{"id":2151,"date":"2026-04-03T19:16:12","date_gmt":"2026-04-03T19:16:12","guid":{"rendered":"https:\/\/rumahpintar.co.id\/blog\/?p=2151"},"modified":"2026-04-09T04:21:21","modified_gmt":"2026-04-09T04:21:21","slug":"contoh-soal-keliling-dan-luas-bangun-datar","status":"publish","type":"post","link":"https:\/\/rumahpintar.co.id\/blog\/contoh-soal-keliling-dan-luas-bangun-datar\/","title":{"rendered":"15 Contoh Soal Keliling dan Luas Bangun Datar, Mulai dari Segi Banyak, Lingkaran dan Gabungan"},"content":{"rendered":"

Bimbel Rumah Pintar<\/strong><\/a> –<\/span> Soal keliling dan luas bangun datar adalah konsep geometri yang paling sering muncul dalam kehidupan sehari-hari sekaligus menjadi langganan soal TKA Matematika SMP.<\/p>\n

Prinsip dasar yang harus dikuasai adalah hafal rumus keliling dan luas berbagai bangun datar. Kunci keberhasilan terletak pada ketelitian dalam substitusi nilai, konsistensi satuan, serta kemampuan memecah bangun gabungan menjadi bagian-bagian sederhana.<\/p>\n

Mengapa latihan soal keliling dan luas bangun datar sangat penting?<\/strong>\u00a0Soal TKA Matematika SMP sering menyajikan masalah kontekstual seperti taman dengan kolam di tengah, dinding yang akan dicat, perubahan luas akibat perubahan ukuran, hingga pemilihan merek cat yang paling efisien.<\/p>\n

Tanpa latihan yang cukup, siswa sering keliru dalam memilih rumus (misal luas segitiga lupa dikali 1\/2), salah menggunakan nilai\u00a0\u03c0<\/span>\u03c0<\/span><\/span><\/span><\/span>\u00a0(harus 22\/7 jika jari-jari kelipatan 7), atau kesulitan menghitung luas daerah gabungan.<\/p>\n

Dengan menguasai 15 variasi soal berikut (dengan konsep sama tetapi angka dan konteks berbeda dari contoh sebelumnya), Sahabat Rumah Pintar akan mampu menghitung keliling dan luas dengan tepat, membandingkan luas, menganalisis perubahan luas, dan menjawab soal TKA dengan percaya diri.<\/p>\n

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15 Contoh Soal Keliling dan Luas Bangun Datar dan Pembahasannya<\/h2>\n
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Soal 1 (Taman dengan Kolam \u2013 Tipe PGK)<\/h3>\n

Sebuah taman berbentuk persegi panjang berukuran 24 m \u00d7 18 m. Di tengahnya terdapat kolam berbentuk lingkaran dengan diameter 14 m. Gunakan\u00a0\u03c0=227<\/span>\u03c0<\/span>=<\/span><\/span>7<\/span><\/span>22<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>. Tentukan Benar\/Salah pernyataan berikut:<\/p>\n

    \n
  • \n

    A. Luas taman adalah 432 m\u00b2<\/p>\n<\/li>\n

  • \n

    B. Luas kolam adalah 154 m\u00b2<\/p>\n<\/li>\n

  • \n

    C. Luas daerah yang ditanami rumput kurang dari 280 m\u00b2<\/p>\n<\/li>\n<\/ul>\n

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

      \n
    • \n

      Luas taman =\u00a0p\u00d7l=24\u00d718=432<\/span>p<\/span>\u00d7<\/span><\/span>l<\/span>=<\/span><\/span>24<\/span>\u00d7<\/span><\/span>18<\/span>=<\/span><\/span>432<\/span><\/span><\/span><\/span>\u00a0m\u00b2 \u2192\u00a0A = BENAR<\/strong><\/p>\n<\/li>\n

    • \n

      Jari-jari kolam =\u00a014\u00f72=7<\/span>14<\/span>\u00f7<\/span><\/span>2<\/span>=<\/span><\/span>7<\/span><\/span><\/span><\/span>\u00a0m<\/p>\n<\/li>\n

