in the figure of triangle ABC,d divides ca in the ratio of 4:3 .if DE// BC,then find the area of BCDE:are of triangle ABC.(diagram is not provided)




Since, ED  BC  and AB is a transversal, thenAED = ABC    Corresponding anglesSince, ED  BC and AC is a transversal, thenADE = ACB    Corresponding anglesIn  AED and  ABC,AED = ABC   Proved aboveADE = ACB   Proved above    AED ~  ABC   AA AEEB = EDBC = ADAC    Corresponding sides of smilar 's are proportional EDBC = ADAC  .............1Now, ADDC = 43 DCAD = 34 DCAD + 1 = 34 + 1 DC + ADAD = 74 ACAD = 74 ADAC = 47Now, From 1,EDBC = 47

Now, we know that ratio of area of 2 similar 's is equal to squares of ratio of corresponding sides. ar.  AEDar.  ABC = EDBC2 ar.  ABC - ar. BCDEar. ABC = 472 ar.  ABC - ar.  BCDE ar.  ABC = 1649 49 ar.  ABC - 49 ar. BCDE = 16 ar.  ABC 33 ar.  ABC = 49 ar. BCDE ar. BCDEar.  ABC = 3349

  • 40

Your question seems incorrect.I drew the diagram as it was described but I BCED is not a triangle,nor even quadrilateral.It forms like Z if I read.

  • -3

no its right,when de is drawn parallel to third side,triangle AED and BCDE WHICH SEEMS LIKE TRAPEZUIM IS FORMED IT IS CORRECT

  • 17
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