Question

Given ABCD is congruent EFGH. solve for x. Round the answers to the nearest hundredth

Answer 1

As ABCD is congruent withEFGH, it means that both figures have the same mesarurements, even if their orientation is different.

Then, you have that the angle in A is congruente with the angle in E, the angle B is congruent with the angle F, the angle C is congruent with the angle G, the angle D is congruent with the angle H.

`\begin{gathered} \angle A=\angle E \\ \angle B=\angle F \\ \angle C=\angle G \\ \angle D=\angle H \end{gathered}`

Then:

`3x^2-4x+10=16`

You solve the x from the equation above:

1. Equal the equation to 0

`\begin{gathered} 3x^2-4x+10-16=0 \\ 3x^2-4x-6=0 \end{gathered}`

2. Use the quadratic equation:

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`
`\begin{gathered} x=\frac{-(-4)\pm\sqrt[]{(-4)^2-4(3)(-6)}}{2(3)} \\ x=\frac{4\pm\sqrt[]{16+72}}{6} \\ x=\frac{4\pm\sqrt[]{88}}{6} \\ x_1=\frac{4-\sqrt[]{88}}{6},x_2=\frac{4+\sqrt[]{88}}{6} \\ x_1=-0.896,x_2=2.230 \end{gathered}`

3. As you get two solutions for x. You need to prove which is the right solution:

with x1:

`\begin{gathered} 3x^2-4x+10=16 \\ 3(-0.896^2)-4(-0.869)+10=16 \\ 11.17\\e16 \end{gathered}`

with x2:

`\begin{gathered} 3x^2-4x+10=16 \\ 3(2.23^2)-4(2.23)+10=16 \\ 15.99\approx16 \end{gathered}`

As you can see the right solution is x2, because if you subtitute the x in the equation of the angle A that must be equal to 16, just the x2 gives an approximate value to 16.

Then, the solution for the x is x=2.23