{"id":7804,"date":"2019-09-17T11:00:00","date_gmt":"2019-09-17T04:00:00","guid":{"rendered":"https:\/\/example.com\/?p=7804"},"modified":"2025-03-13T10:29:10","modified_gmt":"2025-03-13T03:29:10","slug":"latihan-soal-penilaian-tengah-semester-2019-matematika-ipa-kelas-xii","status":"publish","type":"post","link":"https:\/\/staging-blog.sirogu.com\/blog\/latihan-soal-penilaian-tengah-semester-2019-matematika-ipa-kelas-xii","title":{"rendered":"Latihan Soal Penilaian Tengah Semester 2019: Matematika IPA Kelas XII"},"content":{"rendered":"<p style=\"text-align: justify;\"><em><img decoding=\"async\" style=\"width: 820px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/UN_-_Latihan_soal_Penilaian_Tengah_Semester_Matematika_Saintek_kelas_12-01.png\" alt=\"Latihan Soal Penilaian Tengah Semester 2019: Matematika IPA Kelas XII\" width=\"820\" \/><\/em><\/p>\n<p><!--more--><\/p>\n<p style=\"text-align: justify;\"><em>Artikel ini memberikan latihan soal sekaligus pembahasan Penilaian Tengah Semester 2019 mata pelajaran Matematika IPA kelas XII<\/em><\/p>\n<p style=\"text-align: center;\">&#8212;<\/p>\n<p style=\"text-align: justify;\">Tak terasa, kamu sudah berada di pertengahan semester ganjil tahun ajaran 2019\/2020. Penilaian Tengah Semester (PTS) pun tinggal menghitung hari. Kamu sudah menyiapakan apa saja <em>nih<\/em>? Apalagi untuk kamu yang duduk di bangku kelas XII. Nilai kamu di PTS ini sangat penting untuk menghadapi SNMPTN sekaligus bisa menjadi persiapan untuk kamu menghadapi ujian-ujian selanjutnya. Ayo, persiapkan diri hadapi PTS dengan latihan soal Matematika IPA kelas XII berikut ini.<\/p>\n<p style=\"text-align: justify;\"><strong><span style=\"color: #990100;\">Baca juga:<\/span> <a href=\"\/latihan-dan-pembahasan-soal-uts-fisika-sma-kelas-12\" rel=\" noopener\">Latihan dan Pembahasan Soal UTS Fisika SMA Kelas 12<\/a><\/strong><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 820px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design%20(1)-24.png\" alt=\"Untitled design (1)-24\" width=\"820\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>TOPIK: BIDANG DATAR<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>SUBTOPIK: DALIL PROYEKSI &amp; DALIL STEWART<\/strong><\/p>\n<ol style=\"text-align: justify;\">\n<li>Perhatikan gambar bangun datar di bawah ini!<\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 286px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/2-92.png\" alt=\"2-92\" width=\"286\" \/><\/p>\n<p style=\"text-align: justify;\">Jika panjang PQ = PS = 4 dm dan QR = 6 dm maka SR = \u2026.<\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<li>5 dm<\/li>\n<li>6 dm<\/li>\n<li>7 dm<\/li>\n<li>8 dm<\/li>\n<li>10 dm<\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><strong>Jawaban B<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan :<\/strong><\/p>\n<p style=\"text-align: justify;\">Gunakan dalil proyeksi untuk menyelesaikan soal di atas.<\/p>\n<p style=\"text-align: justify;\">Maka,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 162px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/1-405577069094.png\" alt=\"1-405577069094\" width=\"162\" \/><\/p>\n<p style=\"text-align: justify;\">Karena PR menyatakan panjang sisi segitiga maka nilai PR yang memenuhi adalah 10.<\/p>\n<p style=\"text-align: justify;\">Sehingga diperoleh SR = PR \u2013 PS = 10 dm \u2013 4 dm = 6 dm.<\/p>\n<p><img decoding=\"async\" style=\"width: 820px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design%20(1)-24.png\" alt=\"Untitled design (1)-24\" width=\"820\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>TOPIK: BIDANG DATAR<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>SUBTOPIK: GARIS ISTIMEWA SEGITIGA II<\/strong><\/p>\n<p style=\"text-align: justify;\">2. Perhatikan gambar di bawah ini.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 323px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/3-55.png\" alt=\"3-55\" width=\"323\" \/><\/p>\n<p style=\"text-align: justify;\">Diketahui garis CD adalah garis bagi segitiga ABC. Panjang CD adalah \u2026.