Karen Campe
@karencampe.bsky.social
1.7K followers 460 following 1.6K posts
Math teacher using technology for understanding Reflections & Tangents Blog: karendcampe.wordpress.com
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karencampe.bsky.social
Ask @mnmmath.bsky.social she knows!
{I won't be at nctm}
karencampe.bsky.social
I used 3 ideas:
-Right triangle with hypotenuse = 10
-Geometric mean** of the 2 parts of diameter
-the question asks for area, not the dimensions separately

**thanks to @howiehua.bsky.social I've had the different means on my mind.
In marked diagram --> y is geometric mean of w and x.
Semicircle and rectangle, with base of rectangle along diameter of semicircle, and top edge of rectangle tangent to semicircle. 

Left side of rectangle divides diameter of semicircle into two parts, labeled w and x. 
Let y be the part of the left side of rectangle from bottom corner up to semicircle intersection. 

Then y is geometric mean of w and x: 
w/y = y/x
karencampe.bsky.social
Cool!!
My usual approach (w/out technology) is rational root theorem and use the remainder theorem to help P(root) = 0 so can do some easy tests with 1, -1, 2 etc.
karencampe.bsky.social
Yes, that's typical phrasing in US.
karencampe.bsky.social
The marked angle is half the sum of the intercepted arcs, so (1/2)•(50+110) --> (1/2)•160 = 80°

This is one of 7 angle-arc cases that can be boiled down to "one big rule"

#iTeachMath #MathsToday #UKMathsChat
Big Rule: Identify the location of the angle’s vertex, which gives the relationship of the angle measure to the intercepted arc(s):

At center: angle = arc
On circle: angle = (1/2)•arc
Inside circle: angle = (1/2)•(arc 1 + arc 2)
Outside circle: angle = (1/2)•(arc 1 – arc 2)

Notes: This condenses 7 rules into 4 rules. Use these varied situations to practice circle vocabulary. In class, I would pantomime an “intercepted arc” with my two arms, as if I were a football player making an interception.
karencampe.bsky.social
Nicely done, David.
At first glance, cubic not easily factored.
Did you use rational root theorem or a graphing calculator to help finding that factorization... or other ideas??
Would love to hear more about your thinking.
#iTeachMath #ProblemSolving
karencampe.bsky.social
Yes the harmonic mean is good for average rates, such as rate of work or rate of speed.
I am waiting for Howie's next video explainer to see if it is also the "average" of divisors or something like that...
karencampe.bsky.social
Dislike... hang in there.
karencampe.bsky.social
Howie, this is absolute gold! Love these explainers of what the different means *mean*
...
So, the arithmetic mean is the average of addends, the geometric mean is the "average" of scale factors, then the harmonic mean must be ... ?
#iTeachMath
karencampe.bsky.social
Number of diagonals? Or diagonals that are bisected/ congruent/ etc.
karencampe.bsky.social
No worries!
Do you have students discover/notice that products of diagonals are the same, or do you just present it as part of the method?
karencampe.bsky.social
Very nice.
The key is that in the area model, the product of diagonals is equal.
[I was momentarily confused by using abcd for those 4 values ... because they aren't same as ABC from quadratic]
The area model can do so much!! Wrote a whole post about it a couple years ago.
karencampe.bsky.social
Huge... and it might have been earlier. I was asked about her work when interviewing for teaching jobs in late 1980s/early 1990s.
karencampe.bsky.social
Yes true! That's why I'm so excited about your "addend first" strategy.

Students who "get" factoring will find the factors either way, but Ss who struggle get another way into it with addends first.
karencampe.bsky.social
^^ yes absolutely.
And your "addend first" idea is accessible to students who can't do numerical factoring readily.

But also yes to @dsrussosusan.bsky.social 's point that sets of addends are infinitely many, but sets of factors are finite.
karencampe.bsky.social
Yes exactly. I was wondering if David's "adding first" method would involve AC.
karencampe.bsky.social
So, I'd try
-10•36, sum = 26
-12•30, sum = 18
-18•20, sum = 2
I must have missed 9...
-15•24, sum = 9 ⭐️

Then
18x^2 – 15x + 24x – 20
3x(6x – 5) + 4(6x – 5)
(6x – 5)(3x + 4)

Done.
karencampe.bsky.social
Thanks for this explanation!
The "AC Method" is to multiply A•C. Then find two numbers whose product is AC and sum is B.
"Split" the B term into the sum you found, then factor by grouping.

18x^2 + 9x – 20
AC = -360
B = 9

We write:
___ • ___ = -360
___ + ___ = 9
then try to fill in blanks