{"id":2159,"date":"2026-04-03T19:29:14","date_gmt":"2026-04-03T19:29:14","guid":{"rendered":"https:\/\/rumahpintar.co.id\/blog\/?p=2159"},"modified":"2026-04-09T04:34:28","modified_gmt":"2026-04-09T04:34:28","slug":"15-contoh-soal-jaring-jaring-bangun-ruang-prisma-tabung-limas-kerucut-lengkap-dengan-pembahasannya","status":"publish","type":"post","link":"https:\/\/rumahpintar.co.id\/blog\/15-contoh-soal-jaring-jaring-bangun-ruang-prisma-tabung-limas-kerucut-lengkap-dengan-pembahasannya\/","title":{"rendered":"15 Contoh Soal Jaring-Jaring Bangun Ruang Prisma, Tabung, Limas, Kerucut, Lengkap dengan Pembahasannya"},"content":{"rendered":"

Bimbel Rumah Pintar<\/a> – <\/span><\/strong>Prinsip dari soal j<\/span>aring-jaring bangun ruang adalah rangkaian bangun datar yang jika dilipat pada garis-garis tertentu akan membentuk bangun ruang tiga dimensi.<\/p>\n

Prinsip dasar yang harus dikuasai adalah setiap bangun ruang memiliki komponen penyusun yang khas. Misalnya saja prisma<\/strong> memiliki dua bangun datar identik sebagai alas dan tutup serta persegi panjang sebagai sisi tegak. Adapun limas<\/strong> memiliki satu alas dan segitiga-segitiga sebagai sisi tegak.<\/p>\n

Lain halnya dengan tabung <\/strong>yang\u00a0memiliki dua lingkaran dan satu persegi panjang (selimut). Sementara kerucut<\/strong>\u00a0memiliki satu lingkaran dan satu juring lingkaran (selimut). Kunci keberhasilan terletak pada kemampuan membayangkan proses melipat, mengidentifikasi pasangan sisi yang akan bertemu, serta menghitung luas jaring-jaring yang sama dengan luas permukaan bangun ruang.<\/p>\n

Mengapa latihan soal jaring-jaring bangun ruang sangat penting?<\/strong> Soal TKA Matematika SMP sering menyajikan masalah kontekstual seperti menentukan jumlah maksimum kemasan yang dapat dibuat dari selembar karton dengan ukuran tertentu, mengidentifikasi jaring-jaring yang benar dari beberapa pilihan, atau menghitung luas minimal karton yang dibutuhkan untuk membuat suatu kemasan.<\/p>\n

\n

15 Contoh Soal Jaring-Jaring Bangun Ruang dan Pembahasannya<\/h2>\n
\n

Soal 1 (Optimalisasi Kemasan dari Karton \u2013 Tipe TKA)<\/h3>\n

Sebuah usaha rumahan mengemas produk makanan menggunakan kemasan berbentuk prisma segitiga yang dibuat dari karton. Setiap kemasan dibuat dari satu jaring-jaring utuh. Karton yang tersedia berukuran 50 cm \u00d7 100 cm. Pemotongan karton dilakukan dengan rapi dan memperhatikan tata letak. Jaring-jaring kemasan dibuat dengan kertas yang saling terhubung dan bukan menggabungkan hasil potongan lain. Jika luas satu jaring-jaring adalah 625 cm\u00b2, berapa kemasan maksimum yang dapat dibuat?<\/p>\n

Pembahasan:<\/strong><\/p>\n

    \n
  • \n

    Luas karton =\u00a050\u00d7100=5.000<\/span>50<\/span>\u00d7<\/span><\/span>100<\/span>=<\/span><\/span>5.000<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

  • \n

    Perkiraan maksimum =\u00a05.000\u00f7625=8<\/span>5.000<\/span>\u00f7<\/span><\/span>625<\/span>=<\/span><\/span>8<\/span><\/span><\/span><\/span>\u00a0kemasan<\/p>\n<\/li>\n