<\/p>\n<p><img decoding=\"async\" style=\"width: 61px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/4-25.png\" alt=\"4-25\" width=\"61\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>Jawaban B<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan :<\/strong><\/p>\n<p style=\"text-align: justify;\">Karena CD adalah garis bagi segitiga ABC, maka<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 76px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/5-26.png\" alt=\"5-26\" width=\"76\" \/><\/p>\n<p style=\"text-align: justify;\">Misalkan\u00a0<img decoding=\"async\" style=\"width: 147px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/6-23.png\" alt=\"6-23\" width=\"147\" \/><\/p>\n<p style=\"text-align: justify;\">Karena AB = 14 cm, maka<\/p>\n<p style=\"text-align: justify;\">\u00a0<img decoding=\"async\" style=\"width: 105px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/7-22.png\" alt=\"7-22\" width=\"105\" \/><\/p>\n<p style=\"text-align: justify;\">Sehingga BD = 10 cm dan AD =\u00a0<img decoding=\"async\" style=\"width: 94px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/8-22.png\" alt=\"8-22\" width=\"94\" \/><\/p>\n<p style=\"text-align: justify;\">Karena CD adalah garis bagi segitiga ABC, maka<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 177px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/9-26.png\" alt=\"9-26\" width=\"177\" \/><\/p>\n<p style=\"text-align: justify;\">Karena panjang sisi tidak mungkin negatif maka\u00a0<img decoding=\"async\" style=\"width: 101px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/10-20.png\" alt=\"10-20\" width=\"101\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 820px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design%20(1)-24.png\" alt=\"Untitled design (1)-24\" width=\"820\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>TOPIK: BIDANG RUANG (KEDUDUKAN DAN PROYEKSI)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>SUBTOPIK: KEDUDUKAN TITIK<\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"background-color: transparent;\">3. Perhatikan kubus ABCD.EFGH di bawah ini.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"background-color: transparent;\"><img decoding=\"async\" style=\"width: 268px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/11-129.png\" alt=\"11-129\" width=\"268\" \/><\/span><\/p>\n<p style=\"text-align: justify;\">Jika titik P, Q dan R berturut-turut terletak pada rusuk BF, AE dan BC maka pernyataan yang benar adalah \u2026.<\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<li>Titik P berada pada bidang CDG<\/li>\n<li>Titik Q berada pada bidang ABCD<\/li>\n<li>Titik Q dan R sebidang<\/li>\n<li>Titik R berada pada bidang ABD<\/li>\n<li>Titik P, Q dan R sebidang<\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><strong>Jawaban D<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan :<\/strong><\/p>\n<ul style=\"text-align: justify;\">\n<li>Titik P terletak pada bidang ABFE dan bidang BCGF<\/li>\n<li>Titik Q terletak pada bidang ABFE dan bidang ADHE<\/li>\n<li>Titik R terletak pada bidang ABCD dan bidang BCGF<\/li>\n<li>Titik P dan Q sebidang (pada bidang ABFE)<\/li>\n<li>Titik P dan R sebidang (pada bidang BCGF)<\/li>\n<\/ul>\n<p style=\"text-align: justify;\">Jadi dari beberapa pernyataan di atas maka pernyataan yang benar pada pilihan jawaban adalah titik R berada pada bidang ABD, karena jika bidang ABD diperluas maka diperoleh bidang ABCD.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 820px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design%20(1)-24.png\" alt=\"Untitled design (1)-24\" width=\"820\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>TOPIK: BIDANG RUANG (KEDUDUKAN DAN PROYEKSI)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>SUBTOPIK: KEDUDUKAN GARIS TERHADAP BIDANG<\/strong><\/p>\n<ol style=\"text-align: justify;\" start=\"4\">\n<li>Perhatikan kubus ABCD.EFGH di bawah ini.<\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 208px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/12-19.png\" alt=\"12-19\" width=\"208\" \/><\/p>\n<p style=\"text-align: justify;\">Misalkan P adalah titik tengah ED. Bidang yang tegak lurus dengan garis AP adalah bidang \u2026.<\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<li>BCHE<\/li>\n<li>ADGF<\/li>\n<li>CDEF<\/li>\n<li>BDHF<\/li>\n<li>ABGH<\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><strong>Jawaban C<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan :<\/strong><\/p>\n<p style=\"text-align: justify;\">Perhatikan bahwa karena P adalah titik tengah dari diagonal ED maka P juga menjadi titik tengah dari diagonal AH.