Geometry Problem 1537. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

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Let S(CHD) = X, S(AEG) = Y. Let the height between AD & BC be h.

ReplyDeleteLet AF = a and EH = p

So S(AEF) = 10 + Y = ah/2 ......(1)

and S(AEH) = 3 +Y = ap/2 ........(2)

(1) - (2); X = (a-p).h/2 = 10 +Y - (3 + Y) = 7

Sumith Peiris

Moratuwa

Sri Lanka

https://photos.app.goo.gl/hsSdBqPvrrEzUCPTA

ReplyDeleteLet a= AF=EC

Let h= height of parallelogram ABCD

Let h1 and h2 are heights shown in the stretch

Let [XYZ]= area of triangle XYZ

Let u= EH/AF= h1/h2=u

We have h1= ụh/(1+u)

And h2= h/(1+u)

Calculate [AGF]=½.a .h2= ½. (ah)/(1+u)

Calculate [GEH]= ½. h1.EH= 1/2/ (au^2)/(1+u)

Calculate [HCD]=½(a-EH)(h)=1/2a.h.(1-u)

Verify that with any the value of u and h we always have [AGF]=[GEH]+[HCD]

So [HCD]= [AGF]-[GEH]= 10- 3= 7