    • \n

      Luas kolam =\u00a0\u03c0r2=227\u00d772=227\u00d749=22\u00d77=154<\/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>72<\/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>49<\/span>=<\/span><\/span>22<\/span>\u00d7<\/span><\/span>7<\/span>=<\/span><\/span>154<\/span><\/span><\/span><\/span>\u00a0m\u00b2 \u2192\u00a0B = BENAR<\/strong><\/p>\n<\/li>\n

    • \n

      Luas rumput =\u00a0432\u2212154=278<\/span>432<\/span>\u2212<\/span><\/span>154<\/span>=<\/span><\/span>278<\/span><\/span><\/span><\/span>\u00a0m\u00b2 \u2192 Apakah kurang dari 280 m\u00b2? 278 < 280 \u2192\u00a0C = BENAR<\/strong><\/p>\n<\/li>\n<\/ul>\n

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

      \n

      Soal 2 (Persegi Panjang \u2013 Keliling dan Luas)<\/h3>\n

      Sebuah lapangan berbentuk persegi panjang dengan panjang 32 m dan lebar 15 m. Berapa keliling dan luas lapangan tersebut?<\/p>\n

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

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      • \n

        Keliling =\u00a02(p+l)=2(32+15)=2\u00d747=94<\/span>2<\/span>(<\/span>p<\/span>+<\/span><\/span>l<\/span>)<\/span>=<\/span><\/span>2<\/span>(<\/span>32<\/span>+<\/span><\/span>15<\/span>)<\/span>=<\/span><\/span>2<\/span>\u00d7<\/span><\/span>47<\/span>=<\/span><\/span>94<\/span><\/span><\/span><\/span>\u00a0m<\/p>\n<\/li>\n

      • \n

        Luas =\u00a0p\u00d7l=32\u00d715=480<\/span>p<\/span>\u00d7<\/span><\/span>l<\/span>=<\/span><\/span>32<\/span>\u00d7<\/span><\/span>15<\/span>=<\/span><\/span>480<\/span><\/span><\/span><\/span>\u00a0m\u00b2<\/p>\n<\/li>\n<\/ul>\n

        Jawaban: Keliling = 94 m, Luas = 480 m\u00b2<\/strong><\/p>\n

        \n

        Soal 3 (Lingkaran \u2013 Keliling dan Luas)<\/h3>\n

        Sebuah roda memiliki diameter 70 cm. Berapa keliling dan luas permukaan roda tersebut? (\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

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        • \n

          Jari-jari =\u00a070\u00f72=35<\/span>70<\/span>\u00f7<\/span><\/span>2<\/span>=<\/span><\/span>35<\/span><\/span><\/span><\/span>\u00a0cm<\/p>\n<\/li>\n

        • \n

          Keliling =\u00a0\u03c0\u00d7d=227\u00d770=22\u00d710=220<\/span>\u03c0<\/span>\u00d7<\/span><\/span>d<\/span>=<\/span><\/span>7<\/span><\/span>22<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>70<\/span>=<\/span><\/span>22<\/span>\u00d7<\/span><\/span>10<\/span>=<\/span><\/span>220<\/span><\/span><\/span><\/span>\u00a0cm<\/p>\n<\/li>\n

        • \n

          Luas =\u00a0\u03c0r2=227\u00d7352=227\u00d71225=22\u00d7175=3.850<\/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>3<\/span>52<\/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>1225<\/span>=<\/span><\/span>22<\/span>\u00d7<\/span><\/span>175<\/span>=<\/span><\/span>3.850<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n<\/ul>\n

          Jawaban: Keliling = 220 cm, Luas = 3.850 cm\u00b2<\/strong><\/p>\n

          \n

          Soal 4 (Luas Gabungan \u2013 Persegi dan Segitiga)<\/h3>\n

          Sebuah bangun terdiri dari persegi dengan sisi 12 cm dan segitiga di atasnya dengan alas sama dengan sisi persegi dan tinggi 9 cm. Berapa luas total bangun?<\/p>\n

