DfwRevolution From United States of America, joined Jan 2010, 977 posts, RR: 51
Reply 1, posted (9 years 5 months 2 weeks 6 days 20 hours ago) and read 1091 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

Jfkaua From United States of America, joined Aug 2004, 1000 posts, RR: 3
Reply 2, posted (9 years 5 months 2 weeks 6 days 20 hours ago) and read 1084 times:

no the answer is (D).. just trying to figure out why. E is actually the first I eliminated because since the diagram is not drawn to scale the line could tangent the circle at any angle..

Jfkaua From United States of America, joined Aug 2004, 1000 posts, RR: 3
Reply 3, posted (9 years 5 months 2 weeks 6 days 20 hours ago) and read 1080 times:

hmmm your answer does seem to be logical.. any idea how it could be d?

Jhooper From United States of America, joined Dec 2001, 6204 posts, RR: 12
Reply 4, posted (9 years 5 months 2 weeks 6 days 20 hours ago) and read 1078 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.

Jfkaua From United States of America, joined Aug 2004, 1000 posts, RR: 3
Reply 5, posted (9 years 5 months 2 weeks 6 days 20 hours ago) and read 1077 times:

ooppss i read the answer key wrong!! you were absolutley right..

DfwRevolution From United States of America, joined Jan 2010, 977 posts, RR: 51
Reply 6, posted (9 years 5 months 2 weeks 6 days 20 hours ago) and read 1074 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

Jfkaua From United States of America, joined Aug 2004, 1000 posts, RR: 3
Reply 7, posted (9 years 5 months 2 weeks 6 days 20 hours ago) and read 1070 times:

yep E is certainly correct.. I have 2 more ?'s that I got wrong on the math portion.. I'll post them tommorow as I am do lazy to make the question up now..

Mir From United States of America, joined Jan 2004, 21626 posts, RR: 55
Reply 8, posted (9 years 5 months 2 weeks 6 days 20 hours ago) and read 1068 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..

DfwRevolution did the work mathematically and so proved that the angle BAD is a 90-degree angle.

A tangent line will always be at a right angle to a radius that touches the circle at the same point.

So since angle ABC is a right angle, and angle BAD is a right angle, then lines BC and AD would be parallel and never intersect.

His work seems pretty sound to me. Are you sure that it's not E?

-Mir

7 billion, one nation, imagination...it's a beautiful day

Jhooper From United States of America, joined Dec 2001, 6204 posts, RR: 12
Reply 9, posted (9 years 5 months 2 weeks 6 days 20 hours ago) and read 1064 times:

sorry DfwRevolution for repeating exactly what you said. I guess I should have refreshed my browser before posting because I didn't notice your reply...

Last year 1,944 New Yorkers saw something and said something.

DfwRevolution From United States of America, joined Jan 2010, 977 posts, RR: 51
Reply 10, posted (9 years 5 months 2 weeks 6 days 20 hours ago) and read 1063 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 Indonesia, joined Sep 2000, 4504 posts, RR: 71
Reply 11, posted (9 years 5 months 2 weeks 4 days 21 hours ago) and read 994 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.