<\/p>\n<p style=\"text-align: justify;\">Garis AP dapat diperpanjangan menjadi garis AH.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 224px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/13-17.png\" alt=\"13-17\" width=\"224\" \/><\/p>\n<p style=\"text-align: justify;\"><u>Garis AP dan bidang BCHE<\/u><\/p>\n<p style=\"text-align: justify;\">Perhatikan bahwa garis EF terletak pada bidang BCHE. Namun, garis AP tidak tegak lurus garis EH. Sehingga AP tidak tegak lurus bidang BCHE.<\/p>\n<p style=\"text-align: justify;\"><u>Garis AP dengan bidang ADGF<\/u><\/p>\n<p style=\"text-align: justify;\">Perhatikan bahwa garis AD terletak pada bidang ADGF. Namun, garis AP tidak tegak lurus dengan garis AD. Sehingga AP tidak tegak lurus bidang ADGF.<\/p>\n<p style=\"text-align: justify;\"><u>Garis AP dan bidang CDEF<\/u><\/p>\n<p style=\"text-align: justify;\">Perhatikan gambar di bawah ini.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 194px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/14-5577069081.png\" alt=\"14-5577069081\" width=\"194\" \/><\/p>\n<p style=\"text-align: justify;\">Perhatikan bahwa AH tegak lurus ED maka AP juga tegak lurus ED. Bidang yang memuat ED adalah bidang CDEF. Sehingga garis AP tegak lurus bidang CDEF.<\/p>\n<p style=\"text-align: justify;\"><u>Garis AP dan bidang BDHF<\/u><\/p>\n<p style=\"text-align: justify;\">Perhatikan bahwa garis DH berada pada bidang BDHF. Namun, garis AH (sebagai perpanjangan garis AP) tidak tegak lurus garis DH. Sehingga AP tidak tegak lurus bidang BDHF.<\/p>\n<p style=\"text-align: justify;\"><u>Garis AP dan bidang ABGH<\/u><\/p>\n<p style=\"text-align: justify;\">Garis AP tidak tegak lurus bidang ABGH karena gari AP terletak pada bidang ABGH.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 820px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design%20(1)-24.png\" alt=\"Untitled design (1)-24\" width=\"820\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>TOPIK: BIDANG RUANG (KEDUDUKAN DAN PROYEKSI)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>SUBTOPIK: PROYEKSI<\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"background-color: transparent;\">5. Perhatikan kubus ABCD.EFGH berikut ini.<\/span><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 233px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/15-14.png\" alt=\"15-14\" width=\"233\" \/><\/p>\n<p style=\"text-align: justify;\">Proyeksi bidang BFC ke bidang CHE adalah bidang\u2026.<\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<li>BCE<\/li>\n<li>BCD<\/li>\n<li>BCM dengan M titik tengan BE<\/li>\n<li>BCN dengan N titik tengah CH<\/li>\n<li>BCO dengan O titik tengah EH<\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><strong>Jawaban C<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan :<\/strong><\/p>\n<p style=\"text-align: justify;\">Perhatikan gambar di bawah ini.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 233px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/15-14.png\" alt=\"15-14\" width=\"233\" \/><\/p>\n<p style=\"text-align: justify;\">Bidang BFC jika diperluas maka akan menjadi bidang BCHE. Garis potong bidang BCHE dengan bidang BCE adalah garis BC. Kemudian proyeksikan titik F ke bidang BCHE.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 222px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/16-13.png\" alt=\"16-13\" width=\"222\" \/><\/p>\n<p style=\"text-align: justify;\">Misalkan titik M adalah titik potong diagonal sisi BE dan diagonal sisi AF. Maka titik M berada tepat pada pertengahan garis BE.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 209px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/17-11.png\" alt=\"17-11\" width=\"209\" \/><\/p>\n<p style=\"text-align: justify;\">Ingat kembali bahwa diagonal sisi pada kubus saling tegak lurus. Sehingga, proyeksi titik F ke bidang BCHE sama dengan proyeksi titik F ke garis BE, yaitu titik M.