  • \n

    Dengan tata letak yang rapi (jaring-jaring berbentuk persegi panjang 25 cm \u00d7 25 cm misalnya), dapat disusun:<\/p>\n

      \n
    • \n

      Arah 50 cm: muat 2 \u00d7 25 cm = 50 cm (pas)<\/p>\n<\/li>\n

    • \n

      Arah 100 cm: muat 4 \u00d7 25 cm = 100 cm (pas)<\/p>\n<\/li>\n<\/ul>\n<\/li>\n

    • \n

      Total =\u00a02\u00d74=8<\/span>2<\/span>\u00d7<\/span><\/span>4<\/span>=<\/span><\/span>8<\/span><\/span><\/span><\/span>\u00a0kemasan<\/p>\n<\/li>\n<\/ul>\n

      Jawaban: 8 kemasan<\/strong><\/p>\n

      \n

      Soal 2 (Mengidentifikasi Jaring-Jaring Kubus yang Benar)<\/h3>\n

      Manakah susunan berikut yang merupakan jaring-jaring kubus yang benar? (Pilihlah yang mungkin)
      \nA. 6 persegi berjajar dalam satu baris
      \nB. 6 persegi berbentuk salib (tengah dikelilingi 4, plus 1 di salah satu sisi)
      \nC. 5 persegi saja
      \nD. 6 persegi dengan 2 persegi berdempetan di luar<\/p>\n

      Pembahasan:<\/strong><\/p>\n

        \n
      • \n

        A: 6 persegi berjajar \u2192 saat dilipat, persegi ujung tidak akan bertemu dengan tepat \u2192\u00a0BUKAN<\/strong><\/p>\n<\/li>\n

      • \n

        B: Bentuk salib (tengah dikelilingi 4, plus 1 di salah satu sisi) \u2192 dapat dilipat menjadi kubus \u2192\u00a0BENAR<\/strong><\/p>\n<\/li>\n

      • \n

        C: Hanya 5 persegi \u2192 kurang 1 sisi \u2192\u00a0BUKAN<\/strong><\/p>\n<\/li>\n

      • \n

        D: Tergantung susunan, beberapa bentuk dengan 2 persegi di luar bisa valid \u2192\u00a0MUNGKIN<\/strong><\/p>\n<\/li>\n<\/ul>\n

        Jawaban: B (dan beberapa kemungkinan D tergantung susunan)<\/strong><\/p>\n

        \n

        Soal 3 (Luas Jaring-Jaring Balok)<\/h3>\n

        Sebuah balok berukuran panjang 8 cm, lebar 5 cm, tinggi 4 cm. Berapa luas jaring-jaringnya (luas permukaan balok)?<\/p>\n

        Pembahasan:<\/strong><\/p>\n

          \n
        • \n

          Luas permukaan balok =\u00a02(pl+pt+lt)<\/span>2<\/span>(<\/span>pl<\/span>+<\/span><\/span>pt<\/span>+<\/span><\/span>lt<\/span>)<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

        • \n

          =\u00a02(8\u00d75+8\u00d74+5\u00d74)<\/span>2<\/span>(<\/span>8<\/span>\u00d7<\/span><\/span>5<\/span>+<\/span><\/span>8<\/span>\u00d7<\/span><\/span>4<\/span>+<\/span><\/span>5<\/span>\u00d7<\/span><\/span>4<\/span>)<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

        • \n

          =\u00a02(40+32+20)=2\u00d792=184<\/span>2<\/span>(<\/span>40<\/span>+<\/span><\/span>32<\/span>+<\/span><\/span>20<\/span>)<\/span>=<\/span><\/span>2<\/span>\u00d7<\/span><\/span>92<\/span>=<\/span><\/span>184<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n<\/ul>\n

