Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Thursday, April 13, 2017

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

Labels:
area,
measurement,
square,
triangle

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Quadrilateral AEGH is cyclic so

ReplyDelete∠ (EAD)= ∠ (KHG)

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

Problem 1333

ReplyDeleteLet (ACE) is area triangle ACE then (ACE)=CE.AD/2=GH.AD/2=(ADH)=42.

(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

Name: AD=a, DG=b, S(AED)=m, S(EDK)=n

ReplyDelete(a²)/2=42+m, (b²)/2=x+n, (AE²)/2=56+m+n,

Using AE²=a²+b² => x=14

Let AD = a and DG = b

ReplyDeleteAEGH 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

Three triangle are similar.

ReplyDeleteby pytagorian theorem

x=56-42=14

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

ReplyDeletetr ACE, EGK and AKH are all similar.

ReplyDeleteApply pytagorian theorem in KGH to get x = 56 - 42 = 14

Nice problem.

ACE,AHK & GEK are similar Triangles.

ReplyDeleteA(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