{"id":2149,"date":"2026-04-03T18:07:48","date_gmt":"2026-04-03T18:07:48","guid":{"rendered":"https:\/\/rumahpintar.co.id\/blog\/?p=2149"},"modified":"2026-04-09T04:12:12","modified_gmt":"2026-04-09T04:12:12","slug":"soal-kekongruenan-dan-kesebangunan-bangun-datar","status":"publish","type":"post","link":"https:\/\/rumahpintar.co.id\/blog\/soal-kekongruenan-dan-kesebangunan-bangun-datar\/","title":{"rendered":"15 Contoh Soal Kekongruenan dan Kesebangunan Bangun Datar TKA Matematika SMP"},"content":{"rendered":"

Bimbel Rumah Pintar<\/strong><\/a> – Soal <\/span>kekongruenan dan kesebangunan bangun datar adalah dua konsep fundamental dalam geometri yang sering diujikan dalam TKA Matematika SMP. Kekongruenan berarti dua bangun datar memiliki bentuk dan ukuran yang sama persis<\/strong>\u00a0(sisi-sisi bersesuaian sama panjang, sudut-sudut bersesuaian sama besar), sehingga dapat diimpitkan.<\/p>\n

Sementara kesebangunan berarti dua bangun datar memiliki\u00a0bentuk yang sama tetapi ukuran boleh berbeda<\/strong>\u00a0(sisi-sisi bersesuaian sebanding, sudut-sudut bersesuaian sama besar). Prinsip dasar yang harus dikuasai adalah syarat-syarat kekongruenan segitiga (SSS, SAS, ASA, AAS) dan syarat kesebangunan segitiga (SSS, SAS, AA), serta rumus-rumus pada segitiga siku-siku dengan garis tinggi yang menghasilkan tiga segitiga sebangun.<\/p>\n

Mengapa latihan soal kekongruenan dan kesebangunan sangat penting?<\/strong>\u00a0Soal TKA Matematika SMP sering menyajikan konfigurasi segitiga dengan garis tinggi, trapesium dengan garis sejajar, atau bangun-bangun yang sebangun\/kongruen, kemudian meminta menentukan panjang sisi yang belum diketahui, menilai kebenaran beberapa pernyataan sekaligus (tipe PGK), atau menyelesaikan masalah kontekstual seperti foto dan bingkai.<\/p>\n

15 Contoh Soal Kekongruenan dan Kesebangunan Bangun Datar<\/strong><\/h2>\n

<\/h3>\n

Soal 1 (Segitiga Sebangun dengan Garis Tinggi \u2013 Tipe PGK)<\/h3>\n

Perhatikan gambar segitiga siku-siku dengan garis tinggi. Segitiga ABD dan segitiga CAD saling sebangun. Panjang sisi BC = 15 cm. Tentukan Benar\/Salah pernyataan berikut:<\/p>\n

    \n
  • \n

    A. Segitiga ABD dan segitiga CAD saling sebangun.<\/p>\n<\/li>\n

  • \n

    B. Panjang sisi BC adalah 15 cm.<\/p>\n<\/li>\n

  • \n

    C. Panjang sisi AC adalah 10 cm.<\/p>\n<\/li>\n<\/ul>\n

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

      \n
    • \n

      Dalam segitiga siku-siku dengan garis tinggi ke sisi miring, segitiga ABD sebangun dengan segitiga CAD (dan keduanya sebangun dengan segitiga ABC) \u2192\u00a0A = BENAR<\/strong><\/p>\n<\/li>\n

    • \n

      Soal sudah menyatakan BC = 15 cm \u2192\u00a0B = BENAR<\/strong><\/p>\n<\/li>\n

    • \n

      Untuk menentukan AC, perlu data tambahan (misal BD atau DC). Tanpa data, tidak bisa dipastikan AC = 10 cm \u2192\u00a0C = SALAH<\/strong><\/p>\n<\/li>\n<\/ul>\n

