In a triangle ABC,pt. D is on side AB nd point E is on side AC. BECD is a trapezium.. DE:BC=3:5. calculate the ratio of the area of ADE and BCED ? a)9/25 b)9/16 c)27/125 d)3/5 e)12/25

thephoenix wrote:In a triangle ABC,pt. D is on side AB nd point E is on side AC. BECD is a trapezium.. DE:BC=3:5. calculate the ratio of the area of ADE and BCED ? a)9/25 b)9/16 c)27/125 d)3/5 e)12/25 B

Ajith thanks for the drawing and the explanation below it reminds me of school days But could u please help me understand this = 1/(BC/DE*AF/AG -1) how did u derive it ??? Thanks! must say good work!

bhumika.k.shah wrote:Ajith thanks for the drawing and the explanation below it reminds me of school days But could u please help me understand this = 1/(BC/DE*AF/AG -1) how did u derive it ??? Thanks! must say good work! The area of a triangle = 1/2 base*height area of triangle DEA = 1/2* DE* AG are of triangle ABC = 1/2*BC*AF Area of the trapezium = Area of ABC - Area of DEA Ratio of area of DEA/ area of trapezium = DE*AG/ (BC*AF-DE*AG) I was dividing both numerator and denominator by DE*AG to get 1/(BC/DE*AF/AG -1)

Got it ! ajith wrote:bhumika.k.shah wrote:Ajith thanks for the drawing and the explanation below it reminds me of school days But could u please help me understand this = 1/(BC/DE*AF/AG -1) how did u derive it ??? Thanks! must say good work! The area of a triangle = 1/2 base*height area of triangle DEA = 1/2* DE* AG are of triangle ABC = 1/2*BC*AF Area of the trapezium = Area of ABC - Area of DEA Ratio of area of DEA/ area of trapezium = DE*AG/ (BC*AF-DE*AG) I was dividing both numerator and denominator by DE*AG to get 1/(BC/DE*AF/AG -1)