Question

suppose a triangle is drawn on a coordinate plane. using vertices with the coordinates (1, 1), (4, 1), and (5, 4), compute the perimeter. round your answer to the nearest tenth. (1 point)
o 10.6 units
o 11 units
o 11.2 units
o 12 units

Explanation:

Step1: Identify the sides

First, we need to find the lengths of the sides of the triangle using the distance formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) (and \(z\)-coordinate if needed, but here \(z = 1\) for the first two points, and \(z = 4\) for the third, wait, actually the coordinates are \((1,1,1)\), \((4,1,1)\), and \((5,4,1)\)? Wait, maybe it's a typo, but looking at the options, let's assume the third point is \((5,4,1)\) or maybe 2D? Wait, the first two points: \((1,1)\) and \((4,1)\) (assuming 2D, \(z = 1\) is same, so distance is \(|4 - 1| = 3\) units. Then between \((4,1)\) and \((5,4)\): distance is \(\sqrt{(5 - 4)^2 + (4 - 1)^2} = \sqrt{1 + 9} = \sqrt{10} \approx 3.16\). Then between \((1,1)\) and \((5,4)\): \(\sqrt{(5 - 1)^2 + (4 - 1)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\)? Wait, no, maybe the third point is \((5,4)\) (2D). Wait, maybe the problem is to find the perimeter? Wait, the options are 10.5, 11, 11.2, 12. Let's recalculate.

Wait, maybe the coordinates are \((1,1)\), \((4,1)\), and \((5,4)\) (2D, \(z = 1\) is a typo or 3D with same \(z\)). So:

Side 1: between \((1,1)\) and \((4,1)\): \(d_1 = \sqrt{(4 - 1)^2 + (1 - 1)^2} = \sqrt{9 + 0} = 3\)

Side 2: between \((4,1)\) and \((5,4)\): \(d_2 = \sqrt{(5 - 4)^2 + (4 - 1)^2} = \sqrt{1 + 9} = \sqrt{10} \approx 3.16\)

Side 3: between \((5,4)\) and \((1,1)\): \(d_3 = \sqrt{(1 - 5)^2 + (1 - 4)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\)? No, that can't be. Wait, maybe the third point is \((5,4)\) and the first two are \((1,1)\) and \((4,1)\), and we need to find the perimeter? Wait, 3 + \(\sqrt{10}\) + 5 ≈ 3 + 3.16 + 5 = 11.16 ≈ 11.2? Wait, no, maybe I misread the coordinates. Wait, the original problem says: vertices with coordinates (1,1,1), (4,1,1), and (5,4,1). So 3D, but \(z\)-coordinate is 1 for first two, 1 for first two, 1 for third? Wait, no, (5,4,1) has \(z = 1\), same as others. So 2D in \(x\)-\(y\) plane, \(z = 1\) is constant. So:

Distance between (1,1,1) and (4,1,1): \(d_1 = \sqrt{(4 - 1)^2 + (1 - 1)^2 + (1 - 1)^2} = \sqrt{9 + 0 + 0} = 3\)

Distance between (4,1,1) and (5,4,1): \(d_2 = \sqrt{(5 - 4)^2 + (4 - 1)^2 + (1 - 1)^2} = \sqrt{1 + 9 + 0} = \sqrt{10} \approx 3.16\)

Distance between (5,4,1) and (1,1,1): \(d_3 = \sqrt{(1 - 5)^2 + (1 - 4)^2 + (1 - 1)^2} = \sqrt{16 + 9 + 0} = \sqrt{25} = 5\)? No, that's 5, but 3 + 3.16 + 5 ≈ 11.16, which is approximately 11.2. Wait, maybe the third point is (5,4, something else)? Wait, maybe the third point is (5,4,1) and the first two are (1,1,1) and (4,1,1). Then the sides are 3, \(\sqrt{10}\), and \(\sqrt{(5 - 1)^2 + (4 - 1)^2} = \sqrt{16 + 9} = 5\)? No, that's 3, ~3.16, 5. Sum is 3 + 3.16 + 5 = 11.16 ≈ 11.2.

Step2: Calculate the perimeter

Sum the lengths: \(3 + \sqrt{10} + 5 \approx 3 + 3.16 + 5 = 11.16 \approx 11.2\)

Answer:

11.2 units (Option C: 11.2 units)