For given figure, the formula for Pythagoras theorem is:

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

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

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

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

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Solution

The correct option is C $a^{2} + b^{2} = c^{2}$
The formula of Pythagoras theorem for the right angled triangles is $a^{2} + b^{2} = c^{2}$ .
The diagram shows a right angled triangle which is right angeled at C. So, we use $a^{2} + b^{2} = c^{2}$ to find the unknown length.
Where,
$a =$ Height
$b =$ base
$c =$ Hypotenuse

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

In the adjoining figure, if BC = a, AC = b, AB = c and $∠ C A B = 120^{\circ}$ , then the correct relation is-

\frac{a^{2} - {(b - c)}^{2}}{{(a + c)}^{2} - b^{2}} + \frac{b^{2} - {(a - c)}^{2}}{{(a + b)}^{2} - c^{2}} + \frac{c^{2} - {(a - b)}^{2}}{{(b + c)}^{2} - a^{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)

|\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}

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