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Sum of Two Consecutive Triangular Numbers Is a Square Number

If three numb...

Question

If three numbers a, b, c form a Pythagorean triplet, then

a² + b² = c²

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a² + b² + c²= 0

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a² x b² = c²

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a² - b² = c²

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Solution

The correct option is A
a² + b² = c²

If the numbers a, b, c form a Pythagorean triplet then, a² + b² = c².

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Similar questions

(a^{2} - b^{2}) (b^{2} + c^{2}) = (b^{2} - c^{2}) (a^{2} + b^{2})

a (b^{2} + c^{2}) = c (a^{2} + b^{2})

a, b, c

a^{2} = b^{2} + c^{2} - 2 b c \sqrt{1 - a^{2}}, b^{2} = c^{2} + a^{2} - 2 c a \sqrt{1 - b^{2}}, c^{2} = a^{2} + b^{2} - 2 a b \sqrt{1 - c^{2}}

a = c \sqrt{1 - b^{2}} + b \sqrt{1 - c^{2}}

a, b, c

A, B, C \in R - (0}

a A + b B + c C + \sqrt{(a^{2} + b^{2} + c^{2}) (A^{2} + B^{2} + C^{2})} = 0

\frac{a B}{b A} + \frac{b C}{c B} + \frac{c A}{a C}

a + b + c = 0

\frac{1}{b^{2} + c^{2} - a^{2}} + \frac{1}{c^{2} + a^{2} - b^{2}} + \frac{1}{a^{2} + b^{2} - c^{2}}

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

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