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 210px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/18-11.png\" alt=\"18-11\" width=\"210\" \/><\/p>\n<p style=\"text-align: justify;\">Jadi proyeksi bidang BCF ke bidang BCHE adalah bidang BCM dimana M adalah titik tengah BE.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 820px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design%20(1)-24.png\" alt=\"Untitled design (1)-24\" width=\"820\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>TOPIK: BIDANG RUANG (JARAK)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>SUBTOPIK: JARAK TITIK KE BIDANG<\/strong><\/p>\n<p style=\"text-align: justify;\">6. Diketahui sebuah kubus ABCD.EFGH dengan rusuk 6 cm. Jarak titik E ke bidang ACH adalah \u2026. cm.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 223px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/19-10.png\" alt=\"19-10\" width=\"223\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 78px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/20-9.png\" alt=\"20-9\" width=\"78\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>JAWABAN : B<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>PEMBAHASAN :<\/strong><\/p>\n<p style=\"text-align: justify;\">Jika kita tambahkan satu buah kubus lagi disebelah kiri kubus ABCD.EFGH kemudian perluas bidang ACH seperti pada gambar berikut ini.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 340px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/21-8.png\" alt=\"21-8\" width=\"340\" \/><\/p>\n<p style=\"text-align: justify;\">Sehingga jarak titik E ke bidang ACH sama dengan jarak titik E ke garis AP\u2019, atau sama juga dengan jarak titik D ke titik O.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 217px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/22-9.png\" alt=\"22-9\" width=\"217\" \/><\/p>\n<p style=\"text-align: justify;\">Sehingga dengan persamaan luas segitiga DHP diperoleh :<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 196px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/23-6.png\" alt=\"23-6\" width=\"196\" \/><\/p>\n<p style=\"text-align: justify;\">Jadi Jarak titik E ke bidang ACH adalah\u00a0\u00a0<img decoding=\"async\" style=\"width: 55px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/24-5.png\" alt=\"24-5\" width=\"55\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 820px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design%20(1)-24.png\" alt=\"Untitled design (1)-24\" width=\"820\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>TOPIK: BIDANG RUANG (JARAK)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>SUBTOPIK: JARAK DUA BIDANG SEJAJAR<\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"background-color: transparent;\">7. Diketahui kubus ABCD.EFGH dengan panjang rusuk 10 cm. Jarak bidang ACH dengan bidang BEG adalah \u2026. cm<\/span><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 234px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/25-5.png\" alt=\"25-5\" width=\"234\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 67px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/26-4.png\" alt=\"26-4\" width=\"67\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>JAWABAN : A<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>PEMBAHASAN :<\/strong><\/p>\n<p style=\"text-align: justify;\">Untuk memudahkan buat bidang diagonal BDHF dengan titik M dan titik N adalah titik tengah perpotongan diagonal atap dan alas seperti gambar berikut.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 226px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/27-4.png\" alt=\"27-4\" width=\"226\" \/><\/p>\n<p style=\"text-align: justify;\">Bidang BEG berptongan dengan bidang BDHF di garis BM. Sedangkan bidang ACH berpotongan dengan bidang BDHF di garis HN.<\/p>\n<p style=\"text-align: justify;\">Jika kita buat segitiga BMN pada bidang diagonal BDHF maka jarak bidang ACH ke bidang BEG sama dengan jarak titik N ke garis BM atau sama dengan NO.