The Pythagoras theorem , In the right triangle, the square of the hypotenuse is equal to the sum of other two sides. What are we proving here?

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

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

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

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a^{2} x b^{2} = c^{2}

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Solution

The correct option is C $a^{2} + b^{2} = c^{2}$
Let $a, b$ and $c$ be the sides of triangle with $c$ as hypotenuse.

Then according to Pythagoras theorem,

$a^{2} + b^{2} = c^{2}$

Option $C$ is correct.

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Similar questions

\frac{a^{2} + b^{2} + c^{2}}{R^{2}}

8

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

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

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A B C

\frac{tan A}{tan B} = \frac{c^{2} + a^{2} - b^{2}}{c^{2} + b^{2} - a^{2}}

If a, b, c, d are in G.p., prove that :

(i) $(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d)^{2}$ are in G.P.

(ii) $(a^{2} - b^{2}), (b^{2} - c^{2}), (c^{2} - d)^{2}$ are in G.P.

(iii) $\frac{1}{a^{2} + b^{2}}, \frac{1}{b^{2} + c^{2}}, \frac{1}{c^{2} + d^{2}}$ are in G.P.

(iv) $(a^{2} + b^{2} + c^{2}), (a b + b c + c d), (b^{2} + c^{2} + d^{2})$

\frac{a^{2} + b^{2} + c^{2}}{R^{2}}

8

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

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