          Jawaban: 184 cm\u00b2<\/strong><\/p>\n

          \n

          Soal 4 (Ukuran Selimut Tabung pada Jaring-Jaring)<\/h3>\n

          Sebuah tabung memiliki jari-jari 10,5 cm dan tinggi 15 cm. Tentukan ukuran persegi panjang yang menjadi selimut tabung dalam jaring-jaringnya! (\u03c0=227<\/span>\u03c0<\/span>=<\/span><\/span>7<\/span><\/span>22<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>)<\/p>\n

          Pembahasan:<\/strong><\/p>\n

            \n
          • \n

            Panjang selimut = keliling lingkaran alas =\u00a02\u03c0r=2\u00d7227\u00d710,5=2\u00d722\u00d71,5=66<\/span>2<\/span>\u03c0<\/span>r<\/span>=<\/span><\/span>2<\/span>\u00d7<\/span><\/span>7<\/span><\/span>22<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>10<\/span>,<\/span>5<\/span>=<\/span><\/span>2<\/span>\u00d7<\/span><\/span>22<\/span>\u00d7<\/span><\/span>1<\/span>,<\/span>5<\/span>=<\/span><\/span>66<\/span><\/span><\/span><\/span>\u00a0cm<\/p>\n<\/li>\n

          • \n

            Lebar selimut = tinggi tabung = 15 cm<\/p>\n<\/li>\n<\/ul>\n

            Jawaban: 66 cm \u00d7 15 cm<\/strong><\/p>\n

            \n

            Soal 5 (Luas Jaring-Jaring Limas Segi Empat)<\/h3>\n

            Sebuah limas segi empat memiliki alas persegi dengan sisi 10 cm dan tinggi segitiga sisi tegak 12 cm. Berapa luas jaring-jaringnya?<\/p>\n

            Pembahasan:<\/strong><\/p>\n

              \n
            • \n

              Luas alas =\u00a010\u00d710=100<\/span>10<\/span>\u00d7<\/span><\/span>10<\/span>=<\/span><\/span>100<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

            • \n

              Luas satu sisi tegak =\u00a012\u00d710\u00d712=60<\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>10<\/span>\u00d7<\/span><\/span>12<\/span>=<\/span><\/span>60<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

            • \n

              Luas 4 sisi tegak =\u00a04\u00d760=240<\/span>4<\/span>\u00d7<\/span><\/span>60<\/span>=<\/span><\/span>240<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

            • \n

              Luas total jaring-jaring =\u00a0100+240=340<\/span>100<\/span>+<\/span><\/span>240<\/span>=<\/span><\/span>340<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n<\/ul>\n

              Jawaban: 340 cm\u00b2<\/strong><\/p>\n

              \n

              Soal 6 (Jaring-Jaring Kerucut \u2013 Luas dan Ukuran)<\/h3>\n

              Sebuah kerucut memiliki jari-jari alas 5 cm dan garis pelukis 13 cm. Berapa luas jaring-jaringnya? (\u03c0=3,14<\/span>\u03c0<\/span>=<\/span><\/span>3<\/span>,<\/span>14<\/span><\/span><\/span><\/span>)<\/p>\n

              Pembahasan:<\/strong><\/p>\n

                \n
              • \n

                Luas alas =\u00a0\u03c0r2=3,14\u00d752=3,14\u00d725=78,5<\/span>\u03c0<\/span>r<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>3<\/span>,<\/span>14<\/span>\u00d7<\/span><\/span>52<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>3<\/span>,<\/span>14<\/span>\u00d7<\/span><\/span>25<\/span>=<\/span><\/span>78<\/span>,<\/span>5<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

              • \n

                Luas selimut =\u00a0\u03c0rs=3,14\u00d75\u00d713=3,14\u00d765=204,1<\/span>\u03c0<\/span>rs<\/span>=<\/span><\/span>3<\/span>,<\/span>14<\/span>\u00d7<\/span><\/span>5<\/span>\u00d7<\/span><\/span>13<\/span>=<\/span><\/span>3<\/span>,<\/span>14<\/span>\u00d7<\/span><\/span>65<\/span>=<\/span><\/span>204<\/span>,<\/span>1<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

