Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 838.

## Friday, December 28, 2012

### Problem 838: Parallelogram, Perpendicular, Diagonal, Metric Relations

Labels:
diagonal,
metric relations,
parallelogram,
perpendicular,
triangle

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cosBAD = cosEBA = 3/4

ReplyDeleteBy cosine law,

x^2 = 4^2 + 5^2 - 2(4)(5)cosBAD

x = sqrt11

sqrt(11)

ReplyDeleteUsing Pythagoras Theorem,

ReplyDeleteAE^2 = 4^2 − 3^2 = 7

AC^2 = 8^2 + 7 = 71

By Parallelogram Law,

2×(4^2 + 5^2) = x^2 + 71

x^2 = 11

x = √11

Notam cu P punctul in care paralela la BD prin E intalneste pe AD.Vom obtine

ReplyDeleteEP=BD=X;EB=DP=5-3=2.Aplicam teorema lui Pitagora

AE^2 = 4^2 − 3^2 = 7

X=EP^2=AE^2+AP^2=7+2^2=11 =>x = √11

Problem 838

ReplyDeleteDraw BF perpendicular in AD,then AF=3 and FD=5-3=2.Is BD^2=BF^2+FD^2=(4^2-3^2)+2^2=16-9+4=11.So x=√11.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE