a) The ratio of the part of the triangle inside the circle and outside stays the same no matter what. So, I can set up an equality:
[tex] \frac{4}{6} = \frac{x}{5} [/tex]
Solving for x gives us:
[tex]20=6x[/tex]
[tex]x = \frac{20}{6}= \frac{10}{3} [/tex]
And that's your answer.
Same thing for part b:
[tex] \frac{5}{3} = \frac{x}{4} [/tex]
[tex]20=3x[/tex]
[tex]x = \frac{20}{3} [/tex]
And there ya go. If you have any queries, I'd be happy to help.
