Question

question 5 of 10
is the following shape a right triangle? how do you know?
a. no, the side lengths do not fit the pythagorean theorem.
b. there is not enough information to determine.
c. yes, two sides are perpendicular, and the side lengths fit the pythagorean theorem.
d. no, there is no right angle.

Explanation:

Step1: Identify coordinates of points

From the grid: $A(-3, 4)$, $B(-1, 2)$, $C(-4, 0)$

Step2: Calculate side lengths (distance formula)

Distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

• $AB$: $\sqrt{(-1-(-3))^2+(2-4)^2}=\sqrt{(2)^2+(-2)^2}=\sqrt{4+4}=\sqrt{8}$
• $BC$: $\sqrt{(-4-(-1))^2+(0-2)^2}=\sqrt{(-3)^2+(-2)^2}=\sqrt{9+4}=\sqrt{13}$
• $AC$: $\sqrt{(-4-(-3))^2+(0-4)^2}=\sqrt{(-1)^2+(-4)^2}=\sqrt{1+16}=\sqrt{17}$

Step3: Check Pythagorean theorem

Test if $(\sqrt{8})^2+(\sqrt{13})^2=(\sqrt{17})^2$:
$8+13=21$, $17
eq21$
Test other pairs: $(\sqrt{8})^2+(\sqrt{17})^2=8+17=25
eq13$; $(\sqrt{13})^2+(\sqrt{17})^2=13+17=30
eq8$

Step4: Check slopes for perpendicularity

Slope formula: $m=\frac{y_2-y_1}{x_2-x_1}$

• Slope of $AB$: $\frac{2-4}{-1-(-3)}=\frac{-2}{2}=-1$
• Slope of $BC$: $\frac{0-2}{-4-(-1)}=\frac{-2}{-3}=\frac{2}{3}$
• Slope of $AC$: $\frac{0-4}{-4-(-3)}=\frac{-4}{-1}=4$

No pair of slopes multiply to $-1$, so no perpendicular sides.

Answer:

A. No, the side lengths do not fit the Pythagorean theorem.
D. No, there is no right angle.