Monday, January 22, 2007

Problems for 1-23-07

Above are 4 more problems. Questions 1 and 4 can be attempted by Grades 7 and up. The technique for the probability question is worth demonstrating in middle school (imo).


Anonymous said...

Do you intend to show #2 qualitatively? ie, one whole minus (half of a half plus half of a half plus half of a half of a half) ?

Dave Marain said...

some students have a knack for cutting these figures into equal parts and recognizing how the parts can be rearranged or combined - all students need to be exposed to this spatial approach; however, some, like myself, think quantitatively with greater comfort whether by assigning a length of '2x' to each side or choosing a convenient numerical value for each side, and obtaining expressions for the areas of the 3 'corner' triangles, etc. Shouldn't all middle schoolers be exposed to multiple strategies like these?

Anonymous said...

I know little about middle schoolers, but yes, that makes sense.

I think when I was a kid I would have considered both of these solutions too slick. I would have assigned the side of the square length x and calculated the area of the triangle in terms of x. No subtracting.

Or maybe not. I don't know if I encountered any probability before the 11th grade.

Dave Marain said...

Answers to 1-23-07:
1) 40
2) 3/8 or 37.5%
3) 36
4) 29
Did you find the wording of the #1 and #4 ambiguous?

Anonymous said...

I just started to read this blog, and work through some of the problems. I'm lost on #3. Could you show the work on the solution to this problem?


Dave Marain said...

'Proof' for #3:
1. Angle ACB also = 75 degrees from AB = AC.
2. Therefore angle A = 30 degrees.
3. Therefore triangle ADC is a 30-60-90!
4. Since CD = 6 (side opposite 30), AC =12.
5. Therefore AB = 12 aslo.
6. Take AB to be the base and CD the height!
7. Area = (12)(6)/2 = 36.

Of course, you could argue that 30-60 arguments really are trig but...
This is a classic problem requiring the student to look at the diagram in a variety of different ways. The key was looking at AB as the base and obtaining AB from AC. Nasty but beautiful imo!!

druin said...

I wish I had paid attention to that on #3! I found the area of the two smaller triangles and added them -
*sigh* that's what I get for doing them when I'm sleepy! :)