Question 1118301
The hour hand of a clock is half the length of the minute hand.
 How many more times area will the minute hand swept than the hour hand if they both move through the same angle? 
:
let r = the radius of the hr hand
then
2r = radius of the minute hand
:
{{{(Minhand A)/(hrhand A)}}} = {{{(pi*(2r)^2)/(pi*r^2)}}} = {{{(pi*4r^2)/(pi*r^2)}}}
Cancel {{{pi*r^2}}}
4 times