Problem based on Triangle | PRMO-2012| Problem 7

In \(\triangle ABC\) we have \(AC=BC=7\) and \(AB=2\).Suppose that \(D\) is a point on line \(AB\) such that \(B\) lies between \(A\) and \(D\) and \(CD=8\) .what is the length of the segment \(DB\)?

Given that \(AC=BC=7\) & \(CD=8\).we have to find out \(BD\).Let \(BD=x\).we draw a perpendicular from \(C\) to \(AB\) at the point \(M\).Therefore clearly \(\triangle CMB\) & \(\triangle CMD\) are right angle.so if we can find out the value of \(CM\) from the \(\triangle CMB\) then we can find out the value \(BD\) from the \(\triangle CMD\)

Can you now finish the problem ..........

From the above diagram,In \(\triangle CMB\) we can say that \(CM=\sqrt{49-1}=4\sqrt 3\)

Given \(AB=2\) and \(M\) is the mid point of \(\triangle ABC\) (As AC=BC=7,Isosceles triangle),

Therefore \(BM=1\), So \(MD=x+1\)

From the \(\triangle CMD\), \((X+1)^2+(4\sqrt 3)^2=64\) \(\Rightarrow x=3,-5\)

we will take the positive value ,so \(BD=3\)

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