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

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          • \n

            Luas persegi =\u00a0s2=122=144<\/span>s<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>1<\/span>22<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>144<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

          • \n

            Luas segitiga =\u00a012\u00d7a\u00d7t=12\u00d712\u00d79=6\u00d79=54<\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>a<\/span>\u00d7<\/span><\/span>t<\/span>=<\/span><\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>12<\/span>\u00d7<\/span><\/span>9<\/span>=<\/span><\/span>6<\/span>\u00d7<\/span><\/span>9<\/span>=<\/span><\/span>54<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

          • \n

            Luas total =\u00a0144+54=198<\/span>144<\/span>+<\/span><\/span>54<\/span>=<\/span><\/span>198<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n<\/ul>\n

            Jawaban: 198 cm\u00b2<\/strong><\/p>\n

            \n

            Soal 5 (Trapesium \u2013 Luas)<\/h3>\n

            Sebuah trapesium memiliki panjang sisi sejajar 18 cm dan 10 cm, serta tinggi 7 cm. Berapa luas trapesium tersebut?<\/p>\n

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

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            • \n

              L=12\u00d7(a+b)\u00d7t=12\u00d7(18+10)\u00d77=12\u00d728\u00d77=14\u00d77=98<\/span>L<\/span>=<\/span><\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>(<\/span>a<\/span>+<\/span><\/span>b<\/span>)<\/span>\u00d7<\/span><\/span>t<\/span>=<\/span><\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>(<\/span>18<\/span>+<\/span><\/span>10<\/span>)<\/span>\u00d7<\/span><\/span>7<\/span>=<\/span><\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>28<\/span>\u00d7<\/span><\/span>7<\/span>=<\/span><\/span>14<\/span>\u00d7<\/span><\/span>7<\/span>=<\/span><\/span>98<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n<\/ul>\n

              Jawaban: 98 cm\u00b2<\/strong><\/p>\n

              \n

              Soal 6 (Jajar Genjang \u2013 Luas)<\/h3>\n

              Sebuah jajar genjang memiliki alas 16 cm dan tinggi 9 cm. Berapa luasnya?<\/p>\n

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

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              • \n

                L=a\u00d7t=16\u00d79=144<\/span>L<\/span>=<\/span><\/span>a<\/span>\u00d7<\/span><\/span>t<\/span>=<\/span><\/span>16<\/span>\u00d7<\/span><\/span>9<\/span>=<\/span><\/span>144<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n<\/ul>\n

                Jawaban: 144 cm\u00b2<\/strong><\/p>\n

                \n

                Soal 7 (Belah Ketupat \u2013 Luas)<\/h3>\n

                Belah ketupat memiliki panjang diagonal 14 cm dan 20 cm. Berapa luasnya?<\/p>\n

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

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                • \n

                  L=12\u00d7d1\u00d7d2=12\u00d714\u00d720=7\u00d720=140<\/span>L<\/span>=<\/span><\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>d<\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>d<\/span>2<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>14<\/span>\u00d7<\/span><\/span>20<\/span>=<\/span><\/span>7<\/span>\u00d7<\/span><\/span>20<\/span>=<\/span><\/span>140<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n<\/ul>\n

                  Jawaban: 140 cm\u00b2<\/strong><\/p>\n

                  \n

                  Soal 8 (Layang-Layang \u2013 Luas)<\/h3>\n

                  Sebuah layang-layang memiliki diagonal 24 cm dan 15 cm. Berapa luasnya?<\/p>\n

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

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                  • \n

                    L=12\u00d7d1\u00d7d2=12\u00d724\u00d715=12\u00d715=180<\/span>L<\/span>=<\/span><\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>d<\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>d<\/span>2<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>24<\/span>\u00d7<\/span><\/span>15<\/span>=<\/span><\/span>12<\/span>\u00d7<\/span><\/span>15<\/span>=<\/span><\/span>180<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n<\/ul>\n

