Question

If the length of one leg of a right triangle is 16 and the length of the hypotenuse is 2 more than the length of the other leg, what are the lengths of all the sides of the triangle? Other leg = a0 Hypotenuse = a1

Answer 1

Answer:

All sides are : 16,63,65 units

Step-by-step explanation:

Let the other leg be = x

So, hypotenuse will be = x+2

And one leg is = 16

So, Pythagoras theorem is :

$a^{2} +b^{2}=c^{2}$ , where c is hypotenuse.

Now we can calculate this as:

$16^{2} +x^{2} =(x+2)^{2}$

=> $256+x^{2} =x^{2} +4x+4$

Clubbing the like terms together;

=> $x^{2} -x^{2} -4x=-256+4$

$-4x=-252$

x = 63

So, hypotenuse is = $63+2=65$ units

Therefore, all the lengths are = 16,63 and 65 units.

Answer 2

Refer to brainly.com/question/9480049

The lengths of the sides of the triangle are 16, 63, and 65.

a0 = 63
a1 = 65

_____
You have
.. a0² +16² = a1²
You want
.. a1 = a0 +2
so
.. a0² +256 = (a0 +2)² = a0² +4a0 +4 . . . . substitute and expand
.. 252 = 4a0 . . . . . . . . . subtract a0² +4
.. 63 = a0 . . . . . . . . . . . .divide by 4
.. a1 = 63 +2 = 65