There is no escape...believe it or not. Mathematics is everywhere.
The two following problem permit to know how you work your way in maths.
Explanation came after the two problems
Don't read bellow before having answered both problems. |
This problem will trap people used to "recognise and apply" approach.
When solving problem, you first attempt to classify it then use a predefined algorithm/set of tools.
This problem is classified as "Optimisation kind problem" thus you're going to use "Integral and algebra".
It's possible but very long and error prone :-)
The easiest way is to wonder if you can find some usable properties. Let's simplify the problem by fixing two points and moving the last one.
You will quickly see that the area is maximum when the third point is fare away of base (wich is fixed length since the two point are fixed). To be far away, the point must be on base bisector.
Since you can fix any two point of the 3, it should be true for any chosen couple ! The only triangle having his vertices on his sides bisector is the equilateral one.
This one trap people using short-cuts :-)
The exhibited trapezoid does NOT exist
Answer is then simple : Area does not exist
Proving it is a little harder
First let's notice that g=h, I'll refer to g as h
|DA| = 5 and |BC| = 10
|DC| = 8
|AB| = |EF| = 4
Let's define |DE|=x then |FC|=4-x
Now some relations that use Pythagoras theorem :
25=h²+x²
100=h²+(4-x)²
Let's isolate h² so we can try to find a suitable "x" value.
h²=25-x²
h²=100-(16+x²-8x)
Search for "x" value :
25-x²=100-16-x²+8x
25=84+8x
So x is : -7.375
Let's replace that value in first identity :
h²=25-( -7.375)²
Not possible in R...
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