                    Jawaban: 180 cm\u00b2<\/strong><\/p>\n

                    \n

                    Soal 9 (Soal Memilih Cat \u2013 Efisiensi)<\/h3>\n

                    Pak Ahmad memiliki dinding berbentuk persegi panjang dengan ukuran 10 m \u00d7 6 m. Ia akan membeli cat. Merek A: 1 kaleng untuk 12 m\u00b2 dengan harga Rp50.000. Merek B: 1 kaleng untuk 15 m\u00b2 dengan harga Rp60.000. Merek mana yang lebih hemat?<\/p>\n

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

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                    • \n

                      Luas dinding =\u00a010\u00d76=60<\/span>10<\/span>\u00d7<\/span><\/span>6<\/span>=<\/span><\/span>60<\/span><\/span><\/span><\/span>\u00a0m\u00b2<\/p>\n<\/li>\n

                    • \n

                      Merek A: butuh\u00a060\u00f712=5<\/span>60<\/span>\u00f7<\/span><\/span>12<\/span>=<\/span><\/span>5<\/span><\/span><\/span><\/span>\u00a0kaleng \u00d7 Rp50.000 = Rp250.000<\/p>\n<\/li>\n

                    • \n

                      Merek B: butuh\u00a060\u00f715=4<\/span>60<\/span>\u00f7<\/span><\/span>15<\/span>=<\/span><\/span>4<\/span><\/span><\/span><\/span>\u00a0kaleng \u00d7 Rp60.000 = Rp240.000<\/p>\n<\/li>\n

                    • \n

                      Merek B lebih hemat Rp10.000<\/p>\n<\/li>\n<\/ul>\n

                      Jawaban: Merek B lebih hemat<\/strong><\/p>\n

                      \n

                      Soal 10 (Perubahan Luas Akibat Perubahan Ukuran)<\/h3>\n

                      Sebuah persegi memiliki sisi 6 cm. Jika sisi diperbesar menjadi 3 kali lipat, berapa kali lipat luasnya?<\/p>\n

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

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                      • \n

                        Luas awal =\u00a062=36<\/span>62<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>36<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

                      • \n

                        Sisi baru =\u00a06\u00d73=18<\/span>6<\/span>\u00d7<\/span><\/span>3<\/span>=<\/span><\/span>18<\/span><\/span><\/span><\/span>\u00a0cm, luas baru =\u00a0182=324<\/span>1<\/span>82<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>324<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

                      • \n

                        Perbandingan =\u00a0324:36=9<\/span>324<\/span>:<\/span><\/span>36<\/span>=<\/span><\/span>9<\/span><\/span><\/span><\/span>\u00a0kali lipat<\/p>\n<\/li>\n<\/ul>\n

                        Jawaban: 9 kali lipat<\/strong><\/p>\n

                        \n

                        Soal 11 (Luas Daerah Arsiran \u2013 Persegi dan Lingkaran)<\/h3>\n

                        Sebuah persegi dengan sisi 28 cm. Di dalamnya terdapat lingkaran yang menyinggung keempat sisinya. Berapa luas daerah di luar lingkaran? (\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

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                        • \n

                          Luas persegi =\u00a0282=784<\/span>2<\/span>82<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>784<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

                        • \n

                          Diameter lingkaran = 28 cm \u2192 jari-jari = 14 cm<\/p>\n<\/li>\n

                        • \n

                          Luas lingkaran =\u00a0227\u00d7142=227\u00d7196=22\u00d728=616<\/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 arsiran =\u00a0784\u2212616=168<\/span>784<\/span>\u2212<\/span><\/span>616<\/span>=<\/span><\/span>168<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n<\/ul>\n

                          Jawaban: 168 cm\u00b2<\/strong><\/p>\n

                          \n

                          Soal 12 (Luas Gabungan \u2013 Persegi Panjang dan Setengah Lingkaran)<\/h3>\n

                          Sebuah bangun terdiri dari persegi panjang (panjang 20 cm, lebar 14 cm) dan setengah lingkaran pada sisi lebarnya (diameter = lebar = 14 cm). Berapa luas total? (\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

