DfwRevolution From United States of America, joined Jan 2010, 846 posts, RR: 51 Reply 1, posted (8 years 8 months 1 week 1 day 3 hours ago) and read 950 times:
E... they will never intersect
Very simple solution:
The radius is 4, so the full area of the circle is (4^2)(Pi) = 50.26
Now find the area of (4)(pi) = 12.56
This means that 4(pi) is exactly 25% of the circle, thus, the angle from BA to AD is a 90 degree angle.
If line BC is tangiental to point B, and AD is perpendicular BA... AD and BC are parallel.
Never overanaylize SAT math... look for easy solutions, don't work the problem. Pick the answer you think is right and work it backwards. That's how I worked this one... took me about 15 seconds once I grabbed by TI-83
Jhooper From United States of America, joined Dec 2001, 6197 posts, RR: 13 Reply 4, posted (8 years 8 months 1 week 1 day 3 hours ago) and read 937 times:
Hmmmm.......I'm no math expert by any stretch of the imagination, but I'd say the lines won't intersect (E).
This is because we know the area of the entire circle is pi X r X r, or 16(pi). Divide that by 4 and you get 4(pi). Therefore, we know the shaded area represents 90 degrees of the circle (even though the figure isn't "drawn to scale"). Since that's true, AB and AD make a right angle and therefore the lines won't intersect.
Am I smoking crack here or does my logic work?
Last year 1,944 New Yorkers saw something and said something.
DfwRevolution From United States of America, joined Jan 2010, 846 posts, RR: 51 Reply 6, posted (8 years 8 months 1 week 1 day 2 hours ago) and read 933 times:
Quoting Jfkaua (Reply 2): E is actually the first you can eliminate because since the diagram is not drawn to scale the line could tangent the circle at any angle
If Angle-BAD is perpendicular, BA is tangental, then segment AD will always be parallel. If the solution manual says D then I will shut-up and let someone else have a go... I can't see any other solution than E
DfwRevolution From United States of America, joined Jan 2010, 846 posts, RR: 51 Reply 10, posted (8 years 8 months 1 week 1 day 2 hours ago) and read 922 times:
Quoting Jfkaua (Reply 5): ooppss i read the answer key wrong!! you were absolutley right..
Lol.. no problem, it happens. One thing to remember about circle geometry is that there is only a single tangental line at a single angle from at a given point. Try to find another tangental line and you'll cross over another point of the circle.
HB-IWC From Greece, joined Sep 2000, 4450 posts, RR: 73 Reply 11, posted (8 years 8 months 6 days 3 hours ago) and read 853 times:
Quoting DfwRevolution (Reply 10): One thing to remember about circle geometry is that there is only a single tangental line at a single angle from at a given point. Try to find another tangental line and you'll cross over another point of the circle.
I'm not sure what you are trying to say here, so I might be misunderstanding your line, but one can draw exactly TWO tangental lines to a circle from any point outside of that circle.
As for the original problem, a nice variation and typical SAT question could be:
In the figure above, line BC is tangent to the circle with center A and radius 4 at point B. The area of the shaded region is (8/3)(pi). What is the distance from A to the intersection of lines AD and BC?
The answer to this question would be 8, and you could for instance use special formulas for 30-60-90 right triangles to easily solve it.
As for preparation for the SAT Math Sections, don't forget that the New SAT 1 is quite a bit more difficult than the old one, and that, although the Quantitative Comparisons have been eliminated, additional topics including Pre Calculus, Counting Problems (General Counting Principle, Combinations,...) and Elementary Probability and Statistics have been included.