Question

if the sides of a right angles triangle are x, x+2, x+4 (where x > 0 then what is
the value of x​

Answer 1

Step-by-step explanation:

use Pythagoras Theorem

Note : longest side is hypotenuse which is x+4

$\: \: \: \: \: \: \: \: \: \: \: \: \: {(x + 4)}^{2} = {x}^{2} + {(x + 2)}^{2} \\\implies \: {x}^{2} - 4x - 12 = 0 \\ \implies \: (x - 6)(x + 2) = 0 \\ \implies \: x > 0\implies \: x = 6$

Answer 2

Answer:

x =6

Step-by-step explanation:

In right angle triangle, larger side is hypotenuse and left are prep and base respectively

Thus applying pythagoras theorem

${(x + 4)}^{2} = {x}^{2} + {(x + 2)}^{2} \\ {x}^{2} +16 + 8x = {x}^{2} + {x}^{2} + 4 + 4x \\ {x}^{2} - 12 - 4x = 0 \\ {x}^{2} - 4x - 12 = 0 \\ {x }^{2} - 6x + 2x - 12 = 0 \\ x(x - 6) + 2(x - 6) = 0 \\ (x - 6)(x + 2) = 0 \\ x = 6 \: or - 2$

As length cannot be negative

thus x = 6 (neglective negative value,-2)

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