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In given figure BD: DF: FC = 7:3:2, AE:EC=5:4. FG is Angle bisector of angle EFC such that EG: GC=2:1. if ar△ABC=2430cm², then find area of triangle FGC?
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In given figure BD: DF: FC = 7:3:2, AE:EC=5:4. FG is Angle bisector of...
Given Information:
- BD: DF: FC = 7:3:2
- AE: EC = 5:4
- EG: GC = 2:1
- Area of △ABC = 2430 cm²
Calculating the Lengths:
- Let BD = 7x, DF = 3x, FC = 2x
- Let AE = 5y, EC = 4y
- Let EG = 2z, GC = z
Calculating the Areas:
- Area of △ABC = Area of △ABD + Area of △DFC + Area of △BEC
- Let K be the area of △BEC
- Using the formula for area of a triangle, we can write:
K = 1/2 * BE * EC
- Since AE:EC = 5:4, BE = 5z, EC = 4z
- Therefore, K = 1/2 * 5z * 4z = 10z²
- Area of △BEC = 10z²
Finding the Area of △FGC:
- Since FG is the angle bisector of ∠EFC, we have EF/FC = EG/GC
- Substituting the values, we get (7x + 3x)/2x = 2z/z
- Solving, we get x = 2z
- Therefore, FG = EF = 10x = 20z
- Area of △FGC = 1/2 * FG * GC = 1/2 * 20z * z = 10z²
Calculating the Final Area:
- Since EG:GC = 2:1, we have EG = 2z, GC = z
- Area of △FGC = 10z²
- Area of △FGC: Area of △BEC = 10z² : 10z² = 1:1
- Area of △FGC = 1/2 * Area of △ABC = 1/2 * 2430 cm² = 1215 cm²
Therefore, the area of triangle FGC is 1215 cm².
- BD: DF: FC = 7:3:2
- AE: EC = 5:4
- EG: GC = 2:1
- Area of △ABC = 2430 cm²
Calculating the Lengths:
- Let BD = 7x, DF = 3x, FC = 2x
- Let AE = 5y, EC = 4y
- Let EG = 2z, GC = z
Calculating the Areas:
- Area of △ABC = Area of △ABD + Area of △DFC + Area of △BEC
- Let K be the area of △BEC
- Using the formula for area of a triangle, we can write:
K = 1/2 * BE * EC
- Since AE:EC = 5:4, BE = 5z, EC = 4z
- Therefore, K = 1/2 * 5z * 4z = 10z²
- Area of △BEC = 10z²
Finding the Area of △FGC:
- Since FG is the angle bisector of ∠EFC, we have EF/FC = EG/GC
- Substituting the values, we get (7x + 3x)/2x = 2z/z
- Solving, we get x = 2z
- Therefore, FG = EF = 10x = 20z
- Area of △FGC = 1/2 * FG * GC = 1/2 * 20z * z = 10z²
Calculating the Final Area:
- Since EG:GC = 2:1, we have EG = 2z, GC = z
- Area of △FGC = 10z²
- Area of △FGC: Area of △BEC = 10z² : 10z² = 1:1
- Area of △FGC = 1/2 * Area of △ABC = 1/2 * 2430 cm² = 1215 cm²
Therefore, the area of triangle FGC is 1215 cm².
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In given figure BD: DF: FC = 7:3:2, AE:EC=5:4. FG is Angle bisector of angle EFC such that EG: GC=2:1. if ar△ABC=2430cm², then find area of triangle FGC?
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In given figure BD: DF: FC = 7:3:2, AE:EC=5:4. FG is Angle bisector of angle EFC such that EG: GC=2:1. if ar△ABC=2430cm², then find area of triangle FGC? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about In given figure BD: DF: FC = 7:3:2, AE:EC=5:4. FG is Angle bisector of angle EFC such that EG: GC=2:1. if ar△ABC=2430cm², then find area of triangle FGC? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In given figure BD: DF: FC = 7:3:2, AE:EC=5:4. FG is Angle bisector of angle EFC such that EG: GC=2:1. if ar△ABC=2430cm², then find area of triangle FGC?.
In given figure BD: DF: FC = 7:3:2, AE:EC=5:4. FG is Angle bisector of angle EFC such that EG: GC=2:1. if ar△ABC=2430cm², then find area of triangle FGC? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about In given figure BD: DF: FC = 7:3:2, AE:EC=5:4. FG is Angle bisector of angle EFC such that EG: GC=2:1. if ar△ABC=2430cm², then find area of triangle FGC? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In given figure BD: DF: FC = 7:3:2, AE:EC=5:4. FG is Angle bisector of angle EFC such that EG: GC=2:1. if ar△ABC=2430cm², then find area of triangle FGC?.
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