Given that Upper Delta ABC is a triangle with sides of lengths​ a, b, and c such that csquaredgreater thanasquaredplusbsquared​, prove Upper Delta ABC is an obtuse triangle.

Question

A
B
C
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c
Question content area bottom
Part 1
Draw a right triangle Upper Delta EFG with legs of lengths a and b. Label the hypotenuse x. Determine what needs to be proved to prove Upper Delta ABC is an obtuse triangle.
​Prove: mangleC
greater than
mangleG
A
B
C
a
b
c
E
F
G
a
b
x
Part 2
Given that csquaredgreater thanasquaredplusbsquared use
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for the Upper Delta EFG to substitute
  
enter your response here for a squared plus b squared to get c
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enter your response here.

Answer 1

To prove that triangle \( \Delta ABC \) is an obtuse triangle given that \( c^2 > a^2 + b^2 \), we will take two steps as outlined in the question.

Part 1: Determine the Proof

In triangle \( \Delta ABC \), let \( c \) be the side opposite angle \( C \), and let \( a \) and \( b \) be the lengths of the other two sides adjacent to angle \( C \).

To prove that \( \Delta ABC \) is an obtuse triangle, we need to show that:

\[ m \angle C > 90^\circ \]

This means that angle \( C \) is greater than a right angle.

Part 2: Substitute using the Right Triangle \( \Delta EFG \)

Now, consider a right triangle \( \Delta EFG \) with legs \( a \) and \( b \) (where \( EF = a \) and \( FG = b \)), and hypotenuse \( x \) (where \( EG = x \)).

According to the Pythagorean theorem, we have:

\[ x^2 = a^2 + b^2 \]

Now, given that \( c^2 > a^2 + b^2 \), we can substitute \( a^2 + b^2 \) with \( x^2 \) in the inequality:

\[ c^2 > x^2 \]

This implies that the square of the side \( c \) (opposite angle \( C \) in triangle \( \Delta ABC \)) is greater than the square of the hypotenuse \( x \) of triangle \( \Delta EFG \).

Since \( x^2 \) is the sum of the squares of the legs \( a \) and \( b \), having \( c^2 > x^2 \) means that \( c \) is longer than the hypotenuse of triangle \( \Delta EFG \).

By the properties of triangles, if a side opposite an angle exceeds the length of the hypotenuse formed by the other two sides (as in a right triangle), the angle opposite this side must be obtuse. Thus:

\[ m \angle C > 90^\circ \]

In summary for the completion of Part 2, if we fill in the blanks:

Given that \( c^2 > a^2 + b^2 \) use \( x^2 \) for the triangle \( \Delta EFG \) to substitute \( a^2 + b^2 \) to get \( c \) greater than \( x \).

This completes the proof that triangle \( \Delta ABC \) is an obtuse triangle.