1.

P,Q and R are on AB, BC and AC of the equilateral triangle ABC respectively. AP:PB=CQ:QB=1:2. G is the centroid of the triangle PQB and R is the mid-points of AC. Find BG:GR.

Answer»

`1:2`
`2:3`
`3:4`
`4:5`

Solution :
LET `AB=BC=AC=3x`
`therefore BP=BQ=PQ=2x`
`(because AP:BP=CQ:BQ=1:2)`
As `Delta PRQ and Delta BAC` are equilateral triangle, the centroid of `Delta BPQ` LIES on BR (where BR is median drawn on to AC).
We KNOW that centroid divides the median in the ratio `2:1`.
`therefore BC=(2)/(3)([SQRT(3)(2x)])/(2)=(2sqrt(3x))/(3)`
But `BR=(sqrt((3)))/(2)=(3sqrt(3x))/(2)`
Now `GR=BR-BG`
`=(3sqrt(3x))/(2)-(2sqrt(3x))/(3)=(5sqrt(3x))/(6)`
Now `BC :GR =(2sqrt(3x))/(3):(5sqrt(3x))/(6) =4:5`.


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