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Area of a Triangle

In the follow...

Question

In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, Then
$a r (F E D) = \frac{1}{8} a r (A F C)$

a r (F E D) = \frac{1}{2} a r (A F C)

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a r (F E D) = \frac{3}{8} a r (A F C)

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a r (F E D) = \frac{1}{4} a r (A F C)

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a r (F E D) = \frac{1}{8} a r (A F C)

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Solution

The correct option is D $a r (F E D) = \frac{1}{8} a r (A F C)$

Area (AFC) = area (AFD) + area (ADC)
$= a r (B E F) + \frac{1}{2} a r (A B C) [I n (i v), a r (B F E) = a r (A F D); A D i s m e d i a n o f Δ A B C]$
$= a r (B F E) + \frac{1}{2} \times 4 a r (B D E) [I n (i), a r (B D E) = \frac{1}{4} a r (A B C)]$
=ar(BEF)+2ar(BDE)....(5)
Now, by (v),
ar (BFE) = 2ar(FED) …….(6)
ar(BDE)=ar(BFE)+ar(FED)
=2ar(FED)+ar(FED)
=3ar(FED) ……(7)
Therefore, from equations (5), (6), and (7),we get:
$a r (A F C) = 2 a r (F E D) + 2 \times 3 a r (F E D) = 8 a r (F E D)$
$∴ a r (A F C) = 8 a r (F E D)$
Hence, $a r (F E D) = \frac{1}{8} = a r (A F C)$

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Similar questions

A B C

B D E

D

B C

A E

B C

F

a r (B D E) = \frac{1}{4} a r (A B C)

a r (B D E) = \frac{1}{2} a r (B A E)

a r (A B C) = 2 a r (B E C)

a r (B F E) = a r (A F D)

a r (B F E) = 2 a r (F E D)

a r (F E D) = \frac{1}{8} a r (A F C)

In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

(i)

(ii)

(iii)

(iv)

(v)

(vi)

[Hint: Join EC and AD. Show that BE || AC and DE || AB, etc.]

a r (B D E) = \frac{1}{2} a r (B A E)

a r (F E D) = \frac{1}{8} a r (A F C)

a r (B D E) = \frac{1}{2} a r (B A E)

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