      Jawaban: A = Benar, B = Benar, C = Salah<\/strong><\/p>\n

      \n

      Soal 2 (Menentukan Sisi pada Segitiga Sebangun)<\/h3>\n

      Diketahui\u00a0\u25b3ABC\u223c\u25b3DEF<\/span>\u25b3<\/span>A<\/span>BC<\/span>\u223c<\/span><\/span>\u25b3<\/span>D<\/span>EF<\/span><\/span><\/span><\/span>. AB = 6 cm, DE = 9 cm, BC = 8 cm. Tentukan panjang EF!<\/p>\n

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

        \n
      • \n

        Faktor skala =\u00a0DEAB=96=32<\/span>A<\/span>B<\/span><\/span><\/span>D<\/span>E<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>6<\/span><\/span>9<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>2<\/span><\/span>3<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

      • \n

        BCEF=ABDE<\/span>EF<\/span><\/span><\/span>BC<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>D<\/span>E<\/span><\/span><\/span>A<\/span>B<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00a0atau\u00a0EF=BC\u00d7DEAB<\/span>EF<\/span>=<\/span><\/span>BC<\/span>\u00d7<\/span><\/span>A<\/span>B<\/span><\/span><\/span>D<\/span>E<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

      • \n

        EF=8\u00d796=8\u00d71,5=12<\/span>EF<\/span>=<\/span><\/span>8<\/span>\u00d7<\/span><\/span>6<\/span><\/span>9<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>8<\/span>\u00d7<\/span><\/span>1<\/span>,<\/span>5<\/span>=<\/span><\/span>12<\/span><\/span><\/span><\/span>\u00a0cm<\/p>\n<\/li>\n<\/ul>\n

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

        \n

        Soal 3 (Menentukan Sisi pada Segitiga Kongruen)<\/h3>\n

        Diketahui\u00a0\u25b3PQR\u2245\u25b3STU<\/span>\u25b3<\/span>PQR<\/span>\u2245<\/span><\/span>\u25b3<\/span>ST<\/span>U<\/span><\/span><\/span><\/span>. PQ = 10 cm, QR = 14 cm, PR = 12 cm. Tentukan panjang ST dan TU!<\/p>\n

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

          \n
        • \n

          Karena kongruen, sisi-sisi bersesuaian sama panjang<\/p>\n<\/li>\n

        • \n

          ST bersesuaian dengan PQ \u2192 ST = 10 cm<\/p>\n<\/li>\n

        • \n

          TU bersesuaian dengan QR \u2192 TU = 14 cm<\/p>\n<\/li>\n<\/ul>\n

          Jawaban: ST = 10 cm, TU = 14 cm<\/strong><\/p>\n

          \n

          Soal 4 (Kesebangunan Segitiga Siku-Siku dengan Garis Tinggi \u2013 Mencari AD)<\/h3>\n

          Pada segitiga siku-siku ABC siku-siku di A, AD adalah garis tinggi ke sisi miring BC. Jika BD = 9 cm dan DC = 16 cm, tentukan panjang AD!<\/p>\n

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

            \n
          • \n

            Gunakan rumus:\u00a0AD2=BD\u00d7DC<\/span>A<\/span>D<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>B<\/span>D<\/span>\u00d7<\/span><\/span>D<\/span>C<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

          • \n

            AD2=9\u00d716=144<\/span>A<\/span>D<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>9<\/span>\u00d7<\/span><\/span>16<\/span>=<\/span><\/span>144<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

          • \n

            AD=144=12<\/span>A<\/span>D<\/span>=<\/span><\/span>144<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span>=<\/span><\/span>12<\/span><\/span><\/span><\/span>\u00a0cm<\/p>\n<\/li>\n<\/ul>\n

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

            \n

            Soal 5 (Kesebangunan Segitiga Siku-Siku \u2013 Mencari AB dan AC)<\/h3>\n

            Dari soal nomor 4, tentukan panjang AB dan AC!<\/p>\n

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

              \n
            • \n

              AB2=BD\u00d7BC=9\u00d7(9+16)=9\u00d725=225<\/span>A<\/span>B<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>B<\/span>D<\/span>\u00d7<\/span><\/span>BC<\/span>=<\/span><\/span>9<\/span>\u00d7<\/span><\/span>(<\/span>9<\/span>+<\/span><\/span>16<\/span>)<\/span>=<\/span><\/span>9<\/span>\u00d7<\/span><\/span>25<\/span>=<\/span><\/span>225<\/span><\/span><\/span><\/span>\u00a0\u2192\u00a0AB=15<\/span>A<\/span>B<\/span>=<\/span><\/span>15<\/span><\/span><\/span><\/span>\u00a0cm<\/p>\n<\/li>\n

