Respuesta :
Answer:
[tex]4,10,11[/tex] not Right Triangle
[tex]4, 4, \sqrt{32}[/tex] Right Triangle
[tex]8, 7, \sqrt{113}[/tex] Right Triangle.
[tex]4, 4, \sqrt{24}[/tex] not Right Triangle.
[tex]7, 8, 15[/tex] not Right Triangle.
Step-by-step explanation:
To determine if a set of side lengths corresponds to a right triangle, the sides must satisfy the Pitagorean identity:
[tex]side1^{2} + side2^{2} = hypotenuse^{2}[/tex]
Notice that the hypotenuse is always the greatest side of the set...
Let's plug the given values into the idenity above.
Option 1
[tex]4^{2} + 10^{2} =_{?} 11^{2}[/tex]
[tex]16 + 100 =_{?} 121[/tex]
[tex]116 \neq 121[/tex]
Option 2
[tex]4^{2} + 4^{2} =_{?} (\sqrt{32})^{2}[/tex]
[tex]16 + 16 =_{?} 32[/tex]
[tex]32 = 32[/tex]
Option 3
[tex]8^{2} + 7^{2} =_{?} (\sqrt{113})^{2}[/tex]
[tex]64 + 49 =_{?} 113[/tex]
[tex]113 = 113[/tex]
Option 4
[tex]4^{2} + 4^{2} =_{?} (\sqrt{24})^{2}[/tex]
[tex]16 + 16 =_{?} 24[/tex]
[tex]32 \neq 24[/tex]
Option 5
[tex]7^{2} + 8^{2} =_{?} 15^{2}[/tex]
[tex]49 + 64 =_{?} 225[/tex]
[tex]113 \neq 121[/tex]