Geometry Problem 1516. Post your solution in the comment box below.

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

Details: Click on the figure below.

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ReplyDeleteForm equilateral Triangle LMN, L on AB extended, M on CD extended and N on EF extended. Let AF = NF = x and DE = EM = y

ReplyDeleteSince LN = MN,

x + 7 + 3 = x + 2 + y from which y = 8 = DE

Sumith Peiris

Moratuwa

Sri Lanka

Solution 2

ReplyDeleteIt is easy to see that CD//AF and BC//EF.

Complete parallelogram with sides CDM, MEF, FAT, TBC

So TC = FM and so 7 + 3 = 2 + DE since Tr.s ABT and DEM are both equilateral

Hence DE = 8

Sumith Peiris

Moratuwa

Sri Lanka