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In the figure, ABC and BDE are two equilateral triangles such that D is the midpoint of BC. If AE intersects BC at F, then
  • a)
    ar(FED) = ar(AFC)
  • b)
    ar(FED) = ar(AFC)/2
  • c)
    ar(FED) = ar(AFC)/4
  • d)
    ar(FED) = ar(AFC)/8
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
In the figure, ABC and BDE are two equilateral triangles such that D i...
Let the altitude of the triangle ABC is h
Then altitude of the triangle BED = h/2
Now, ar(Δ FED) = (1/2)*FD*(h/2) = (FD  * h)/2
and ar(Δ AFC) = (1/2)*FC*h
 = (1/2)*(FD + DC)*h  
= (1/2)*(FD + BD)*h
= (1/2)*(FD + BF +  FD)*h
= (1/2)*(2FD + BF)*h
= (1/2)*(2FD + 2FD)*h
= 2FD*H
⇒  ar(Δ AFC)/8 = 2FD*H/8
⇒  ar(Δ AFC)/8 = 2FD*H/4
⇒  ar(Δ AFC)/8 = ar(Δ FED)
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Community Answer
In the figure, ABC and BDE are two equilateral triangles such that D i...
Let the altitude of the triangle ABC is h
Then altitude of the triangle BED = h/2
Now, ar(Δ FED) = (1/2)*FD*(h/2) = (FD  * h)/2
and ar(Δ AFC) = (1/2)*FC*h
 = (1/2)*(FD + DC)*h  
= (1/2)*(FD + BD)*h
= (1/2)*(FD + BF +  FD)*h
= (1/2)*(2FD + BF)*h
= (1/2)*(2FD + 2FD)*h
= 2FD*H
⇒  ar(Δ AFC)/8 = 2FD*H/8
⇒  ar(Δ AFC)/8 = 2FD*H/4
⇒  ar(Δ AFC)/8 = ar(Δ FED)
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In the figure, ABC and BDE are two equilateral triangles such that D is the midpoint of BC. If AE intersects BC at F, thena)ar(FED) = ar(AFC)b)ar(FED) = ar(AFC)/2c)ar(FED) = ar(AFC)/4d)ar(FED) = ar(AFC)/8Correct answer is option 'D'. Can you explain this answer? for Class 9 2025 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about In the figure, ABC and BDE are two equilateral triangles such that D is the midpoint of BC. If AE intersects BC at F, thena)ar(FED) = ar(AFC)b)ar(FED) = ar(AFC)/2c)ar(FED) = ar(AFC)/4d)ar(FED) = ar(AFC)/8Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 9 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the figure, ABC and BDE are two equilateral triangles such that D is the midpoint of BC. If AE intersects BC at F, thena)ar(FED) = ar(AFC)b)ar(FED) = ar(AFC)/2c)ar(FED) = ar(AFC)/4d)ar(FED) = ar(AFC)/8Correct answer is option 'D'. Can you explain this answer?.
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