Question

Solve these question.....
I will mark your answer as brainliest answer

Answer 1

Let the shortest side be x

Hypotenuse be x+10

Third side be (x+10-6)

=x+4

By applying Pythagoras Tgeorem

i.e

${hypotenuse}^{2} = {perpendicular}^{2} + { base}^{2}$

${(x + 10)}^{2} = {(x + 4)}^{2} + {x}^{2}$

We know that

${(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}$

Now

${x}^{2} + 100 + 20x = {x}^{2} + 16 + 8x + {x}^{2}$

$100 - 16 = {x}^{2} + {x}^{2} - {x}^{2} + 8x - 20$

$84 = {x}^{2} - 12x$

Or

${x}^{2} - 12x - 84 = 0$

Here

a=1 b= -12 and c= -84

We know that

$d = {b}^{2} - 4ac$

$d = { ( - 12)}^{2} - (4 \times 1 \times - 84)$

$d = 144 + ( - 336)$

$d = 144 - 336$

$d = - 192$

Therefore,

D is less than 0

This pair of equation has no roots