Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Details: Click on the figure below.
Tuesday, March 12, 2019
Geometry Problem 1424: Triangle, Parallelogram, Trapezoid, Area, Diagonal, Measurement
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BCF is similar to EDF in a 3:1 ratio so BF:FD = 3:1 also therefore the area of BEF is also 3x area of EDF = 3. Then since the BC:ED = 3:1 => AE:ED 2:1 and the area of ABE is 2x AED = 2(3 + 1) = 8.
ReplyDeleteSo the total is AED + BEF = 8 + 3 = 11
Triangle BFC is similar to DFE with ratio of similarity= 1/3
ReplyDeleteSo ratio of 2 altitudes from F of these 2 triangles =1/3
So Area(BFC)/Area(ABD)= ration of altitudes from F and B= ¾
So area(ABD)=12 and area(ABFE)=11
From the given: BC = 3 ED, => S(FDC) = 3
ReplyDelete=> S(x) = 11
The 2 triangles BCF and DEF are similar and hence the squares of the corresponding sides are proportional to the respective areas
ReplyDeleteSo if DE = p, BC = AD = 3p
Also if DF = q, BF = 3q
Hence S(DEF) = p.q.sin<EDF and S(ADB) = 3p.4q.sin<EDF = 12p.q.sin<EDF = 12.S(DEF) = 12
Therefore S(ABFE) = 12 - 1 = 11
Sumith Peiris
Moratuwa
Sri Lanka
Solution 2
ReplyDeleteUsing the same notations as before S(BEF) = 3.S(DEF) = 3
S(ABE) = 2.S(BED) = 2 X (1 + 3) = 8
So S(ABFE) = 8 + 3 = 11
Sumith Peiris
Moratuwa
Sri Lanka