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  • In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that (vi) ar (FED) = 1/8 ar (AFC)

Q : 5    In Fig. 9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

(vi) ar (FED) = \frac{1}{8} ar (AFC)

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

Answers (1)

best_answer


In right angled triangle \DeltaABD, we get

\\AB^2 = AD^2+BD^2\\ AD^2 = AB^2-BD^2
            =a^2 - \frac{a^2}{4}=\frac{3a^2}{4}
AD=\frac{\sqrt{3}a}{2}

So, in \DeltaPED,

PE^2 = DE^2-DP^2
             \\=(\frac{a}{2})^2-(\frac{a}{4})^2\\ =\frac{3a^2}{16}
So,  PE=\frac{\sqrt{3}a}{4}

Therefore, the Area of \Delta AFD =1/2 (FD)\frac{\sqrt{3}a}{2}..........(i)

And, Area of triangle \Delta EFD =1/2 (FD)\frac{\sqrt{3}a}{4}...........(ii)

From eq (i) and eq (ii), we get
ar(\DeltaAFD) = 2. ar(\DeltaEFD) 
Since ar(\DeltaAFD) = ar(\DeltaBEF)

\Rightarrow\small ar(BFE)=2ar(FED) --------(1)

ar(\Delta AFC) =
\\= ar(\Delta AFD) + ar(\Delta ADC)\\ = ar(\Delta BFE) + 1/2. ar(\Delta ABC)\\ = ar(\Delta BFE) + 1/2\times [4\times ar(\Delta BDE)]\\ = ar(\Delta BFE) + 2\times ar(\Delta BDE)
= 2\times ar (\Delta FED) + 2\times [ar(\Delta BFE) + ar(\Delta FED)] .....[from (1)] 
 \\=2ar(\Delta FED) + 2[3\times ar(\Delta FED)]\\ =8\times ar(\Delta FED)

ar (FED) = \frac{1}{8} ar (AFC)
Hence proved.

Posted by

manish painkra

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