/strong><\/p>\n

Demikian pembahasan kunci jawaban matematika kelas 8 halaman 319 320 321 soal Uji Kompetensi Semester 2 bagian esai.***<\/p>\n","protected":false},"excerpt":{"rendered":"

PIKIRANACEH.COM – Hallo siswa! 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