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In figure $$ABC$$ and $$BDE$$ are two equilateral triangles such thta $$D$$ is the mid-point of $$BC.AE$$ intersects $$BC$$ in $$F$$. Prove that
$$ar(\triangle FED)=\dfrac {1}{8} ar(\triangle AFC)$$
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Q1
In figure $$ABC$$ and $$BDE$$ are two equilateral triangles such thta $$D$$ is the mid-point of $$BC.AE$$ intersects $$BC$$ in $$F$$. Prove that
$$ar(\triangle FED)=\dfrac {1}{8} ar(\triangle AFC)$$
$$ar(\triangle FED)=\dfrac {1}{8} ar(\triangle AFC)$$
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Q2
In figure $$ABC$$ and $$BDE$$ are two equilateral triangles such thta $$D$$ is the mid-point of $$BC.AE$$ intersects $$BC$$ in $$F$$. Prove that
$$ar(\triangle BDE)=\dfrac {1}{4}ar(\triangle ABC)$$
$$ar. (\triangle ABC)=2 ar.(\triangle BEC)$$
$$ar(\triangle BDE)=\dfrac {1}{4}ar(\triangle ABC)$$
$$ar. (\triangle ABC)=2 ar.(\triangle BEC)$$
View Solution
Q3
In figure $$ABC$$ and $$BDE$$ are two equilateral triangles such thta $$D$$ is the mid-point of $$BC.AE$$ intersects $$BC$$ in $$F$$. Prove that
$$ar(\triangle BDE)=\dfrac {1}{4}ar(\triangle ABC)$$
$$ar. (\triangle FED)=\dfrac 18 ar.(\triangle AFC)$$
$$ar(\triangle BDE)=\dfrac {1}{4}ar(\triangle ABC)$$
$$ar. (\triangle FED)=\dfrac 18 ar.(\triangle AFC)$$
View Solution
Q4
In figure $$ABC$$ and $$BDE$$ are two equilateral triangles such thta $$D$$ is the mid-point of $$BC.AE$$ intersects $$BC$$ in $$F$$. Prove that
$$ar(\triangle BDE)=\dfrac {1}{4}ar(\triangle ABC)$$
$$ar. (\triangle BEF)=ar.(\triangle AFD)$$
$$ar(\triangle BDE)=\dfrac {1}{4}ar(\triangle ABC)$$
$$ar. (\triangle BEF)=ar.(\triangle AFD)$$
View Solution
Q5
In figure $$ABC$$ and $$BDE$$ are two equilateral triangles such thta $$D$$ is the mid-point of $$BC.AE$$ intersects $$BC$$ in $$F$$. Prove that
$$ar(\triangle BDE)=\dfrac {1}{4}ar(\triangle ABC)$$
$$ar. (\triangle BDE)=\dfrac 12 ar.(\triangle BAE)$$
$$ar(\triangle BDE)=\dfrac {1}{4}ar(\triangle ABC)$$
$$ar. (\triangle BDE)=\dfrac 12 ar.(\triangle BAE)$$
View Solution