5. Find the perimeter of \( \triangle A B C \) with vertices \( A(2,4), B(8,12) \), and \( C(24,0) \). 6. Determine whether \( \triangle D E F \) with vertices \( D(6,-6), E(39,-12) \), and \( F(24,18) \) is scalene, isosceles, or equilateral.
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To find the perimeter of \( \triangle ABC \), calculate the lengths of its sides using the distance formula. The lengths of \( AB \), \( BC \), and \( CA \) are given by: 1. \( AB = \sqrt{(8-2)^2 + (12-4)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \) 2. \( BC = \sqrt{(24-8)^2 + (0-12)^2} = \sqrt{256 + 144} = \sqrt{400} = 20 \) 3. \( CA = \sqrt{(2-24)^2 + (4-0)^2} = \sqrt{484 + 16} = \sqrt{500} = 10\sqrt{5} \) Adding these lengths gives the perimeter: \( P = 10 + 20 + 10\sqrt{5} \). For \( \triangle DEF \), compute the lengths of its sides: 1. \( DE = \sqrt{(39-6)^2 + (-12+6)^2} = \sqrt{1089 + 36} = \sqrt{1125} \) 2. \( EF = \sqrt{(24-39)^2 + (18+12)^2} = \sqrt{225 + 900} = \sqrt{1125} \) 3. \( FD = \sqrt{(24-6)^2 + (18+6)^2} = \sqrt{324 + 576} = \sqrt{900} = 30 \) Since \( DE = EF \) and are equal, while \( FD \) is different, \( \triangle DEF \) is classified as isosceles!
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