in a rectangle ABCD , AB=6,BC=3.point E between B and C,and f between E and C are such that BE=EF=FC intersect BD at P and Q respectively.the ratio of bp:pq:qd can be written as r:s:t where the greatest common factor of r,s,t is 1.find r+s+t=?
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since triangle APD is similar to traingle EPB,DP/PB=AD/BE=3.
THIS MEANS THAT DQ=3.BD/5.
AS TRIANGLE APD AND TRIANGLE BEP ARE SIMILAR,WE SEE THAT PD/PB=3/1=THUS PB=BD/4
∴R/S/T=1/4÷2/5-1/4÷3/5=5/3/12=r+s+t=5+3+12=20
∴E=20
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