In given figure PQR is a right angled triangle, right angled at Q and QS`bot`PR. If PQ=6 cm and PS=4cm, then find QS, RS and QR.

Explanation

Given `DeltaPQR` in which `angleQ=90^(@)`,QS`bot`PR and PQ=6 cm, PS=4cm
In `DeltaSQP and DeltaSRQ`,
`anglePSQ=angleRSQ` [each equal to `90^(@)`]
`angleSPQ=angleSQR` [each equal to `90^(@)-angleR`]
`therefore DeltaSQP~DeltaSRQ`
Then, `(SQ)/(PS)=(SR)/(SQ)`
`rArr SQ^(2)=PSxxSR` ...(i)
In right angled `DeltaPSQ, PQ^(2)=PS^(2)+QS^(2)` [by Pythagoras theorem]
`rArr (6)^(2)=(4)^(2)=QS^(2)`
`rArr 36=16+QS^(2)`
`rArr QS^(2)=36-16=20`
`therefore QS=sqrt(20)=2sqrt5cm`
On puttting the value of QS in Eq. (i) we get
`(2sqrt5)^(2)=4xxSR`
`rArr SR=(4xx5)/4=5 cm`
In right angled `DeltaQSR, QR^(2)=QS^(2)+SR^(2)`
`rArr QR^(2)=(2sqrt5)^(2)+(5)^(2)`
`rArr QR^(2)+20+25`
`therefore QR=sqrt(45)=3sqrt5`cm
Hence, QS=`2sqrt5` cm RS= 5 cm and QR= `3sqrt5` cm