Szczegółowe wyjaśnienie:
pole =(a+b)×h÷2
[tex](4 \sqrt{3 } + 5 \sqrt{3} + \sqrt{12} ) \times \sqrt{24} \div 2 = (8 \sqrt{3} + 2 \sqrt{3}) \times 2 \sqrt{6} \div 2 = 10 \sqrt{3} \times 2 \sqrt{6} \div 2 = 20 \sqrt{18} \div 2 = 10 \sqrt{18 } = 10 \times 3 \sqrt{2} = 30 \sqrt{2} [/tex]
obwod
[tex]4 \sqrt{3 } + \sqrt{27} + 5 \sqrt{3} + \sqrt{12} + \sqrt{36} = 4 \sqrt{3} + 3 \sqrt{3} + 5 \sqrt{3} + 2 \sqrt{3 } + 6 = 14 \sqrt{3} + 6[/tex]