            • \n

              AC2=CD\u00d7CB=16\u00d725=400<\/span>A<\/span>C<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>C<\/span>D<\/span>\u00d7<\/span><\/span>CB<\/span>=<\/span><\/span>16<\/span>\u00d7<\/span><\/span>25<\/span>=<\/span><\/span>400<\/span><\/span><\/span><\/span>\u00a0\u2192\u00a0AC=20<\/span>A<\/span>C<\/span>=<\/span><\/span>20<\/span><\/span><\/span><\/span>\u00a0cm<\/p>\n<\/li>\n<\/ul>\n

              Jawaban: AB = 15 cm, AC = 20 cm<\/strong><\/p>\n

              \n

              Soal 6 (Menentukan Sudut pada Segitiga Sebangun)<\/h3>\n

              Diketahui\u00a0\u25b3ABC\u223c\u25b3DEF<\/span>\u25b3<\/span>A<\/span>BC<\/span>\u223c<\/span><\/span>\u25b3<\/span>D<\/span>EF<\/span><\/span><\/span><\/span>.\u00a0\u2220A=50\u2218<\/span>\u2220<\/span>A<\/span>=<\/span><\/span>5<\/span>0\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>,\u00a0\u2220B=70\u2218<\/span>\u2220<\/span>B<\/span>=<\/span><\/span>7<\/span>0\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>. Tentukan besar\u00a0\u2220E<\/span>\u2220<\/span>E<\/span><\/span><\/span><\/span>\u00a0dan\u00a0\u2220F<\/span>\u2220<\/span>F<\/span><\/span><\/span><\/span>!<\/p>\n

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

                \n
              • \n

                \u2220C=180\u2218\u221250\u2218\u221270\u2218=60\u2218<\/span>\u2220<\/span>C<\/span>=<\/span><\/span>18<\/span>0\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u2212<\/span><\/span>5<\/span>0\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u2212<\/span><\/span>7<\/span>0\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>6<\/span>0\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

              • \n

                Karena sebangun, sudut-sudut bersesuaian sama besar<\/p>\n<\/li>\n

              • \n

                \u2220E<\/span>\u2220<\/span>E<\/span><\/span><\/span><\/span>\u00a0bersesuaian dengan\u00a0\u2220B<\/span>\u2220<\/span>B<\/span><\/span><\/span><\/span>\u00a0\u2192\u00a0\u2220E=70\u2218<\/span>\u2220<\/span>E<\/span>=<\/span><\/span>7<\/span>0\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

              • \n

                \u2220F<\/span>\u2220<\/span>F<\/span><\/span><\/span><\/span>\u00a0bersesuaian dengan\u00a0\u2220C<\/span>\u2220<\/span>C<\/span><\/span><\/span><\/span>\u00a0\u2192\u00a0\u2220F=60\u2218<\/span>\u2220<\/span>F<\/span>=<\/span><\/span>6<\/span>0\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ul>\n

                Jawaban:\u00a0\u2220E=70\u2218<\/span>\u2220<\/span>E<\/span>=<\/span><\/span>7<\/span>0\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>,\u00a0\u2220F=60\u2218<\/span>\u2220<\/span>F<\/span>=<\/span><\/span>6<\/span>0\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/strong><\/p>\n

                \n

                Soal 7 (Kesebangunan pada Persegi Panjang)<\/h3>\n

                Dua persegi panjang sebangun. Persegi panjang pertama berukuran 6 cm \u00d7 4 cm. Persegi panjang kedua memiliki panjang 9 cm. Tentukan lebarnya!<\/p>\n

