## Thursday, April 13, 2017

### Geometry Problem 1333: Two Squares Side by Side, Three Triangles, Area, Measurement

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

1. Quadrilateral AEGH is cyclic so
And triangle EKG similar to AKH ( case AA)
Ratio of similarity= EK/AK= sqrt( area(EKG)/area(AKH))
Triangle ADE congruent to HFE so AEH is isosceles right triangle
Triangle AEC similar to AKH ( case AA)
Ratio = AE/AK = sqrt( area(AEC)/area(AKH))= sqrt(42/56)= sqrt(3)/2 => AEK is 30-60-90 triangle
EK/AK= ½ => area(EKG)/area(AKH)= ¼ => Area(EKG)= ¼ * 56= 14

2. Problem 1333
(KDH)=(KAH)-(ADH)=56-42=14. But EH//CG (AE perpendicular CG) so GH=CE.
Is EDHG trapezium so x=(KEG)=(KDH)=14.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

3. Name: AD=a, DG=b, S(AED)=m, S(EDK)=n
(a²)/2=42+m, (b²)/2=x+n, (AE²)/2=56+m+n,
Using AE²=a²+b² => x=14

4. Let AD = a and DG = b
AEGH is concyclic and so Tr. AEH is right isosceles.
So Tr.s AED and EFH are congruent and
GH = a-b.
S(AGH)-S(HEG) = 56-x = 1/2((a+b)(a-b) - b(a-b))

So 56-x = 1/2a(a-b) = S(AEC) = 42

Hence x = 56-42 = 14

Sumith Peiris
Moratuwa
Sri Lanka

5. Three triangle are similar.
by pytagorian theorem
x=56-42=14

6. More generally: Area(ACG) + Area(EKG) = Area(AKH)

7. tr ACE, EGK and AKH are all similar.
Apply pytagorian theorem in KGH to get x = 56 - 42 = 14
Nice problem.

8. ACE,AHK & GEK are similar Triangles.
A(ACE)/A(AHK)=42/56=3/4
=>AK=2AE/Sqrt(3)
Let AE=x,EK=y and AK=2x/Sqrt(3)
Apply Pythagoras to AEK
=>y2=x2/3
=>A(GEK)=42/3=14