Given:
The length of the triangle is: 27, 3, x, y, and z.
Find: Value of "y"
Sol:
In triangle ACD is:
Use Pythagoras theorem:
[tex]\text{ Hypotenuse}^2=\text{ Base}^2+\text{ Perpendicular}^2[/tex]
Apply for triangle ACD then:
[tex]\begin{gathered} AC^2=AD^2+CD^2 \\ \\ (27+3)^2=z^2+x^2 \\ \\ 30^2=z^2+x^2...........(1) \end{gathered}[/tex]
In triangle CBD
Apply theorem then:
[tex]x^2=27^2+y^2.........(2)[/tex]
In Triangle ABD
Apply theorem then:
[tex]\begin{gathered} z^2=3^2+y^2...........(3) \\ \end{gathered}[/tex]
From eq (2) and eq(3) put the value in eq(1) then:
[tex]\begin{gathered} 30^2=z^2+x^2 \\ \\ 30^2=3^2+y^2+27^2+y^2 \\ \\ 900=9+2y^2+729 \\ \\ 900-(729+9)=2y^2 \\ \\ 162=2y^2 \\ \\ y^2=\frac{162}{2} \\ \\ y^2=81 \\ \\ y=9 \end{gathered}[/tex]
So the value of "y" is 9.
