Question

If the triangles third vertex has a coordinate (0,y) which is possible value of Y

Answer 1

$y=4\sqrt[]{3}$

Explanation

the coordinate (0,y) makes a triangle

then

$\begin{gathered} \text{hypotenuse}=8 \\ \text{side}_1=\Delta x=4-0=4 \\ \text{side}_{2_{}}=\Delta y=y-0=y \end{gathered}$

now, we can use the Pythagorean theorem to find y. P.T states that the sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse,so in this case.

$\text{side}^2_1+side^2_2=hypotenuse^2$

replace.

$\begin{gathered} \text{side}^2_1+side^2_2=hypotenuse^2 \\ 4^2+y^2=8^2 \\ 16+y^2=64 \\ y^2=64-16(\text{subtract 16 in both sides)} \\ y^2=48 \\ \text{square root in both sides} \\ y=\sqrt[]{48} \\ \text{apply properties of the root} \\ y=\sqrt[]{16\cdot3} \\ y=\sqrt[]{16}\cdot\sqrt[]{3} \\ y=4\sqrt[]{3} \end{gathered}$

therefore, a posible value for y is

$y=4\sqrt[]{3}$

I hope this helps you