June 12, 2020

Finding side of Triangle | PRMO-2014 | Problem 15

Contents

Try this beautiful problem from PRMO, 2014 based on Finding side of Triangle.

Finding side of Triangle | PRMO | Problem 15

Let XOY be a triangle with angle XOY=90 degrees. Let M and N be the midpoints of the legs OX and OY, respectively. Suppose that XN=19 and YM=22. what is XY?

\(28\)
\(26\)
\(30\)

Key Concepts

Geometry

Triangle

Pythagoras

Check the Answer

But try the problem first...

Answer:\(26\)

Source

Try with Hints

First hint

Given that \(\angle XOY=90^{\circ}\) .Let M and N be the midpoints of the legs OX and OY, respectively, and that XN=19 and YM=22. Now \(\triangle XON\) & \(\triangle MOY\) are Right angle Triangle. Use Pythagoras theorem .......

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

Second Step

Let \(XM=MO=p\) and \(ON=NY=q\).Now using Pythagoras theorm on \(\triangle XON\) & \(\triangle MOY\) we have...

\(OX^2 +ON^2=XN^2\) \(\Rightarrow 4p^2 +q^2=19^2\) \(\Rightarrow 4p^2 +q^2=361\)...........(1) and \(OM^2 +OY^2=MY^2\) \(\Rightarrow p^2 +4q^2=22^2\) \(\Rightarrow p^2 +4q^2=484\)......(2)

Final Step

Now Adding (1)+(2)=\((4p^2 +q^2=361)\)+\((p^2 +4q^2=484\) \(\Rightarrow 5(p^2+q^2)=845\) \(\Rightarrow (p^2+q^2)=169\) \(\Rightarrow 4(p^2+q^2)=676\) \(\Rightarrow (OX)^2+(OY)^2=(26)^2\) \(\Rightarrow (XY)^2=(26)^2\) \(\Rightarrow XY=26\).