In given figure, ABC is a triangle right angled at B and BD`bot`AC. If AD=4 cm and CD= 5 cm, then find BD and AB.

Explanation

Given`DeltaABC` in which`angleB=90^(@) and BDbotAC`
Also, AD=4 cm and CD= 5 cm
In `DeltaADB and DeltaCDB, angleADB=angleCDB` [each equal to `90^(@)`]
and `angleBAD=angleDBC`[each equal to `90^(@)-angleC`]
`therefore DeltaDBA~DeltaDCB` [by AAA similarity criterion]
Then, `(DB)/(DA)=(DC)/(DB)`
`rArr DB^(2)=DAxxDC`
`rArrDB^(2)=4xx5`
`rArrDB=2sqrt5cm`
In right angled `DeltaBDC, BC^(2)=BD^(2)+CD^(2)` [by phythogoras therom]
`rArrBC^(2)=(2sqrt5)^(2)+(5)^(2)`
=20+25=45
`rArrBC=sqrt(45)=3sqrt5`
Again, `DeltaDBA~DeltaDCB`
`therefore(DB)/(DC)=(BA)/(BC)`
`rArr(2sqrt5)/5=(BA)/(3sqrt5)`
`therefore BA=(2sqrt5xx3sqrt5)/5=6 cm`
Hence, BD=`2sqrt5` cm and AB= 6 cm