Question

in the given figure of triangle ABC d divides CA in the ratio 4 is to 3 and if de parallel BC then find area quadrilateral bcde is 2 area triangle ABC​

Answer 1

Answer:

33:49

Step-by-step explanation:

ED||BC

∠AED=∠ABC

∠ADE=∠ACB

INΔABC AND ΔAED

WE HAVE

∠AED=∠ABC

∠ADE=∠ACB

SO ΔAED~ΔABC

ae/ab=ed/bc=ad/ac

ed/bc=ad/ac...........(1)

ad/dc=4/3

dc/ad=3/4

dc/ad+1=3/4+1

ac/ad=7/4

ad/ac=4/7

froam e(1)

ed/bc=4/7

ratio of areas of two similar triangle is equal to the square of ratios of their corresponding sides

ar(aed)/ar(abc)=(4/7)^2

ar(Abc)-ar(bcde)/ar(abc)=16/49

49ar(abc)-49ar(bcde)=16ar(abc)

33ar(abc)=49ar(bcde)

ar (bcde)/ar(abc)=33/49

ar (bcde):ar(abc)=33:49

HOPE IT'S HELP YOU