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

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

Click the figure below to view more details of problem 1188.

## Monday, February 15, 2016

### Geometry Problem 1188: Triangle, Parallelogram, Parallel Lines, Trapezoid, Area

Subscribe to:
Post Comments (Atom)

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