There are at least two ways to cope with this problem each of tham leading to two results. 1)Directly apply Pytagorean theorem, or use coordinates, what reduces to the same :x^2+y^2=r, while x=r-8 and y=r-4 Therefor:(r-8)^2+(r-4)^2=r^2 I solved this square equation through the discriminante called delta,to receive two values of r: 4 and 20. Being r=4-4=0, this result is out of the game. 2)Another way to solve this problem by applying the intersecting cords theorem :(2r-4)*4 =(r-8)^2 After simplyfaing we receive the same square equation than before :r^2-24r+80 =0,to be solved by one of the comon ways to solve square equations. As final conclusion we have got 20 units. Greetings!
I just love the simplicity with which this channel shows the joy of solving maths. Great video as always . Lots of love. ♥️
There are at least two ways to cope with this problem each of tham leading to two results. 1)Directly apply Pytagorean theorem, or use coordinates, what reduces to the same :x^2+y^2=r, while x=r-8 and y=r-4 Therefor:(r-8)^2+(r-4)^2=r^2 I solved this square equation through the discriminante called delta,to receive two values of r: 4 and 20. Being r=4-4=0, this result is out of the game. 2)Another way to solve this problem by applying the intersecting cords theorem :(2r-4)*4 =(r-8)^2 After simplyfaing we receive the same square equation than before :r^2-24r+80 =0,to be solved by one of the comon ways to solve square equations. As final conclusion we have got 20 units. Greetings!
Very difficult question indeed. Loved the solution.
Pythagoras theorem when used properly works wonders man.