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What is the perimeter of the triangle with the vertices A(–4, 2), B(0, –1) and C(3, 3) ?(a) \(7+3\sqrt2\)(b) \(10+5\sqrt2\)(c) \(11+6\sqrt2\)(d) \(5+\sqrt2\) |
Answer» (b)10 + \(5\sqrt2\) Perimeter of ΔABC = AB + BC + CA = \(\sqrt{(0+4)^2+(-1-2)^2}\) + \(\sqrt{(3-0)^2+(3+1)^2}\) + \(\sqrt{(3-4)^2+(3-2)^2}\) = \(\sqrt{16+9}\) + \(\sqrt{9+16}\) +\(\sqrt{49+1}\) = \(\sqrt{25}\) + \(\sqrt{25}\) + \(\sqrt{50}\) = 5 + 5 + \(5\sqrt2\) = 10 + \(5\sqrt2\) |
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