SOLUTION
![\begin{gathered} EI=4x+7 \\ EG=2(EI) \\ EG=2(4x+7) \\ EG=8x+14 \\ \\ FI=5x+5 \\ FH=2(FI) \\ FH=2(5x+5) \\ FH=10x+10 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/b3qiv60wy213cm7h82675wqpvugfpxm1ue.png)
To find x, equate both diagonals FH and EG as they are equal
![\begin{gathered} FH=EG \\ 10x+10=8x+14 \\ 10x-8x=14-10 \\ 2x=4 \\ x=2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/jkkk16wkrmv09pt45g02chdwshriapina1.png)
To find EG
![\begin{gathered} EG=8x+14 \\ =8(2)+14 \\ =16+14 \\ =30 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/kk97jrye55310ma42budg5lh0vbv4hgea2.png)
EG = 30
Since lenght IE and IH are equal (15) , triangle IHE is an iscoceless triangle
with equal base angles
x+x+54=180
2x+54=180
2x=180-54
2x=126
x=126/2
x=63degrees.
[tex]\begin{gathered} \text{Therefore} \\