Geometry problem solutions4 (30..39)

30.
Calculate angle x in picture below:



Solution


Shift BC parallel such that point B is on D.
(Picture enlarged for clearity)



E is shifted point C.
The angles colored red were calculated from the fact that
ΔADE is isosceles.
Shift ΔADE horizontaly until D reaches B and E is on C.
G is the shifted point A.



Add line BF with LCBF=24o.

AF is a straight line.
LAFB=78o, so BF = BC.

G,C,F are angles of a regular 15 polygon with center B.
First an intermezzo, see picture below:



MP rotates over angle 2α to MQ.
As a result SP rotates around S by angle α

Now the last step (picture reduced).
H is center of CF.
HK is diameter of the crcle.



HB rotates over angle 12o to CB.
KH rotates 6o with center K to KC.
x = 6o

Postscript
My impression is that a more elegant solution must be possible.
Who helps?
31.
Calculate angle x in picture below.



Solution

A quick guess seems x = 18 degrees, but this is just an approximation.
A geometrical solution using compass and ruler seems impossible.
Below a trigonometrical solution using the sine rule.


32.
In picture below is
    AB=BC=CA=CD=AE.
Prove that AB = 3.AF



Solution

Draw CG || BF
Draw line through C || AD.
Use similarity of triangles.
See :


33.
Given is a 180 degree circle arc and rectangle ABCD.
BD=10.
Calculate the area of the rectangle.



Solution

First this:



Let r be the radius of the arc.
Let BC=s.
See before:


34.
See picture below with given angles in square ABCD:



Question:
Calculate angle AEF.

Solution




Rotate ΔAFD around point A.
This results in ΔAGB.
Draw GF.
LBAE=25.
LGAE=20+25=45.
AE is bissectrix in isosceles triangle AGF.
So, GF is perpendicular to AE.
LFGB=70-45=25.
LGEA=LAEF=90-25=65.
35.
Calculate angle x in picture below.



Solution (1)

This solution is my own.

LBAC=50+30=80.
LBEA=10+70=80.
so:
BA=BE

Draw perpendicular bisector of AE.
Extend DE, F is the intersection with the perpendicaular bisector.



BF intersects AD in G.
Draw GE.
LFGE=60.
LBDG=60.
So:
LDGE=60.
GE is bisector of exterior angle of ΔBDG.
BE is bisector of LFBC.
Bisectors of a triangle intersect at one point.
So DE also is bisector of LADC=80.
x = 70-40 = 30 degrees.

Solution (2)

This solution is of Nelson Tunala and was found in Facebook group "classical mathematics".



BA=BE (see solution 1)
Extend AD, point F is such that BF=BA.
LBAF=LBFA=50
BE=BF, LEBF=60 so ΔBEF is regular.
LDBF=LDFB=50 so BD=DF.
so:
ΔBED is congruent with ΔFED
LBED=LFED.
x = 60 / 2 = 30 degrees.

Note

The picture has several similar triangles.
I tried to prove that DA : DC = EA: EC
which is the case when DE is a bisector.
36.
Prove that r2 = ab.



Solution

First this:



There we go:


37.
Calculate distance x in rectangle below:



Solution


38.


ABCD is a rectangle, AB=4 and AD=3.
Point E is positioned randomly at line CD.
Calculate the area of triangle EFG.

This problem is interesting for two reasons:
1. the area of triangle EFG is independent of the position of E.
2. I describe two solutions, one with some algebra, a second without.

Notation: the area of triangle ABC is written as [ABC]

Solution (1)





Solution (2)


[AFG] = [ABG]..........result of equal base and height.

[ABE] = [ABG] - [AEG]
[EFG] = [AFG] - [AEG]

[EFG] = [ABE] = 3*4/2 = 6
39.
See picture below.
The triangles other than ΔABC are equilateral.
LA = 60 degrees.
Prove that area S + S2 = S1 + S3.



Preface.
The area of a triangle equals base * height/2.
An equilateral triangle with sides z has area:



See above piture , let :
- AB=c
- AC=b
- BC=a

Apply cosine rule in ΔABC:
    a2 = b2 + c2 - 2bc.cos(60)
    cos(60) = 0,5....
    a2 = b2 + c2 - bc
    vermenigvuldig met x:
    a2x = b2x + c2x - bcx

    a2x = S2
    b2x = S1
    c2x = S3
    bcx = S....

    S2 = S1 + S3 - S

    S + S2 = S1 + S3