I'm currently doing a lot of self study with mathematics. I live in The Netherlands and hope to be admitted to Leiden University somewhere in 2016.
Now, I have encountered a problem in my workbook which I can't seem to be able to solve, even after I looked at the worked solutions..
So I made a picture of it, and hope some of you can help me pinpoint what knowledge I'm missing and what I'm not getting. (I will translate the problem completely below.)
Translation : 5.4 Working with mathematical models.
The sheet of paper ABCD of 20cm by 30cm is folded in such a way that point D is laying on side AB of the sheet of paper. The point where D touches AB is called D' and the fold line which is created is called PQ, look at the figure on the right.
a* (You can ignore this one)
b Take AP = 5 cm
why is D'P = 15 cm?
Show that this is true with a calculation and
that in this case AD' is about 14.1 cm.
The Ans. to b.
Take AP = 5.
D'P = AD' - AP = 20 - 5 = 15 ©
Pythagoras: AP ² + AD' ² = D'P ²
-------------- 25 + AD' = 225
-------------- AD' ² = 200
-------------- AD' = √200 ≈ 14.1 cm.
Now, what do I not understand.
I placed a copyright sign to show what I'm talking about
This part is conflicting with my understanding of Pythagoras..
How can AD'(20) be longer than the hypothenuse? shouldn't it be as follows:
D'P = AD' + AP ? What's going on? With the rest of it I'm oké.
c Take AP = 8 cm © and calculate AD' in mm's accurate.
We ask ourselves how long should we make AP such that AP = AD'
Put AP = x cm ©
Ans. to c.
Take AP = 8 ©.
D'P = AD' - AP = 20 - 8 = 12 ©
Pythagoras: AP ² + AD' ² = D'P ²
-------------- 64 + AD' = 114
-------------- AD' ² = 80
-------------- AD' = √80 ≈ 8.9 cm.
Now, what do I not understand.
The overall character of the problem asked to solve..
So you take AP = 8 cm, AP has to equal AD', but, AP is also equal to x??
Could someone walk me through the reasoning process?
problems d, e and f build further upon b and c.
Hope this is clear enough.
Thanks, Bowser.
For the first : Note that $AD'\not=20$ and that $PD'=PD=AD-AP=20-5$.
For the second : Note that $AP\not=AD'$ and that $PD'=PD=AD-AP=20-8$.
The point is to see $PD'=PD=AD-AP$ and to consider a right triangle $APD'$.