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                          • \n

                            Luas persegi panjang =\u00a020\u00d714=280<\/span>20<\/span>\u00d7<\/span><\/span>14<\/span>=<\/span><\/span>280<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

                          • \n

                            Jari-jari setengah lingkaran =\u00a014\u00f72=7<\/span>14<\/span>\u00f7<\/span><\/span>2<\/span>=<\/span><\/span>7<\/span><\/span><\/span><\/span>\u00a0cm<\/p>\n<\/li>\n

                          • \n

                            Luas setengah lingkaran =\u00a012\u00d7\u03c0r2=12\u00d7227\u00d749=12\u00d722\u00d77=12\u00d7154=77<\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>\u03c0<\/span>r<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>7<\/span><\/span>22<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>49<\/span>=<\/span><\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>22<\/span>\u00d7<\/span><\/span>7<\/span>=<\/span><\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>154<\/span>=<\/span><\/span>77<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

                          • \n

                            Luas total =\u00a0280+77=357<\/span>280<\/span>+<\/span><\/span>77<\/span>=<\/span><\/span>357<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n<\/ul>\n

                            Jawaban: 357 cm\u00b2<\/strong><\/p>\n

                            \n

                            Soal 13 (Perbandingan Luas Dua Lingkaran)<\/h3>\n

                            Lingkaran A memiliki jari-jari 7 cm. Lingkaran B memiliki jari-jari 21 cm. Berapa perbandingan luas lingkaran A : B? (\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 A =\u00a0227\u00d772=227\u00d749=154<\/span>7<\/span><\/span>22<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>72<\/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>49<\/span>=<\/span><\/span>154<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

                            • \n

                              Luas B =\u00a0227\u00d7212=227\u00d7441=22\u00d763=1.386<\/span>7<\/span><\/span>22<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>2<\/span>12<\/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>441<\/span>=<\/span><\/span>22<\/span>\u00d7<\/span><\/span>63<\/span>=<\/span><\/span>1.386<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

                            • \n

                              Perbandingan A : B =\u00a0154:1.386=1:9<\/span>154<\/span>:<\/span><\/span>1.386<\/span>=<\/span><\/span>1<\/span>:<\/span><\/span>9<\/span><\/span><\/span><\/span><\/p>\n<\/li>\n<\/ul>\n

                              Jawaban: 1 : 9<\/strong><\/p>\n

                              \n

                              Soal 14 (Luas Daerah Gabungan \u2013 Persegi dan Dua Setengah Lingkaran)<\/h3>\n

                              Sebuah bangun berbentuk persegi dengan sisi 14 cm. Pada dua sisi yang berhadapan, terdapat setengah lingkaran dengan diameter sama dengan sisi persegi. Berapa luas total bangun? (\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 persegi =\u00a0142=196<\/span>1<\/span>42<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>196<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

                              • \n

                                Jari-jari setengah lingkaran =\u00a014\u00f72=7<\/span>14<\/span>\u00f7<\/span><\/span>2<\/span>=<\/span><\/span>7<\/span><\/span><\/span><\/span>\u00a0cm<\/p>\n<\/li>\n

                              • \n

                                Luas satu setengah lingkaran =\u00a012\u00d7227\u00d772=12\u00d722\u00d77=77<\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>7<\/span><\/span>22<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>72<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>22<\/span>\u00d7<\/span><\/span>7<\/span>=<\/span><\/span>77<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

                              • \n

                                Ada 2 setengah lingkaran =\u00a02\u00d777=154<\/span>2<\/span>\u00d7<\/span><\/span>77<\/span>=<\/span><\/span>154<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n

                              • \n

                                Luas total =\u00a0196+154=350<\/span>196<\/span>+<\/span><\/span>154<\/span>=<\/span><\/span>350<\/span><\/span><\/span><\/span>\u00a0cm\u00b2<\/p>\n<\/li>\n<\/ul>\n

