The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
(A) 5 ย (B) 12ย (C) 11 ย (D) 7 + โ5
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Last updated at Oct. 23, 2021 by
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Question 9 The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is (A) 5 (B) 12 (C) 11 (D) 7 + โ5 Let A (0, 4), B (0, 0) and C (3, 0) Thus, Perimeter of triangle = AB + BC + AC Finding AB AB = โ((๐ฅ2 โ๐ฅ1)2+(๐ฆ2 โ๐ฆ1)2) = โ(( 0โ0)2+(0 โ4)2) = โ(02+(โ4)2) = โ16 = โ(4^2 ) = 4 units Finding AC Here, A (0, 4) and C (3, 0) AC = โ((๐ฅ2 โ๐ฅ1)2+(๐ฆ2 โ๐ฆ1)2) = โ((3โ0)2+(0 โ4)2) = โ(32+(โ4)2) = โ(9+ 16) = โ25 = โ(5^2 ) = 5 units Finding BC Here, B (0, 0) and C (3, 0) BC = โ((๐ฅ2 โ๐ฅ1)2+(๐ฆ2 โ๐ฆ1)2) = โ((3โ0)2+(0 โ0)2) = โ(32+(0)2) = โ(9+0) = โ9 = โ(3^2 ) = 3 units Thus, AB = 4, AC = 5, BC = 3 Therefore, Perimeter of triangle = AB + BC + AC = 4 + 3 + 5 = 12 units So, the correct answer is (B)
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