Question

Could someone please solve for X for the third triangle? (One furthest to the right) Thanks!!

Answer 1

Consider the rightmost diagram,

Apply the Pythagoras Theorem in right triangle ADB,

`\begin{gathered} AD^2+BD^2=AB^2 \\ 11^2+z^2=y^2 \\ y^2-z^2=121 \end{gathered}`

This is the first equation.

Now, apply Pythagoras Theorem in right triangle BDC,

`\begin{gathered} BC^2=BD^2+CD^2 \\ BC^2=z^2+9^2 \\ BC^2=z^2+81 \end{gathered}`

Now, apply Pythagoras Theorem in right triangle ABC,

`\begin{gathered} AB^2+BC^2=AC^2 \\ y^2+(z^2+81)=(11+9)^2 \\ y^2+z^2+81=400 \\ y^2+z^2=319 \end{gathered}`

This is the second equation.

Adding the two equations,

`\begin{gathered} (y^2-z^2)+(y^2+z^2)=121+319 \\ 2y^2=440 \\ y^2=220 \\ y\approx14.83 \end{gathered}`

Substitute the value in first equation,

`\begin{gathered} (220)-z^2=121 \\ z^2=220-121 \\ z^2=99 \\ z\approx9.95 \end{gathered}`

Thus, the required values of 'y' and 'z' are 14.83 and 9.95 units, approximately.