Question

!!PLEASE HELP ME ANSWER!!

Answer 1

Answer:

$x = \sqrt{w(w+z)}$

Step-by-step explanation:

By the Pythagorean theorem, triangle BCD has

$y^2 = x^2 - w^2$

and triangle ABD has

$v^2 = y^2 + z^2$.

Substituting the first equation into the second equation for $y^2$ gives the equation

$v^2 = (x^2 - w^2) + z^2$

Now triangle ABC has

$(w+z)^2 = x^2 + v^2$

Expanding out the left side, $w^2 + 2wz + z^2 = x^2 + v^2$.

Now substitute the equation from before, $v^2 = x^2 - w^2 + z^2$, into the new equation.

$w^2 + 2wz + z^2 = x^2 + (x^2 - w^2 + z^2)$

$w^2 + 2wz + z^2 =2x^2 - w^2 + z^2$

We work to isolate $x^2$ by adding $w^2$ to both sides of equation and subtracting $z^2$ from both sides.

$2w^2 + 2wz = 2x^2$

Divide both sides by 2 and you get $w^2 + wz = x^2$

Assume that x is positive and square root both sides to get the final result of

$x = \sqrt{w^2 + wz} = \sqrt{w(w+z)}$