If a right angle triangle has sides of lengths 'a' and 'b' and hypotenuse of length 'c', then

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

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

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

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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Solution

The correct options are
B
$c^{2} - a^{2} = b^{2}$

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

In a right angle triangle, Pythagoras Theorem states that the sum of the squares of the sides containing the right angle is equal to the square of the hypotenuse.

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

Thus, $c^{2} - a^{2} = b^{2}$ and $c^{2} - b^{2} = a^{2}$

Hence (B) and (D)

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Similar questions

2 s = a + b + c

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & (s - a)^{2} & (s - a)^{2} (s - b)^{2} & b^{2} & (s - b)^{2} (s - c)^{2} & (s - c)^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

2 s = a + b + c,

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & {(s - a)}^{2} & {(s - a)}^{2} {(s - b)}^{2} & b^{2} & {(s - b)}^{2} {(s - c)}^{2} & {(s - c)}^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

= 2 s^{3} (s - a) (s - b) (s - c) .

a

b

c

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

a, b, c

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

1 : 2 : 4

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

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