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# Finding side of Triangle | PRMO-2014 | Problem 15

Try this beautiful problem from the Pre-Regional Mathematics Olympiad, PRMO, 2014, based on finding side of Triangle. You may use sequential hints.

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

But try the problem first…

Answer:$$26$$

Source

PRMO-2014, Problem 15

Pre College Mathematics

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

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