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 200px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/28-4.png\" alt=\"28-4\" width=\"200\" \/><\/p>\n<p style=\"text-align: justify;\">Dengan persamaan luas segitiga BMN maka diperoleh :<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 180px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/29-3.png\" alt=\"29-3\" width=\"180\" \/><\/p>\n<p style=\"text-align: justify;\">Jadi jarak bidang ACH ke bidang BEG adalah\u00a0<img decoding=\"async\" style=\"width: 69px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/30-2.png\" alt=\"30-2\" width=\"69\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 820px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design%20(1)-24.png\" alt=\"Untitled design (1)-24\" width=\"820\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>TOPIK: BIDANG RUANG (SUDUT)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>SUBTOPIK: SUDUT DUA GARIS<\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"background-color: transparent;\">8. Diketahui kubus ABCH.EFGH dengan rusuk 4 cm. Titik T membagi dua rusuk BC menjadi dua bagian. Jika <img decoding=\"async\" style=\"width: 17px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/31-2.png\" alt=\"31-2\" width=\"17\" \/>adalah sudut antara garis HT dengan garis BG maka\u00a0<img decoding=\"async\" style=\"width: 82px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/32-1.png\" alt=\"32-1\" width=\"82\" \/><\/span><\/p>\n<p><img decoding=\"async\" style=\"width: 64px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/33-4.png\" alt=\"33-4\" width=\"64\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>Jawaban <\/strong><strong>C<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan :<\/strong><\/p>\n<p style=\"text-align: justify;\">Perhatikan kubus di bawah :<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 317px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/34-1.png\" alt=\"34-1\" width=\"317\" \/><\/p>\n<p style=\"text-align: justify;\">Kemudian pada segitiga ATH, misalkan R adalah titik pada AH sehingga garis TR tegak lurus garis AH.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 174px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/35-1.png\" alt=\"35-1\" width=\"174\" \/><\/p>\n<p style=\"text-align: justify;\">Dengan aturan cos maka diperoleh,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 247px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/36-1.png\" alt=\"36-1\" width=\"247\" \/><\/p>\n<p style=\"text-align: justify;\">Kemudian hitung nilai sin <img decoding=\"async\" style=\"width: 17px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/31-2.png\" alt=\"31-2\" width=\"17\" \/>\u00a0dengan rumus identitas trigonometri berikut :<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 390px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/37-1.png\" alt=\"37-1\" width=\"390\" \/><\/p>\n<p style=\"text-align: justify;\">Jadi nilai dari\u00a0\u00a0<img decoding=\"async\" style=\"width: 157px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/38-1.png\" alt=\"38-1\" width=\"157\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 820px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design%20(1)-24.png\" alt=\"Untitled design (1)-24\" width=\"820\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>TOPIK: BIDANG RUANG (SUDUT)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>SUBTOPIK: SUDUT GARIS DAN BIDANG<\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"background-color: transparent;\">9. Diketahui sebuah kubus ABCD.EFGH dengan panjang sisi kubus <img decoding=\"async\" style=\"width: 33px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/39-1.png\" alt=\"39-1\" width=\"33\" \/>. Sinus sudut antara rusuk BF ke bidang BEG adalah\u2026.<\/span><\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" style=\"width: 68px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/40-1.png\" alt=\"40-1\" width=\"68\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Jawaban B<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan :<\/strong><\/p>\n<p style=\"text-align: justify;\">Perhatikan gambar di bawah ini.