              • \n

                Luas total =\u00a078,5+204,1=282,6<\/span>78<\/span>,<\/span>5<\/span>+<\/span><\/span>204<\/span>,<\/span>1<\/span>=<\/span><\/span>282<\/span>,<\/span>6<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n<\/ul>\n

                Jawaban: 282,6 cm\u00b2<\/strong><\/p>\n

                \n

                Soal 7 (Optimalisasi Karton untuk Kemasan Balok)<\/h3>\n

                Tersedia karton berukuran 40 cm \u00d7 60 cm. Sebuah jaring-jaring balok memiliki luas 150 cm\u00b2 dengan ukuran persegi panjang penyusun 10 cm \u00d7 15 cm. Berapa maksimum kemasan yang dapat dibuat?<\/p>\n

                Pembahasan:<\/strong><\/p>\n

                  \n
                • \n

                  Jaring-jaring balok (dengan asumsi bentuk persegi panjang 15 cm \u00d7 10 cm = 150 cm\u00b2)<\/p>\n<\/li>\n

                • \n

                  Arah 40 cm: muat\u00a040\u00f715=2<\/span>40<\/span>\u00f7<\/span><\/span>15<\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span>\u00a0buah (sisa 10 cm) atau\u00a040\u00f710=4<\/span>40<\/span>\u00f7<\/span><\/span>10<\/span>=<\/span><\/span>4<\/span><\/span><\/span><\/span>\u00a0buah<\/p>\n<\/li>\n

                • \n

                  Arah 60 cm: muat\u00a060\u00f710=6<\/span>60<\/span>\u00f7<\/span><\/span>10<\/span>=<\/span><\/span>6<\/span><\/span><\/span><\/span>\u00a0buah atau\u00a060\u00f715=4<\/span>60<\/span>\u00f7<\/span><\/span>15<\/span>=<\/span><\/span>4<\/span><\/span><\/span><\/span>\u00a0buah<\/p>\n<\/li>\n

                • \n

                  Susunan terbaik: 2 \u00d7 6 = 12 kemasan (dengan orientasi 15 cm di sisi 40 cm? 2\u00d715=30, sisa 10 cm; 6\u00d710=60 pas)<\/p>\n<\/li>\n

                • \n

                  Cek: 2 \u00d7 15 = 30 cm \u2264 40 cm (sisa 10 cm), 6 \u00d7 10 = 60 cm \u2264 60 cm (pas) \u2192\u00a012 kemasan<\/strong><\/p>\n<\/li>\n<\/ul>\n

                  Jawaban: 12 kemasan<\/strong><\/p>\n

                  \n

                  Soal 8 (Menentukan Validitas Jaring-Jaring Tabung)<\/h3>\n

                  Manakah yang merupakan jaring-jaring tabung?
                  \nA. 2 lingkaran + 1 persegi panjang
                  \nB. 2 lingkaran + 2 persegi panjang
                  \nC. 1 lingkaran + 1 persegi panjang
                  \nD. 2 lingkaran + 1 segitiga<\/p>\n

                  Pembahasan:<\/strong><\/p>\n

                    \n
                  • \n

                    Tabung memiliki 2 lingkaran (alas dan tutup) dan 1 persegi panjang (selimut)<\/p>\n<\/li>\n

                  • \n

                    Maka jawaban yang benar adalah\u00a0A<\/strong><\/p>\n<\/li>\n<\/ul>\n

                    Jawaban: A<\/strong><\/p>\n

                    \n

                    Soal 9 (Luas Minimal Karton untuk Prisma Segitiga)<\/h3>\n

                    Sebuah kemasan berbentuk prisma segitiga dengan alas segitiga siku-siku berukuran 6 cm, 8 cm, 10 cm dan tinggi prisma 12 cm. Berapa luas minimal karton yang dibutuhkan untuk membuat satu kemasan?<\/p>\n

