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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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