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 209px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/41-1.png\" alt=\"41-1\" width=\"209\" \/><\/p>\n<p style=\"text-align: justify;\">Diketahui :<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 77px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/42-1.png\" alt=\"42-1\" width=\"77\" \/><\/p>\n<p style=\"text-align: justify;\">Sehingga,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 426px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/43-1.png\" alt=\"43-1\" width=\"426\" \/><\/p>\n<p style=\"text-align: justify;\">Maka,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 269px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/44-1.png\" alt=\"44-1\" width=\"269\" \/><\/p>\n<p style=\"text-align: justify;\">Sinus sudut antara rusuk BF ke bidang BEG adalah\u00a0<img decoding=\"async\" style=\"width: 35px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/45-1.png\" alt=\"45-1\" width=\"35\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 820px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design%20(1)-24.png\" alt=\"Untitled design (1)-24\" width=\"820\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>TOPIK: BIDANG RUANG (SUDUT)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>SUBTOPIK: IRISAN BIDANG<\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"background-color: transparent;\">10. Diketahui kubus ABCD.EFGH dengan titik M adalah titik tengah rusuk CG. Irisan bidang yang melalu titik M, D dan F adalah \u2026.<\/span><\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<li><span style=\"background-color: transparent;\">Segitiga<\/span><\/li>\n<li><span style=\"background-color: transparent;\">Segitiga sama sisi<\/span><\/li>\n<li><span style=\"background-color: transparent;\">Segiempat<\/span><\/li>\n<li><span style=\"background-color: transparent;\">Segilima<\/span><\/li>\n<li><span style=\"background-color: transparent;\">Persegi<\/span><\/li>\n<\/ol>\n<p style=\"text-align: justify;\"><strong>Jawaban C<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Pembahasan :<\/strong><\/p>\n<p style=\"text-align: justify;\">Perhatikan gambar di bawah ini.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 236px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/46-1.png\" alt=\"46-1\" width=\"236\" \/><\/p>\n<p style=\"text-align: justify;\">Karena titik D terletak pada rusuk yang bersebrangan dengan titik F maka titik M akan berseberangan pula dengan titik lain di rusuk AE. Misalkan titik N adalah titik tengah rusuk AE.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 241px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/47-1.png\" alt=\"47-1\" width=\"241\" \/><\/p>\n<p style=\"text-align: justify;\">Sehingga bidang yang terbentuk adalah bidang segiempat NDFM seperti pada gambar di atas.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"width: 820px;\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Untitled%20design%20(1)-24.png\" alt=\"Untitled design (1)-24\" width=\"820\" \/><\/p>\n<p style=\"text-align: justify;\">Gimana, membantu <em>&#8216;kan<\/em>? Mau latihan lebih banyak soal? Yuk, latihan soal-soal terkait materi ujian sekolah di <a href=\"https:\/\/ruangguru.onelink.me\/blPk\/eaff0eb9\" rel=\" noopener\"><strong>ruangbelajar<\/strong><\/a>.<\/p>\n<p style=\"text-align: justify;\"><a class=\"rg-cta\" style=\"text-align: center;\" href=\"https:\/\/link.ruangguru.com\/blPk\/eaff0eb9\" target=\"_blank\" rel=\"noopener\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/cta\/0ec4a2df-db2e-4ae1-8473-692756dcb684.jpeg\" alt=\"IDN CTA Blog ruangbelajar Ruangguru\" width=\"822\" height=\"200\" \/><\/a><\/p>\n","protected":false},"excerpt":{"rendered":"","protected":false},"author":22,"featured_media":7804,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_knawatfibu_url":["https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/UN_-_Latihan_soal_Penilaian_Tengah_Semester_Matematika_Saintek_kelas_12-01.png"],"_edit_last":["1"],"_edit_lock":["1741836408:1"],"_yoast_wpseo_primary_category":["552"],"_yoast_wpseo_content_score":["60"],"_yoast_wpseo_estimated-reading-time-minutes":["18"]},"categories":[552,561],"tags":[76,10,172,37],"class_list":["post-7804","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-latihan-soal","category-sma-kelas-12","tag-kelas-12","tag-konsep-pelajaran","tag-latihan-soal-xii","tag-sma"],"aioseo_notices":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v25.9 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Latihan Soal Penilaian Tengah Semester 2019: Matematika IPA Kelas XII - Belajar Gratis di Rumah Kapan Pun! 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