                    Pembahasan:<\/strong><\/p>\n

                      \n
                    • \n

                      Luas alas =\u00a012\u00d76\u00d78=24<\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>6<\/span>\u00d7<\/span><\/span>8<\/span>=<\/span><\/span>24<\/span><\/span><\/span><\/span>\u00a0cm\u00b2 (2 buah = 48 cm\u00b2)<\/p>\n<\/li>\n

                    • \n

                      Luas sisi tegak:<\/p>\n

                        \n
                      • \n

                        Sisi 1:\u00a06\u00d712=72<\/span>6<\/span>\u00d7<\/span><\/span>12<\/span>=<\/span><\/span>72<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

                      • \n

                        Sisi 2:\u00a08\u00d712=96<\/span>8<\/span>\u00d7<\/span><\/span>12<\/span>=<\/span><\/span>96<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

                      • \n

                        Sisi 3:\u00a010\u00d712=120<\/span>10<\/span>\u00d7<\/span><\/span>12<\/span>=<\/span><\/span>120<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n<\/ul>\n<\/li>\n

                      • \n

                        Luas total =\u00a048+72+96+120=336<\/span>48<\/span>+<\/span><\/span>72<\/span>+<\/span><\/span>96<\/span>+<\/span><\/span>120<\/span>=<\/span><\/span>336<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n<\/ul>\n

                        Jawaban: 336 cm\u00b2<\/strong><\/p>\n

                        \n

                        Soal 10 (Maksimum Kubus dari Karton Persegi)<\/h3>\n

                        Karton berukuran 40 cm \u00d7 40 cm. Jaring-jaring kubus dengan sisi 8 cm. Berapa maksimum kubus yang dapat dibuat? (Asumsikan jaring-jaring kubus berbentuk persegi panjang 24 cm \u00d7 16 cm = 384 cm\u00b2)<\/p>\n

                        Pembahasan:<\/strong><\/p>\n

                          \n
                        • \n

                          Luas satu jaring-jaring kubus =\u00a06\u00d782=384<\/span>6<\/span>\u00d7<\/span><\/span>82<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>384<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

                        • \n

                          Luas karton =\u00a040\u00d740=1.600<\/span>40<\/span>\u00d7<\/span><\/span>40<\/span>=<\/span><\/span>1.600<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

                        • \n

                          Perkiraan =\u00a01.600\u00f7384\u22484,16<\/span>1.600<\/span>\u00f7<\/span><\/span>384<\/span>\u2248<\/span><\/span>4<\/span>,<\/span>16<\/span><\/span><\/span><\/span>\u00a0\u2192 maksimum 4<\/p>\n<\/li>\n

                        • \n

                          Periksa tata letak: susunan 2 \u00d7 2 = 4 kemasan (2 arah 40 cm muat 2 \u00d7 16 = 32 cm, sisa 8 cm; arah lain 2 \u00d7 24 = 48 cm > 40 cm \u2192 tidak muat)<\/p>\n<\/li>\n

                        • \n

                          Alternatif: orientasi berbeda, tetap maksimal 4<\/p>\n<\/li>\n<\/ul>\n

                          Jawaban: 4 kemasan<\/strong><\/p>\n

                          \n

                          Soal 11 (Mengapa 5 Persegi Tidak Bisa Membentuk Kubus?)<\/h3>\n

                          Jelaskan mengapa jaring-jaring yang terdiri dari 5 persegi tidak dapat membentuk kubus!<\/p>\n

                          Pembahasan:<\/strong><\/p>\n

                            \n
                          • \n

                            Kubus memiliki 6 sisi yang berbentuk persegi<\/p>\n<\/li>\n

                          • \n

                            Jika hanya 5 persegi, maka akan kekurangan 1 sisi<\/p>\n<\/li>\n

                          • \n

                            Saat dilipat, akan ada bagian yang terbuka (tidak tertutup)<\/p>\n<\/li>\n<\/ul>\n

