Geometry Problem 1517. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.
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V6 (sqrt of 6)
ReplyDeletehttps://photos.app.goo.gl/CkcFLDMpaKUyBK3a9
DeleteDrop perpendiculars DP and DQ from D to AB and from D to BC, P on AB and Q on BC respectively.
ReplyDeleteSince AD = DC, S(ABD) = S(BDC)
So AB.DP/2 = BC.DQ/2
Hence (c/2)d/(sqrt2) = (1/2).2. (sqrt3./2)d
Simplifying c = sqrt 6
Sumith Peiris
Moratuwa
Sri Lanka
Construct parallelogram BAEC.
ReplyDeleteDrop perpendicular AH, from A to BE.
AEH is a 30-60-90 triangle, and ABH is a 45-45-90 triangle.
BC=AE=2, AH=sqrt 3, AB=sqrt 6.
AB/AD = sin D / sin 45.
ReplyDeleteDC/BC = sin 60 / sin 45.
Multiply side by side
AB/BC = sqrt 3 / sqrt 2 => AB = sqrt 6.