                                Jawaban: 350 cm\u00b2<\/strong><\/p>\n

                                \n

                                Soal 15 (Soal Cerita \u2013 Selisih Luas Dua Taman)<\/h3>\n

                                Taman A berbentuk persegi dengan sisi 25 m. Taman B berbentuk lingkaran dengan diameter 28 m. Taman mana yang lebih luas dan berapa selisihnya? (\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 Taman A =\u00a0252=625<\/span>2<\/span>52<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>625<\/span><\/span><\/span><\/span>\u00a0m\u00b2<\/p>\n<\/li>\n

                                • \n

                                  Jari-jari Taman B =\u00a028\u00f72=14<\/span>28<\/span>\u00f7<\/span><\/span>2<\/span>=<\/span><\/span>14<\/span><\/span><\/span><\/span>\u00a0m<\/p>\n<\/li>\n

                                • \n

                                  Luas Taman B =\u00a0227\u00d7142=227\u00d7196=22\u00d728=616<\/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>\u00a0m\u00b2<\/p>\n<\/li>\n

                                • \n

                                  Taman A lebih luas dengan selisih =\u00a0625\u2212616=9<\/span>625<\/span>\u2212<\/span><\/span>616<\/span>=<\/span><\/span>9<\/span><\/span><\/span><\/span>\u00a0m\u00b2<\/p>\n<\/li>\n<\/ul>\n

                                  Jawaban: Taman A lebih luas, selisih 9 m\u00b2<\/strong><\/p>\n

                                  \n

                                  Ringkasan Rumus Penting Keliling dan Luas Bangun Datar<\/h2>\n
                                  \n
                                  \n
                                  <\/div>\n
                                  <\/div>\n<\/div>\n\n\n\n\n\n\n\n\n\n\n\n\n
                                  Bangun Datar<\/th>\nKeliling<\/th>\nLuas<\/th>\n<\/tr>\n<\/thead>\n
                                  Persegi<\/strong><\/td>\nK=4s<\/span>K<\/span>=<\/span><\/span>4<\/span>s<\/span><\/span><\/span><\/span><\/td>\nL=s2<\/span>L<\/span>=<\/span><\/span>s<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/td>\n<\/tr>\n
                                  Persegi panjang<\/strong><\/td>\nK=2(p+l)<\/span>K<\/span>=<\/span><\/span>2<\/span>(<\/span>p<\/span>+<\/span><\/span>l<\/span>)<\/span><\/span><\/span><\/span><\/td>\nL=p\u00d7l<\/span>L<\/span>=<\/span><\/span>p<\/span>\u00d7<\/span><\/span>l<\/span><\/span><\/span><\/span><\/td>\n<\/tr>\n
                                  Segitiga<\/strong><\/td>\nK=a+b+c<\/span>K<\/span>=<\/span><\/span>a<\/span>+<\/span><\/span>b<\/span>+<\/span><\/span>c<\/span><\/span><\/span><\/span><\/td>\nL=12\u00d7a\u00d7t<\/span>L<\/span>=<\/span><\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>a<\/span>\u00d7<\/span><\/span>t<\/span><\/span><\/span><\/span><\/td>\n<\/tr>\n
                                  Lingkaran<\/strong><\/td>\nK=2\u03c0r=\u03c0d<\/span>K<\/span>=<\/span><\/span>2<\/span>\u03c0<\/span>r<\/span>=<\/span><\/span>\u03c0<\/span>d<\/span><\/span><\/span><\/span><\/td>\nL=\u03c0r2<\/span>L<\/span>=<\/span><\/span>\u03c0<\/span>r<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/td>\n<\/tr>\n
                                  Trapesium<\/strong><\/td>\nK=a+b+c+d<\/span>K<\/span>=<\/span><\/span>a<\/span>+<\/span><\/span>b<\/span>+<\/span><\/span>c<\/span>+<\/span><\/span>d<\/span><\/span><\/span><\/span><\/td>\nL=12\u00d7(a+b)\u00d7t<\/span>L<\/span>=<\/span><\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>(<\/span>a<\/span>+<\/span><\/span>b<\/span>)<\/span>\u00d7<\/span><\/span>t<\/span><\/span><\/span><\/span><\/td>\n<\/tr>\n
                                  Jajar genjang<\/strong><\/td>\nK=2(a+b)<\/span>K<\/span>=<\/span><\/span>2<\/span>(<\/span>a<\/span>+<\/span><\/span>b<\/span>)<\/span><\/span><\/span><\/span><\/td>\nL=a\u00d7t<\/span>L<\/span>=<\/span><\/span>a<\/span>\u00d7<\/span><\/span>t<\/span><\/span><\/span><\/span><\/td>\n<\/tr>\n
                                  Belah ketupat<\/strong><\/td>\nK=4s<\/span>K<\/span>=<\/span><\/span>4<\/span>s<\/span><\/span><\/span><\/span><\/td>\nL=12\u00d7d1\u00d7d2<\/span>L<\/span>=<\/span><\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>d<\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>d<\/span>2<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/td>\n<\/tr>\n
                                  Layang-layang<\/strong><\/td>\nK=2(a+b)<\/span>K<\/span>=<\/span><\/span>2<\/span>(<\/span>a<\/span>+<\/span><\/span>b<\/span>)<\/span><\/span><\/span><\/span><\/td>\nL=12\u00d7d1\u00d7d2<\/span>L<\/span>=<\/span><\/span>2<\/span><\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>d<\/span>1<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>d<\/span>2<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

