Answer:
Therefore, Michael concludes option C)
C)
![(DA)^(2)=(DG)^(2)+(AG)^(2)](https://img.qammunity.org/2020/formulas/mathematics/college/w3ovpeyytho0m3h6ccqnju5ac13mdei7us.png)
Explanation:
Given:
1. DG = 3 and the area of square DEFG is 9.
2. AG = 4 and the area of square GHIA is 16.
3. DA = 5 and the area of square ABCD is 25.
So we have,
![(DG)^(2)=3^(2)=9\\ \\(AG)^(2)=4^(2)=16\\\\(DA)^(2)=5^(2)=25\\](https://img.qammunity.org/2020/formulas/mathematics/college/hvdbg3nd1i242uijusfy2zgpkjmqhwlnux.png)
Now Add DG² and AG² we get
![(DG)^(2)+(AG)^(2)=9+16=25=(DA)^(2)](https://img.qammunity.org/2020/formulas/mathematics/college/iwrruea40qv7nxk8a810072xjfj4vwlo0p.png)
Which is also called as Pythagoras theorem i.e
![(\textrm{Hypotenuse})^(2) = (\textrm{Shorter leg})^(2)+(\textrm{Longer leg})^(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/zzqkt74bo8i0g39uk3gex0h5896knh6wio.png)
Therefore, Michael concludes option C)
C)
![(DA)^(2)=(DG)^(2)+(AG)^(2)](https://img.qammunity.org/2020/formulas/mathematics/college/w3ovpeyytho0m3h6ccqnju5ac13mdei7us.png)