CTET 2 Math and Science 2011 Paper

Question : 39 of 120

Marks: +1, -0

A cyclist at 'C' is cycling towards 'B'. How far will he have to cycle from C before he is equidistant from both A and B ?

Solution:

Let D be a point such that

\DB=\x

and
Also,

\AD=\DB=\x

To find : CD
Since,

\CB=10

∴\CD=\CB–\DB

\CD=10–x

Now, in ΔEDA,

{\ED}=√{x^{2}-16}

and

{\EB}=10-2=8

Now,

\ED+\DB=\EB

⇒√{x^{2}-16}+x=8

squaring both side, we get
x = 5 km

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