Geometry Problem 1520. 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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12V2 (V2 = sqrt of 2)

ReplyDeleteComplete Right Triangle BEF, F on BC extended.

ReplyDeleteComplete Right Triangle DEG, G on DC extended.

Let AD = a and CD = b

Triangles DEG & ACD are congruent ASA

So GE = b = CF and hence EF = a - b and BF = a + b

So in Triangle BEF,

BE^2 = BF^2 + EF^2 = (a+b)^2 + (a-b)^2 = 2 (a^2 + b^2) = 2 AC^2 = 2 X 12^2

Therefore BE = 12. sqrt2

Sumith Peiris

Moratuwa

Sri Lanka

Connect B and D then <BDE=90⁰

ReplyDeleteBE² = BD² + DE²

BE² = 12² + 12² = 12 ROOT2