Show that the points form the vertices of the indicated polygon.
Right triangle: (3, 5), (5, 6), (6, 4)

A) $( \sqrt { 5 } ) ^ { 2 } + ( \sqrt { 5 } ) ^ { 2 } = ( \sqrt { 10 } ) ^ { 2 }$
B) $( \sqrt { 31 } ) ^ { 2 } + ( \sqrt { 5 } ) ^ { 2 } = ( \sqrt { 10 } ) ^ { 2 }$
C) $( \sqrt { 7 } ) ^ { 2 } + ( \sqrt { 5 } ) ^ { 2 } = ( \sqrt { 10 } ) ^ { 2 }$
D) $( \sqrt { 11 } ) ^ { 2 } + ( \sqrt { 5 } ) ^ { 2 } = ( \sqrt { 10 } ) ^ { 2 }$
E) $( \sqrt { 34 } ) ^ { 2 } + ( \sqrt { 5 } ) ^ { 2 } = ( \sqrt { 10 } ) ^ { 2 }$

Question

Show that the points form the vertices of the indicated polygon.  
Right triangle: (3, 5), (5, 6), (6, 4)
 
A) $( \sqrt { 5 } ) ^ { 2 } + ( \sqrt { 5 } ) ^ { 2 } = ( \sqrt { 10 } ) ^ { 2 }$
B) $( \sqrt { 31 } ) ^ { 2 } + ( \sqrt { 5 } ) ^ { 2 } = ( \sqrt { 10 } ) ^ { 2 }$
C) $( \sqrt { 7 } ) ^ { 2 } + ( \sqrt { 5 } ) ^ { 2 } = ( \sqrt { 10 } ) ^ { 2 }$
D) $( \sqrt { 11 } ) ^ { 2 } + ( \sqrt { 5 } ) ^ { 2 } = ( \sqrt { 10 } ) ^ { 2 }$
E) $( \sqrt { 34 } ) ^ { 2 } + ( \sqrt { 5 } ) ^ { 2 } = ( \sqrt { 10 } ) ^ { 2 }$

Quizplus · Accepted Answer

The distances between the points can be calculated using the distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Calculating the distances between each pair of points:- Between (3, 5) and (5, 6), the distance is $\sqrt{(5-3)^2 + (6-5)^2} = \sqrt{4 + 1} = \sqrt{5}$.- Between (5, 6) and (6, 4), the distance is $\sqrt{(6-5)^2 + (4-6)^2} = \sqrt{1 + 4} = \sqrt{5}$.- Between (3, 5) and (6, 4), the distance is $\sqrt{(6-3)^2 + (4-5)^2} = \sqrt{9 + 1} = \sqrt{10}$.Thus, the sides of the triangle are $\sqrt{5}$, $\sqrt{5}$, and $\sqrt{10}$, which satisfies the Pythagorean theorem ($a^2 + b^2 = c^2$), indicating a right triangle.

Show That the Points Form the Vertices of the Indicated (5)2+(5)2=(10)2( \sqrt { 5 } ) ^ { 2 } + ( \sqrt { 5 } ) ^ { 2 } = ( \sqrt { 10 } ) ^ { 2 }(5​)2+(5​)2=(10​)2

Show That the Points Form the Vertices of the Indicated $( \sqrt { 5 } ) ^ { 2 } + ( \sqrt { 5 } ) ^ { 2 } = ( \sqrt { 10 } ) ^ { 2 }$