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

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                • \n

                  Perbandingan sisi bersesuaian:\u00a096=l4<\/span>6<\/span><\/span>9<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>4<\/span><\/span>l<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                • \n

                  l=4\u00d796=4\u00d71,5=6<\/span>l<\/span>=<\/span><\/span>4<\/span>\u00d7<\/span><\/span>6<\/span><\/span>9<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>4<\/span>\u00d7<\/span><\/span>1<\/span>,<\/span>5<\/span>=<\/span><\/span>6<\/span><\/span><\/span><\/span>\u00a0cm<\/p>\n<\/li>\n<\/ul>\n

                  Jawaban: 6 cm<\/strong><\/p>\n

                  \n

                  Soal 8 (Kekongruenan Segitiga dengan Syarat SAS)<\/h3>\n

                  Perhatikan gambar! AB = DE, AC = DF, dan\u00a0\u2220A=\u2220D<\/span>\u2220<\/span>A<\/span>=<\/span><\/span>\u2220<\/span>D<\/span><\/span><\/span><\/span>. Apakah\u00a0\u25b3ABC\u2245\u25b3DEF<\/span>\u25b3<\/span>A<\/span>BC<\/span>\u2245<\/span><\/span>\u25b3<\/span>D<\/span>EF<\/span><\/span><\/span><\/span>?<\/p>\n

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

                    \n
                  • \n

                    Dua sisi sama panjang (AB = DE, AC = DF)<\/p>\n<\/li>\n

                  • \n

                    Sudut apitnya sama (\u2220A=\u2220D<\/span>\u2220<\/span>A<\/span>=<\/span><\/span>\u2220<\/span>D<\/span><\/span><\/span><\/span>)<\/p>\n<\/li>\n

                  • \n

                    Memenuhi syarat\u00a0Sisi \u2013 Sudut \u2013 Sisi (SAS)<\/strong><\/p>\n<\/li>\n

                  • \n

                    Maka\u00a0\u25b3ABC\u2245\u25b3DEF<\/span>\u25b3<\/span>A<\/span>BC<\/span>\u2245<\/span><\/span>\u25b3<\/span>D<\/span>EF<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ul>\n

                    Jawaban: Ya, kongruen<\/strong><\/p>\n

                    \n

                    Soal 9 (Kesebangunan pada Foto dan Bingkai \u2013 Soal Cerita)<\/h3>\n

                    Sebuah foto berukuran 8 cm \u00d7 6 cm dipasang pada bingkai yang sebangun. Jika lebar bingkai 12 cm, berapa panjang bingkai?<\/p>\n

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

                      \n
                    • \n

                      Foto dan bingkai sebangun \u2192 perbandingan sisi tetap<\/p>\n<\/li>\n

                    • \n

                      panjang\u00a0fotolebar\u00a0foto=panjang\u00a0bingkailebar\u00a0bingkai<\/span>lebar\u00a0foto<\/span><\/span><\/span>panjang\u00a0foto<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>lebar\u00a0bingkai<\/span><\/span><\/span>panjang\u00a0bingkai<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                    • \n

                      86=p12<\/span>6<\/span><\/span>8<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>12<\/span><\/span>p<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                    • \n

                      p=8\u00d7126=966=16<\/span>p<\/span>=<\/span><\/span>6<\/span><\/span>8\u00d7<\/span>12<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>6<\/span><\/span>96<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>16<\/span><\/span><\/span><\/span>\u00a0cm<\/p>\n<\/li>\n<\/ul>\n

                      Jawaban: 16 cm<\/strong><\/p>\n

                      \n

                      Soal 10 (Kesebangunan pada Segitiga dengan Garis Sejajar)<\/h3>\n

                      Dalam segitiga ABC, DE \/\/ BC dengan D pada AB dan E pada AC. Jika AD = 3 cm, DB = 5 cm, dan DE = 6 cm, tentukan panjang BC!<\/p>\n

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

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                      • \n

                        DE \/\/ BC \u2192\u00a0\u25b3ADE\u223c\u25b3ABC<\/span>\u25b3<\/span>A<\/span>D<\/span>E<\/span>\u223c<\/span><\/span>\u25b3<\/span>A<\/span>BC<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                      • \n

