In a right angle triangle $A B C$ with sides $a, b, c;$
If pythagoras theorem states that $a^{2} + b^{2} = c^{2}$ then which of the following is true:

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

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

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c = \sqrt{a^{2}} \times b^{2}

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c = \sqrt{a^{2}} \div b^{2}

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Solution

The correct option is A $c = \sqrt{a^{2} + b^{2}}$
If $a^{2} + b^{2} = c^{2}$ , then is $c = \sqrt{a^{2} + b^{2}}$
Therefore, A is the correct answer.

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Similar questions

|\begin{matrix} - a (b^{2} + c^{2} - a^{2}) & 2 b^{3} & 2 c^{3} \\ 2 a^{3} & - b (c^{2} + a^{2} - b^{2}) & 2 c^{3} \\ 2 a^{3} & 2 b^{3} & - c (a^{2} + b^{2} - c^{2}) \end{matrix}| = a b c {(a^{2} + b^{2} + c^{2})}^{3}

A B C

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)

In a triangle ABC, $2 c a s i n \frac{A - B + C}{2}$ is equal to

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}}

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