Question

find the approximate perimeter of △abc plotted below.
a(-4,1), b(-2,3), c(3,-4)
choose 1 answer:
a 20.0
b 21.2
c 22.4
d 23.6

Explanation:

Step1: Calculate AB length

Use distance formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). For \(A(-4,1)\) and \(B(-2,3)\):
\(AB=\sqrt{(-2 - (-4))^2 + (3 - 1)^2}=\sqrt{(2)^2 + (2)^2}=\sqrt{4 + 4}=\sqrt{8}\approx2.83\)

Step2: Calculate BC length

For \(B(-2,3)\) and \(C(3,-4)\):
\(BC=\sqrt{(3 - (-2))^2 + (-4 - 3)^2}=\sqrt{(5)^2 + (-7)^2}=\sqrt{25 + 49}=\sqrt{74}\approx8.60\)

Step3: Calculate AC length

For \(A(-4,1)\) and \(C(3,-4)\):
\(AC=\sqrt{(3 - (-4))^2 + (-4 - 1)^2}=\sqrt{(7)^2 + (-5)^2}=\sqrt{49 + 25}=\sqrt{74}\approx8.60\) Wait, no, recalculate: \(3 - (-4)=7\), \(-4 - 1=-5\), so \(AC=\sqrt{7^2 + (-5)^2}=\sqrt{49 + 25}=\sqrt{74}\approx8.60\)? Wait, no, earlier AB was ~2.83, BC ~8.60, AC: let's recalculate AC. Wait, \(A(-4,1)\), \(C(3,-4)\): \(x\) difference \(3 - (-4)=7\), \(y\) difference \(-4 - 1=-5\). So \(AC=\sqrt{7^2 + (-5)^2}=\sqrt{49 + 25}=\sqrt{74}\approx8.60\)? Wait, but then perimeter would be \(2.83 + 8.60 + 8.60\approx20.03\)? No, that's wrong. Wait, no, I made a mistake. Wait, AB: \(A(-4,1)\), \(B(-2,3)\): \(x\) difference \(-2 - (-4)=2\), \(y\) difference \(3 - 1=2\), so \(AB=\sqrt{2^2 + 2^2}=\sqrt{8}\approx2.83\). BC: \(B(-2,3)\), \(C(3,-4)\): \(x\) difference \(3 - (-2)=5\), \(y\) difference \(-4 - 3=-7\), so \(BC=\sqrt{5^2 + (-7)^2}=\sqrt{25 + 49}=\sqrt{74}\approx8.60\). AC: \(A(-4,1)\), \(C(3,-4)\): \(x\) difference \(3 - (-4)=7\), \(y\) difference \(-4 - 1=-5\), so \(AC=\sqrt{7^2 + (-5)^2}=\sqrt{49 + 25}=\sqrt{74}\approx8.60\)? Wait, no, that can't be, because triangle with two sides ~8.6 and one ~2.83. Wait, maybe I messed up AC. Wait, no, let's check coordinates again. \(A(-4,1)\), \(C(3,-4)\): horizontal distance 7, vertical distance 5 (since 1 to -4 is 5 units down). So \(AC=\sqrt{7^2 + 5^2}=\sqrt{49 + 25}=\sqrt{74}\approx8.60\). Then perimeter is \(AB + BC + AC\approx2.83 + 8.60 + 8.60\approx20.03\)? But the options have 21.2, 22.4, etc. Wait, I must have messed up BC. Wait, \(B(-2,3)\), \(C(3,-4)\): \(x\) difference 5, \(y\) difference -7 (3 to -4 is -7). So \(BC=\sqrt{5^2 + (-7)^2}=\sqrt{25 + 49}=\sqrt{74}\approx8.60\). Wait, maybe AB is wrong? \(A(-4,1)\), \(B(-2,3)\): \(x\) difference 2, \(y\) difference 2, so \(AB=\sqrt{4 + 4}=\sqrt{8}\approx2.83\). Wait, maybe AC is wrong. Wait, \(A(-4,1)\), \(C(3,-4)\): \(x\) from -4 to 3 is 7, \(y\) from 1 to -4 is -5, so squared is 49 + 25 = 74, square root ~8.60. Wait, but then perimeter is ~2.83 + 8.60 + 8.60 = ~20.03, which is option A? But the options have A as 20.0. Wait, maybe I made a mistake in BC. Wait, no, let's recalculate BC: \(B(-2,3)\), \(C(3,-4)\): \(x\) difference 3 - (-2) = 5, \(y\) difference -4 - 3 = -7. So \(5^2=25\), \(-7^2=49\), sum 74, square root ~8.60. AC: \(A(-4,1)\), \(C(3,-4)\): \(x\) difference 7, \(y\) difference -5, sum 49 +25=74, square root ~8.60. AB: ~2.83. So total ~2.83 +8.60 +8.60=20.03, which is ~20.0, option A. But maybe I messed up the coordinates. Wait, the graph: A(-4,1), B(-2,3), C(3,-4). Let me check again. Alternatively, maybe I mixed up B and C? No, the coordinates are given. Wait, maybe the distance formula was applied wrong. Wait, distance formula is \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). So for AB: \(x_2=-2\), \(x_1=-4\), so \(x_2 - x_1=2\); \(y_2=3\), \(y_1=1\), \(y_2 - y_1=2\). So AB=\(\sqrt{2^2 + 2^2}=\sqrt{8}\approx2.83\). For BC: \(x_2=3\), \(x_1=-2\), \(x_2 - x_1=5\); \(y_2=-4\), \(y_1=3\), \(y_2 - y_1=-7\). So BC=\(\sqrt{5^2 + (-7)^2}=\sqrt{25 + 49}=\sqrt{74}\approx8.60\). For AC: \(x_2=3\), \(x_1=-4\), \(x_2 - x_1=7\); \(y_2=-4\), \(y_1=1\), \(y_2 - y_1=-5\). So AC=\(\sqrt{7^2 + (-5)^2}=\sqrt{49 + 25}=\sqrt{74}\approx8.60\). So perimeter is ~2.83 +8.60 +8.60=20.03, which is approximately 20.0, option A.

Answer:

A. 20.0