                            Jawaban: Karena kubus membutuhkan 6 persegi (6 sisi), jaring-jaring dengan 5 persegi tidak cukup untuk menutup semua sisi kubus<\/strong><\/p>\n

                            \n

                            Soal 12 (Perbandingan Jumlah Kemasan Dua Ukuran Karton)<\/h3>\n

                            Karton A: 30 cm \u00d7 60 cm, Karton B: 40 cm \u00d7 45 cm. Luas satu jaring-jaring kemasan = 180 cm\u00b2. Karton mana yang dapat menghasilkan lebih banyak kemasan? (Asumsikan bentuk jaring-jaring memungkinkan penyusunan rapi)<\/p>\n

                            Pembahasan:<\/strong><\/p>\n

                              \n
                            • \n

                              Luas karton A =\u00a030\u00d760=1.800<\/span>30<\/span>\u00d7<\/span><\/span>60<\/span>=<\/span><\/span>1.800<\/span><\/span><\/span><\/span>\u00a0cm\u00b2 \u2192 perkiraan\u00a01.800\u00f7180=10<\/span>1.800<\/span>\u00f7<\/span><\/span>180<\/span>=<\/span><\/span>10<\/span><\/span><\/span><\/span>\u00a0kemasan<\/p>\n<\/li>\n

                            • \n

                              Luas karton B =\u00a040\u00d745=1.800<\/span>40<\/span>\u00d7<\/span><\/span>45<\/span>=<\/span><\/span>1.800<\/span><\/span><\/span><\/span>\u00a0cm\u00b2 \u2192 perkiraan\u00a01.800\u00f7180=10<\/span>1.800<\/span>\u00f7<\/span><\/span>180<\/span>=<\/span><\/span>10<\/span><\/span><\/span><\/span>\u00a0kemasan<\/p>\n<\/li>\n

                            • \n

                              Luas sama, jumlah maksimum sama (10) jika tata letak memungkinkan<\/p>\n<\/li>\n<\/ul>\n

                              Jawaban: Sama-sama dapat 10 kemasan (jika tata letak memungkinkan)<\/strong><\/p>\n

                              \n

                              Soal 13 (Jaring-Jaring Prisma Segi Enam)<\/h3>\n

                              Sebuah prisma segi enam beraturan memiliki sisi alas 4 cm dan tinggi 10 cm. Berapa luas jaring-jaringnya?<\/p>\n

                              Pembahasan:<\/strong><\/p>\n

                                \n
                              • \n

                                Luas alas segi enam =\u00a0332\u00d7s2=332\u00d716=243\u224841,57<\/span>2<\/span><\/span>33<\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>s<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>2<\/span><\/span>33<\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>16<\/span>=<\/span><\/span>24<\/span>3<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span>\u2248<\/span><\/span>41<\/span>,<\/span>57<\/span><\/span><\/span><\/span>\u00a0cm\u00b2 (2 buah \u2248 83,14 cm\u00b2)<\/p>\n<\/li>\n

                              • \n

                                Keliling alas =\u00a06\u00d74=24<\/span>6<\/span>\u00d7<\/span><\/span>4<\/span>=<\/span><\/span>24<\/span><\/span><\/span><\/span>\u00a0cm<\/p>\n<\/li>\n

                              • \n

                                Luas sisi tegak = keliling alas \u00d7 tinggi =\u00a024\u00d710=240<\/span>24<\/span>\u00d7<\/span><\/span>10<\/span>=<\/span><\/span>240<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

                              • \n

                                Luas total \u2248\u00a083,14+240=323,14<\/span>83<\/span>,<\/span>14<\/span>+<\/span><\/span>240<\/span>=<\/span><\/span>323<\/span>,<\/span>14<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n<\/ul>\n

                                Jawaban: \u2248 323,14 cm\u00b2<\/strong><\/p>\n

                                \n

                                Soal 14 (Mengidentifikasi Jaring-Jaring Limas yang Benar)<\/h3>\n

                                Perhatikan gambar jaring-jaring yang terdiri dari 1 persegi dan 4 segitiga. Apakah semua susunan seperti ini pasti dapat membentuk limas segi empat?<\/p>\n

