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In the xy plane, the vertices of a triangle have coordinates(0,0), (3,3), and (7,0). What is the perimeter of the triangle? (A) 13 (B) 34 (C) root 43 (d) 7+root3 (E) 12 +3root2
The formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).
Distance between (0, 0) and (7, 0) is \(d=7\); Distance between (0, 0) and (3, 3) is \(d=\sqrt{(0-3)^2+(0-3)^2}=\sqrt{18}=3\sqrt{2}\); Distance between (3, 3) and (7, 0) is \(d=\sqrt{(3-7)^2+(3-0)^2}=5\);
In the xy plane, the vertices of a triangle have coordinates(0,0), (3,3), and (7,0). What is the perimeter of the triangle? (A) 13 (B) 34 (C) root 43 (d) 7+root3 (E) 12 +3root2
Get distance between (0,0) and (3,3): \(\sqrt{(3 - 0)^2 + (3-0)^2} = \sqrt{(18)} = 3 \sqrt{(2)}\) Get distance between (3,3) and (7,0): \(\sqrt{(7-3)^2 + (3-0)^2} = \sqrt{16 + 9}= \sqrt{25}=5\) Get distance between (0,0) and (7,0): 7
Re: In the xy-plane, the vertices of a triangle have coordinates (0,0),(3,
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31 Oct 2016, 17:09
Relatively straight-forward problem, though I'd suggest plotting it so you can visualize the sides and what two side lengths you will need to calculate.
Base length is 7
(1) Distance between (0,0) & (3,3) --> sqrt[(3^2)+(3^2)] = 3sqrt(2)
Re: In the xy plane, the vertices of a triangle have coordinates
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15 Oct 2019, 03:06
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88% (01:40) correct 
