Question

A right angled triangle $OPQ$ is right angled at $O$. $R$ is the point on $OQ$
If $PQ=17cm,QR=9cm$ and $PR=10cm$. Find $RO$

Answer 1

In this figure there are two triangle $△PORand△POQ$ in both the triangle angle made at O is right angle .

Let the length of OR be x unit.

Apply pythagoras theorm for both the triangle

$Int{\left(PQ\right)}^2={\left(OQ\right)}^2+(OP{)}^2\Rightarrow {17}^2={(x+9)}^2+{\left(OP\right)}^2\Rightarrow {\left(OP\right)}^2={17}^2-{(x+9)}^2$

$(PR{)}^2={\left(OR\right)}^2+{\left(OP\right)}^2\Rightarrow {10}^2=x^2+{\left(OP\right)}^2\Rightarrow {\left(OP\right)}^2={10}^2-x^2$

Since it is clear that the OP is same in both the equations ,so we can equate both the equations

${17}^2-(x+9{)}^2={10}^2-x^2\Rightarrow 289-100=x^2+18x+81-x^2\Rightarrow 189-81=18x\Rightarrow 18x=108\Rightarrow x=6$