                                Pembahasan:<\/strong><\/p>\n

                                  \n
                                • \n

                                  Limas segi empat membutuhkan 1 alas persegi dan 4 segitiga sebagai sisi tegak<\/p>\n<\/li>\n

                                • \n

                                  Namun, tidak semua susunan 1 persegi + 4 segitiga dapat dilipat menjadi limas<\/p>\n<\/li>\n

                                • \n

                                  Syarat: setiap sisi alas harus bertemu dengan alas segitiga yang panjangnya sama dengan sisi alas tersebut<\/p>\n<\/li>\n

                                • \n

                                  Jika ada segitiga yang alasnya tidak sama panjang dengan sisi persegi yang ditempel, maka tidak bisa membentuk limas<\/p>\n<\/li>\n<\/ul>\n

                                  Jawaban: Tidak semua susunan dapat membentuk limas, harus diperhatikan kesesuaian panjang sisi<\/strong><\/p>\n

                                  \n

                                  Soal 15 (Luas Jaring-Jaring Kerucut Terpotong)<\/h3>\n

                                  Sebuah kerucut memiliki jari-jari 14 cm dan tinggi 48 cm. Tentukan luas jaring-jaringnya! (Garis pelukis\u00a0s=r2+t2=142+482=196+2.304=2.500=50<\/span>s<\/span>=<\/span><\/span>r<\/span>2<\/span><\/span><\/span><\/span>+<\/span>t<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span>=<\/span><\/span>142<\/span><\/span><\/span><\/span>+<\/span>482<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span>=<\/span><\/span>196+<\/span>2.304<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span>=<\/span><\/span>2.500<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span>=<\/span><\/span>50<\/span><\/span><\/span><\/span>\u00a0cm;\u00a0\u03c0=227<\/span>\u03c0<\/span>=<\/span><\/span>7<\/span><\/span>22<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>)<\/p>\n

                                  Pembahasan:<\/strong><\/p>\n

                                    \n
                                  • \n

                                    Luas alas =\u00a0\u03c0r2=227\u00d7142=227\u00d7196=22\u00d728=616<\/span>\u03c0<\/span>r<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>7<\/span><\/span>22<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>1<\/span>42<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>7<\/span><\/span>22<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>196<\/span>=<\/span><\/span>22<\/span>\u00d7<\/span><\/span>28<\/span>=<\/span><\/span>616<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

                                  • \n

                                    Luas selimut =\u00a0\u03c0rs=227\u00d714\u00d750=22\u00d72\u00d750=2.200<\/span>\u03c0<\/span>rs<\/span>=<\/span><\/span>7<\/span><\/span>22<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>14<\/span>\u00d7<\/span><\/span>50<\/span>=<\/span><\/span>22<\/span>\u00d7<\/span><\/span>2<\/span>\u00d7<\/span><\/span>50<\/span>=<\/span><\/span>2.200<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

                                  • \n

                                    Luas total =\u00a0616+2.200=2.816<\/span>616<\/span>+<\/span><\/span>2.200<\/span>=<\/span><\/span>2.816<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n<\/ul>\n