                                  Panduan Nilai\u00a0\u03c0<\/span>\u03c0<\/span><\/span><\/span><\/span>:<\/strong><\/p>\n

                                    \n
                                  • \n

                                    Gunakan\u00a0\u03c0=227<\/span>\u03c0<\/span>=<\/span><\/span>7<\/span><\/span>22<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00a0jika jari-jari atau diameter kelipatan 7<\/p>\n<\/li>\n

                                  • \n

                                    Gunakan\u00a0\u03c0=3,14<\/span>\u03c0<\/span>=<\/span><\/span>3<\/span>,<\/span>14<\/span><\/span><\/span><\/span>\u00a0jika jari-jari atau diameter bukan kelipatan 7<\/p>\n<\/li>\n<\/ul>\n

                                    \n

                                    Sahabat Rumah Pintar<\/strong>, dengan menguasai 15 soal di atas, kalian sudah siap menghitung keliling dan luas berbagai bangun datar, menyelesaikan masalah luas gabungan, membandingkan luas, dan memilih opsi paling efisien dalam soal TKA Matematika SMP. Ingat kuncinya:\u00a0hafalkan rumus, perhatikan satuan, dan untuk luas gabungan, pecah menjadi bangun-bangun sederhana<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"

                                    Bimbel Rumah Pintar – Soal keliling dan luas bangun datar adalah konsep geometri yang paling sering muncul dalam kehidupan sehari-hari sekaligus menjadi langganan soal TKA Matematika SMP. Prinsip dasar yang harus dikuasai adalah hafal rumus keliling dan luas berbagai bangun […]<\/p>\n","protected":false},"author":1,"featured_media":2154,"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":[238,241,240,224],"class_list":["post-2151","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-bahas-soal","tag-bangun-datar","tag-keliling-bangun-datar","tag-luas-bangun-datar","tag-soal-tka-matematika-smp"],"_links":{"self":[{"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/posts\/2151","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=2151"}],"version-history":[{"count":2,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/posts\/2151\/revisions"}],"predecessor-version":[{"id":2158,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/posts\/2151\/revisions\/2158"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/media\/2154"}],"wp:attachment":[{"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/media?parent=2151"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/categories?post=2151"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/rumahpintar.co.id\/blog\/wp-json\/wp\/v2\/tags?post=2151"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}