Question

1. triangle abc has vertices at a (3, 5), b (5, 7) and c (7, 5). lisa believes that the perimeter of △abc is 12 units. why is lisa incorrect? a. because ab = 2√2 units, bc = 2√2 units and ac = 4 units. b. because ab = 8 units, bc = 8 units and ac = 16 units. c. because ab = 2 units, bc = 2 units and ac = 2 units. d. because ab = 4 units, bc = 4 units and ac = 8 units.

Explanation:

Answer:

We first need to find the lengths of the sides of the triangle using the distance - formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For side \(AB\) with \(A(3,5)\) and \(B(5,7)\):
\[

$$\begin{align*} AB&=\sqrt{(5 - 3)^2+(7 - 5)^2}\\ &=\sqrt{2^2+2^2}\\ &=\sqrt{4 + 4}\\ &=\sqrt{8}=2\sqrt{2} \end{align*}$$

\]
For side \(BC\) with \(B(5,7)\) and \(C(7,5)\):
\[

$$\begin{align*} BC&=\sqrt{(7 - 5)^2+(5 - 7)^2}\\ &=\sqrt{2^2+( - 2)^2}\\ &=\sqrt{4 + 4}\\ &=\sqrt{8}=2\sqrt{2} \end{align*}$$

\]
For side \(AC\) with \(A(3,5)\) and \(C(7,5)\):
\[

$$\begin{align*} AC&=\sqrt{(7 - 3)^2+(5 - 5)^2}\\ &=\sqrt{4^2+0^2}\\ &=4 \end{align*}$$

\]
The perimeter \(P = AB+BC + AC=2\sqrt{2}+2\sqrt{2}+4=4 + 4\sqrt{2}
eq12\).
Lisa is incorrect because \(AB = 2\sqrt{2}\) units, \(BC = 2\sqrt{2}\) units and \(AC = 4\) units. So the answer is A. Because \(AB = 2\sqrt{2}\) units, \(BC = 2\sqrt{2}\) units and \(AC = 4\) units.