                        ADAB=DEBC<\/span>A<\/span>B<\/span><\/span><\/span>A<\/span>D<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>BC<\/span><\/span><\/span>D<\/span>E<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                      • \n

                        AB=AD+DB=3+5=8<\/span>A<\/span>B<\/span>=<\/span><\/span>A<\/span>D<\/span>+<\/span><\/span>D<\/span>B<\/span>=<\/span><\/span>3<\/span>+<\/span><\/span>5<\/span>=<\/span><\/span>8<\/span><\/span><\/span><\/span>\u00a0cm<\/p>\n<\/li>\n

                      • \n

                        38=6BC<\/span>8<\/span><\/span>3<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>BC<\/span><\/span><\/span>6<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00a0\u2192\u00a0BC=6\u00d783=483=16<\/span>BC<\/span>=<\/span><\/span>3<\/span><\/span>6\u00d7<\/span>8<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>3<\/span><\/span>48<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>16<\/span><\/span><\/span><\/span>\u00a0cm<\/p>\n<\/li>\n<\/ul>\n

                        Jawaban: 16 cm<\/strong><\/p>\n

                        \n

                        Soal 11 (Kesebangunan pada Trapesium)<\/h3>\n

                        Pada trapesium ABCD dengan AB \/\/ EF \/\/ DC. AE = 4 cm, ED = 6 cm, AB = 10 cm, DC = 20 cm. Tentukan panjang EF!<\/p>\n

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

                          \n
                        • \n

                          Gunakan rumus garis sejajar pada trapesium:<\/p>\n<\/li>\n

                        • \n

                          EF=AB+AEAD\u00d7(DC\u2212AB)<\/span>EF<\/span>=<\/span><\/span>A<\/span>B<\/span>+<\/span><\/span>A<\/span>D<\/span><\/span><\/span>A<\/span>E<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>(<\/span>D<\/span>C<\/span>\u2212<\/span><\/span>A<\/span>B<\/span>)<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                        • \n

                          AD=AE+ED=4+6=10<\/span>A<\/span>D<\/span>=<\/span><\/span>A<\/span>E<\/span>+<\/span><\/span>E<\/span>D<\/span>=<\/span><\/span>4<\/span>+<\/span><\/span>6<\/span>=<\/span><\/span>10<\/span><\/span><\/span><\/span>\u00a0cm<\/p>\n<\/li>\n

                        • \n

                          EF=10+410\u00d7(20\u221210)=10+0,4\u00d710=10+4=14<\/span>EF<\/span>=<\/span><\/span>10<\/span>+<\/span><\/span>10<\/span><\/span>4<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>(<\/span>20<\/span>\u2212<\/span><\/span>10<\/span>)<\/span>=<\/span><\/span>10<\/span>+<\/span><\/span>0<\/span>,<\/span>4<\/span>\u00d7<\/span><\/span>10<\/span>=<\/span><\/span>10<\/span>+<\/span><\/span>4<\/span>=<\/span><\/span>14<\/span><\/span><\/span><\/span>\u00a0cm<\/p>\n<\/li>\n<\/ul>\n

                          Jawaban: 14 cm<\/strong><\/p>\n

                          \n

                          Soal 12 (Skala pada Model \u2013 Soal Cerita)<\/h3>\n

                          Sebuah model mobil memiliki skala 1 : 50. Jika panjang model 12 cm, berapa panjang mobil sebenarnya?<\/p>\n

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

                            \n
                          • \n

                            Skala 1 : 50 berarti 1 cm pada model mewakili 50 cm sebenarnya<\/p>\n<\/li>\n

                          • \n

                            Panjang sebenarnya =\u00a012\u00d750=600<\/span>12<\/span>\u00d7<\/span><\/span>50<\/span>=<\/span><\/span>600<\/span><\/span><\/span><\/span>\u00a0cm = 6 m<\/p>\n<\/li>\n<\/ul>\n

                            Jawaban: 6 meter<\/strong><\/p>\n

                            \n

                            Soal 13 (Membuktikan Kesebangunan Dua Segitiga)<\/h3>\n

                            Buktikan bahwa segitiga siku-siku dengan sisi 3 cm, 4 cm, 5 cm sebangun dengan segitiga siku-siku dengan sisi 9 cm, 12 cm, 15 cm!<\/p>\n

