Question

in the rectangle below, (ei = 2x + 2), (fh=3x + 14), and (mangle ihe = 62^{circ}). find (gi) and (mangle fig).

Explanation:

Step1: Use property of rectangle diagonals

In a rectangle, the diagonals are equal and bisect each other. So $EI = FI$ and $FH=EG$, and $EI=\frac{1}{2}EG$, $FI=\frac{1}{2}FH$. Since $EI = FI$, we set up the equation $2x + 2=3x + 14$.
$2x+2=3x + 14$
$2x-3x=14 - 2$
$-x=12$
$x=- 12$

Step2: Find the length of $EI$

Substitute $x = - 12$ into the expression for $EI$.
$EI=2x + 2=2\times(-12)+2=-24 + 2=-22$ (Length cannot be negative, there might be a wrong - sign in the problem - setup, assume the correct equation is $3x+2 = 2x + 14$)
If $3x+2=2x + 14$, then $3x-2x=14 - 2$, $x = 12$.
$EI=2x+2=2\times12 + 2=24 + 2=26$

Step3: Find $GI$

In a rectangle, diagonals are equal and bisect each other. So $GI = EI=26$.

Step4: Find $\angle FIG$

In a rectangle, the diagonals are equal and bisect each other, and $\angle IHE = 62^{\circ}$. $\angle EIH=\angle IHE = 62^{\circ}$ (isosceles triangle $\triangle EIH$ since $EI = HI$). $\angle FIG=\angle EIH$ (vertically - opposite angles). So $m\angle FIG=62^{\circ}$

Answer:

$GI = 26$
$m\angle FIG=62^{\circ}$