                                    Jawaban: 2.816 cm\u00b2<\/strong><\/p>\n

                                    \n

                                    Ringkasan Rumus Penting Jaring-Jaring Bangun Ruang<\/h2>\n
                                    \n
                                    \n
                                    <\/div>\n
                                    <\/div>\n<\/div>\n\n\n\n\n\n\n\n\n\n\n
                                    Bangun Ruang<\/th>\nKomponen Jaring-Jaring<\/th>\nLuas Jaring-Jaring (Luas Permukaan)<\/th>\n<\/tr>\n<\/thead>\n
                                    Kubus<\/strong><\/td>\n6 persegi<\/td>\n6s2<\/span>6<\/span>s<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/td>\n<\/tr>\n
                                    Balok<\/strong><\/td>\n3 pasang persegi panjang<\/td>\n2(pl+pt+lt)<\/span>2<\/span>(<\/span>pl<\/span>+<\/span><\/span>pt<\/span>+<\/span><\/span>lt<\/span>)<\/span><\/span><\/span><\/span><\/td>\n<\/tr>\n
                                    Prisma segitiga<\/strong><\/td>\n2 segitiga + 3 persegi panjang<\/td>\n2\u00d7Lalas+Kalas\u00d7t<\/span>2<\/span>\u00d7<\/span><\/span>L<\/span>a<\/span>l<\/span>a<\/span>s<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>K<\/span>a<\/span>l<\/span>a<\/span>s<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>t<\/span><\/span><\/span><\/span><\/td>\n<\/tr>\n
                                    Limas segi empat<\/strong><\/td>\n1 persegi + 4 segitiga<\/td>\nLalas+4\u00d7Lsisitegak<\/span>L<\/span>a<\/span>l<\/span>a<\/span>s<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>4<\/span>\u00d7<\/span><\/span>L<\/span>s<\/span>i<\/span>s<\/span>i<\/span>t<\/span>e<\/span>g<\/span>ak<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/td>\n<\/tr>\n
                                    Tabung<\/strong><\/td>\n2 lingkaran + 1 persegi panjang<\/td>\n2\u03c0r2+2\u03c0rt<\/span>2<\/span>\u03c0<\/span>r<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>2<\/span>\u03c0<\/span>r<\/span>t<\/span><\/span><\/span><\/span><\/td>\n<\/tr>\n
                                    Kerucut<\/strong><\/td>\n1 lingkaran + 1 juring<\/td>\n\u03c0r2+\u03c0rs<\/span>\u03c0<\/span>r<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>\u03c0<\/span>rs<\/span><\/span><\/span><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

                                    Tips Optimalisasi Karton:<\/strong><\/p>\n

                                      \n
                                    1. \n

                                      Hitung luas karton dan luas satu jaring-jaring<\/p>\n<\/li>\n

                                    2. \n

                                      Perkiraan awal = luas karton \u00f7 luas jaring-jaring<\/p>\n<\/li>\n

                                    3. \n

                                      Perhatikan tata letak (layout) \u2013 ukuran jaring-jaring harus muat pada kedua dimensi karton<\/p>\n<\/li>\n

                                    4. \n

                                      Hasil aktual bisa lebih kecil dari perkiraan karena bentuk jaring-jaring tidak selalu persegi panjang<\/p>\n<\/li>\n<\/ol>\n

                                      \n

                                      Sahabat Rumah Pintar<\/strong>, dengan menguasai 15 soal di atas, kalian sudah siap mengidentifikasi jaring-jaring yang valid, menghitung luas bahan yang dibutuhkan, dan menyelesaikan masalah optimalisasi kemasan di TKA Matematika SMP.<\/p>\n","protected":false},"excerpt":{"rendered":"

                                      Bimbel Rumah Pintar – Prinsip dari soal jaring-jaring bangun ruang adalah rangkaian bangun datar yang jika dilipat pada garis-garis tertentu akan membentuk bangun ruang tiga dimensi. Prinsip dasar yang harus dikuasai adalah setiap bangun ruang memiliki komponen penyusun yang khas. […]<\/p>\n","protected":false},"author":1,"featured_media":2161,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"set","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[4],"tags":[224],"class_list":["post-2159","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-bahas-soal","tag-soal-tka-matematika-smp"],"_links":{"self":[{"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/posts\/2159","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/comments?post=2159"}],"version-history":[{"count":2,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/posts\/2159\/revisions"}],"predecessor-version":[{"id":2167,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/posts\/2159\/revisions\/2167"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/media\/2161"}],"wp:attachment":[{"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/media?parent=2159"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/categories?post=2159"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/tags?post=2159"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}