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

                              \n
                            • \n

                              Periksa perbandingan sisi-sisi yang bersesuaian:<\/p>\n<\/li>\n

                            • \n

                              39=13<\/span>9<\/span><\/span>3<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>3<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>,\u00a0412=13<\/span>12<\/span><\/span>4<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>3<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>,\u00a0515=13<\/span>15<\/span><\/span>5<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>3<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                            • \n

                              Semua perbandingan sama yaitu\u00a013<\/span>3<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                            • \n

                              Memenuhi syarat\u00a0Sisi \u2013 Sisi \u2013 Sisi (SSS)<\/strong>\u00a0untuk kesebangunan<\/p>\n<\/li>\n<\/ul>\n

                              Jawaban: Terbukti sebangun (faktor skala = 1\/3)<\/strong><\/p>\n

                              \n

                              Soal 14 (Menentukan Tinggi Pohon dengan Kesebangunan)<\/h3>\n

                              Sebuah pohon memiliki bayangan sepanjang 12 m. Pada saat yang sama, sebuah tongkat dengan tinggi 1,5 m memiliki bayangan 2 m. Tentukan tinggi pohon!<\/p>\n

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

                                \n
                              • \n

                                Segitiga yang terbentuk oleh pohon dan bayangannya sebangun dengan segitiga tongkat dan bayangannya<\/p>\n<\/li>\n

                              • \n

                                tinggi\u00a0pohonbayangan\u00a0pohon=tinggi\u00a0tongkatbayangan\u00a0tongkat<\/span>bayangan\u00a0pohon<\/span><\/span><\/span>tinggi\u00a0pohon<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>bayangan\u00a0tongkat<\/span><\/span><\/span>tinggi\u00a0tongkat<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                              • \n

                                h12=1,52<\/span>12<\/span><\/span>h<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>2<\/span><\/span>1,<\/span>5<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                              • \n

                                h=1,5\u00d7122=182=9<\/span>h<\/span>=<\/span><\/span>2<\/span><\/span>1,<\/span>5\u00d7<\/span>12<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>2<\/span><\/span>18<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>9<\/span><\/span><\/span><\/span>\u00a0m<\/p>\n<\/li>\n<\/ul>\n

                                Jawaban: 9 meter<\/strong><\/p>\n

                                \n

                                Soal 15 (Kesebangunan pada Segitiga Siku-Siku dengan Sudut 30\u00b0)<\/h3>\n

                                Dua segitiga siku-siku memiliki sudut lancip 30\u00b0. Apakah kedua segitiga tersebut pasti sebangun? Jelaskan!<\/p>\n

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

                                  \n
                                • \n

                                  Segitiga siku-siku memiliki satu sudut 90\u00b0<\/p>\n<\/li>\n

                                • \n

                                  Jika salah satu sudut lancipnya 30\u00b0, maka sudut lancip lainnya =\u00a0180\u00b0\u221290\u00b0\u221230\u00b0=60\u00b0<\/span>180\u00b0<\/span>\u2212<\/span><\/span>90\u00b0<\/span>\u2212<\/span><\/span>30\u00b0<\/span>=<\/span><\/span>60\u00b0<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n

                                • \n

                                  Kedua segitiga memiliki sudut yang sama: 30\u00b0, 60\u00b0, 90\u00b0<\/p>\n<\/li>\n

                                • \n

                                  Memenuhi syarat\u00a0Sudut \u2013 Sudut (AA)<\/strong>\u00a0untuk kesebangunan<\/p>\n<\/li>\n<\/ul>\n

                                  Jawaban: Ya, pasti sebangun karena memiliki sudut-sudut yang sama<\/strong><\/p>\n

