Peter wrote:
IlCreatore wrote:
Let's start from the fact that \(\frac{AB}{AC}=\frac{AC}{BC}\), then given AB = AC+BC, we can rewrite the equality as \(\frac{AC+BC}{AC}=1+\frac{BC}{AC}=\frac{AC}{BC}\). Then, given that BC is smaller than AC, BC/AC is smaller than 1 so that summed to 1 is surely less than 3. Quantity B is greater.
Answer B
thank you for the info: but how do you make sure that BC is smaller than AC given that the picture might not be drawn to scale?
Good question,
But you can try putting the some values,
But remember it should satisfy the equation \(\frac{AB}{AC}=\frac{AC}{BC}\)
Let AB =12
AC= 6
BC = 3
These values satisfy the above equation , and \(\frac{AC}{BC}= 2\) which is less than 3