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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.

Ask by Klein Carroll. in the United States
Feb 25,2025

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5. The perimeter of triangle ABC is 30 + 10√5. 6. Triangle DEF is isosceles.

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Let's solve each problem step by step. =============================================================================== Problem 5: Find the perimeter of triangle A B C with vertices A(2, 4), B(8, 12), and C(24, 0). Step 1. Calculate the length of side AB:   AB = √[(8 − 2)² + (12 − 4)²]     = √[6² + 8²]     = √[36 + 64]     = √100     = 10 Step 2. Calculate the length of side BC:   BC = √[(24 − 8)² + (0 − 12)²]     = √[16² + (−12)²]     = √[256 + 144]     = √400     = 20 Step 3. Calculate the length of side CA:   CA = √[(24 − 2)² + (0 − 4)²]     = √[22² + (−4)²]     = √[484 + 16]     = √500     = 10√5             (since √500 = √(100×5) = 10√5) Step 4. Find the perimeter:   Perimeter = AB + BC + CA         = 10 + 20 + 10√5         = 30 + 10√5 =============================================================================== Problem 6: Determine whether triangle D E F with vertices D(6, -6), E(39, -12), and F(24, 18) is scalene, isosceles, or equilateral. Step 1. Calculate the length of side DE:   DE = √[(39 − 6)² + (−12 − (−6))²]     = √[33² + (−6)²]     = √[1089 + 36]     = √1125     = 15√5    (since √1125 = √(225×5) = 15√5) Step 2. Calculate the length of side EF:   EF = √[(24 − 39)² + (18 − (−12))²]     = √[(−15)² + 30²]     = √[225 + 900]     = √1125     = 15√5 Step 3. Calculate the length of side FD:   FD = √[(6 − 24)² + (−6 − 18)²]     = √[(−18)² + (−24)²]     = √[324 + 576]     = √900     = 30 Step 4. Determine the type of triangle:   Since DE = EF = 15√5 and FD = 30, the triangle has two equal sides.   Therefore, triangle D E F is an isosceles triangle. =============================================================================== Final Answers: 5. The perimeter of triangle A B C is 30 + 10√5. 6. Triangle D E F is isosceles.

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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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