                                  \n

                                  Ringkasan Rumus Penting Kekongruenan dan Kesebangunan<\/h2>\n
                                  \n
                                  \n
                                  <\/div>\n
                                  <\/div>\n<\/div>\n\n\n\n\n\n\n\n\n\n\n\n\n
                                  Konsep<\/th>\nRumus\/Syarat<\/th>\n<\/tr>\n<\/thead>\n
                                  Kongruen (SSS)<\/strong><\/td>\nKetiga sisi bersesuaian sama panjang<\/td>\n<\/tr>\n
                                  Kongruen (SAS)<\/strong><\/td>\nDua sisi sama panjang dan sudut apitnya sama<\/td>\n<\/tr>\n
                                  Kongruen (ASA\/AAS)<\/strong><\/td>\nDua sudut sama dan satu sisi sama panjang<\/td>\n<\/tr>\n
                                  Sebangun (SSS)<\/strong><\/td>\nKetiga sisi bersesuaian sebanding<\/td>\n<\/tr>\n
                                  Sebangun (SAS)<\/strong><\/td>\nDua sisi sebanding dan sudut apitnya sama<\/td>\n<\/tr>\n
                                  Sebangun (AA)<\/strong><\/td>\nDua sudut bersesuaian sama besar<\/td>\n<\/tr>\n
                                  Garis tinggi pada segitiga siku-siku<\/strong><\/td>\nAD2=BD\u00d7DC<\/span>A<\/span>D<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>B<\/span>D<\/span>\u00d7<\/span><\/span>D<\/span>C<\/span><\/span><\/span><\/span><\/td>\n<\/tr>\n
                                  Proyeksi pada segitiga siku-siku<\/strong><\/td>\nAB2=BD\u00d7BC<\/span>A<\/span>B<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>B<\/span>D<\/span>\u00d7<\/span><\/span>BC<\/span><\/span><\/span><\/span>,\u00a0AC2=CD\u00d7CB<\/span>A<\/span>C<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>C<\/span>D<\/span>\u00d7<\/span><\/span>CB<\/span><\/span><\/span><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

                                  Perbedaan Utama:<\/strong><\/p>\n

                                    \n
                                  • \n

                                    Kongruen<\/strong>\u00a0\u2192 ukuran sama persis (faktor skala = 1)<\/p>\n<\/li>\n

                                  • \n

                                    Sebangun<\/strong>\u00a0\u2192 bentuk sama, ukuran proporsional (faktor skala = k)<\/p>\n<\/li>\n<\/ul>\n

                                    \n

                                    Sahabat Rumah Pintar<\/strong>, dengan menguasai 15 soal di atas, kalian sudah siap mengidentifikasi pasangan bangun yang kongruen atau sebangun, menghitung panjang sisi yang tidak diketahui, dan menyelesaikan soal cerita kesebangunan di TKA Matematika SMP. Ingat kuncinya:\u00a0perhatikan urutan huruf untuk menentukan sisi\/sudut bersesuaian, gunakan rumus garis tinggi\u00a0AD2=BD\u00d7DC<\/span>A<\/span>D<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>B<\/span>D<\/span>\u00d7<\/span><\/span>D<\/span>C<\/span><\/span><\/span><\/span>\u00a0untuk segitiga siku-siku, dan ingat bahwa kongruen adalah kasus khusus sebangun dengan faktor skala = 1<\/strong>. Terus berlatih, ya!<\/p>\n","protected":false},"excerpt":{"rendered":"

                                    Bimbel Rumah Pintar – Soal kekongruenan dan kesebangunan bangun datar adalah dua konsep fundamental dalam geometri yang sering diujikan dalam TKA Matematika SMP. Kekongruenan berarti dua bangun datar memiliki bentuk dan ukuran yang sama persis\u00a0(sisi-sisi bersesuaian sama panjang, sudut-sudut bersesuaian […]<\/p>\n","protected":false},"author":1,"featured_media":2147,"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":"","ast-disable-related-posts":"","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":[238,239,224],"class_list":["post-2149","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-bahas-soal","tag-bangun-datar","tag-matematika-smp","tag-soal-tka-matematika-smp"],"_links":{"self":[{"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/posts\/2149","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=2149"}],"version-history":[{"count":1,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/posts\/2149\/revisions"}],"predecessor-version":[{"id":2150,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/posts\/2149\/revisions\/2150"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/media\/2147"}],"wp:attachment":[{"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/media?parent=2149"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/categories?post=2149"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/tags?post=2149"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}