Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Monday, February 15, 2016
Geometry Problem 1188: Triangle, Parallelogram, Parallel Lines, Trapezoid, Area
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SAEDB=SAJHB, SCFGB=SCKHB,
ReplyDeleteDraw AP perpendicular to CK, draw UV//AP (from B)meet etended AJ,CK
SAJKC=K AP, SCKHB=CK BV, SAJHB=AJ BU => SAJKC=SAJHB+SCKHB
Is S1=(AEJ)=(BHD) and S4=(CKF)=(BHG) ,then (ABC)=(JKH) (AJ=BH=CK and paralleles ) ,terefore S+(BLM)=S1+S2+S3+S4+(BLM) or S=S1+S2+S3+S4
DeleteBecause ABHJ and ABDE are both parellelograms easily Tr. AEJ is congruent to Tr. BDH SSS. So S(BDH) = S1
ReplyDeleteSimilarly S(BGH) = S4
Now since ABHJ is a parellelogram AB = JH and similarly BC = HK and AC = JK since ACKH is also a parellelogram ( AJ//KC & AJ = BH = CK)
So Tr.s ABC and HJK are congruent SSS and so S(ABC) = S(HJK)
Subtracting the area of the common triangle BLM,
S = S1+S2+S3+S4
Sumith Peiris
Moratuwa
Sri Lanka
ACKH .....replace with ACKJ
Delete