8/7/2017 10:35:00 am
*** SPOILER ALERT ***
Reply
Simon Geard
8/7/2017 05:06:34 pm
Not sure what 'd' is, I think you mean b=24 (c=12 and a=6) where b == (2r-a).
Reply
Simon Geard
8/7/2017 02:27:48 pm
Consider the right angled triangle with rh node at the centre of the circle. Then
Reply
tom
9/7/2017 10:34:19 pm
Starting at 12**2+(r-6)**2 = r**2, you should get
Reply
Simon Geard
10/7/2017 08:28:06 am
The expansion of (r-6)**2 is r**2-12r+36 so r=15 is the correct solution. Geometrically we are given r>12 and the solution satisfies that constraint.
srk
8/7/2017 06:45:04 pm
(x - r)^2 + y^2 = r^2
Reply
raluca Corneanu
8/7/2017 09:41:42 pm
226.08
Reply
tom
9/7/2017 10:13:31 pm
Call teh diameter of the circle d. Using the three sides of the right-angle triangle whose points are the top of the 12 cm length and the ends of the diameter containing the 6cm length, Pythagoras gives us the following:-
Reply
tom
9/7/2017 10:21:07 pm
In my previous comment, substitute d-6 for both occurrences of d-12; that's what I had in my scribble box, and I carelessly bent it when I used it in my comment - but I copies the next bit of scribble correctly, so at least the final answer was what i intended although the working was somewhat messed up.
Reply
## Leave a Reply. |
## Puzzle Ideas
If you have an idea for puzzle of the month, then please do let me know. ## Archives
July 2017
Don't forget to check out 'Teach Further Maths' - PowerPoints for Teachers and Students of A-Level Further Maths ...and the brilliant NEW educational card game 'Maths Trumps' |