Question

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Answer 1

Hey there

Let's solve your problem number [ 3 ]

Given that

DE || AC

By collory of thales theorem

$\frac{BD}{BA} = \frac{ED }{CA}$
Given that

BD = a

DA = b

CA = y

and ED = x

$\frac{a}{a + b} = \frac{ x}{y} = > x = \frac{ay}{a + b}$

which is the required answer

Now coming to our next question

• solution No. :- [ 4 ]

∆ABC and ∆BDE are two equilateral triangle such that D is the mid point of BC

Let x be the length of the Three sides of triangle ABC

2BD = BC

or BD = x/2 .... ( 1)

We know that every equilateral triangles are similar to each other, then there ratio of the square of corresponding sides will be

$\frac{ ar(ABC)}{( BDE)} = (\frac{AB}{ BD})^{2} = (\frac{ { \frac{x}{x} }}{2})^{2} = ({ \frac{2x}{x} })^{2} = \frac{4}{1} \\ \\ \frac{ar( ABC)}{ar( BDE)} = \frac{4}{1} \\ \: \\ ar(ABC) \: = 4ar \: (BDE) \\ \\ Hence \: ratio \: is \: 4 : \: 1$
Hope this helps you ☺