1.

Determine a point which divides a line segment 6 cm long externally in the ratio 3:5.

Answer»

Solution :STEP of CONSTRUCTION :
1. DRAW a line segment AB=6cm.
2. Draw any ray making an acute ` angleABX` with AB.
3. Along BX, mark off ( LARGER among the ratios) 5 points `B_(1),B_(2),B_(3),B_(4) " and " B_(5)` such that `BB_(1)=B_(1)B_(2)=B_(2)B_(3)=B_(3)B_(4)=B_(4)B_(5)`.
4. Jointhe point `B_(2)(5-3)` with end point A.
5. Draw a line PARALLEL to `B_(2)A` from `B_(5)` (larger among the ratios ) which meets BA produced at P.
The point P so obtained is the required point such that the AP:BP=3:5.
Justification: In `DeltaPBB_(5)`,
Since , `B_(2)A||B_(5)P` ( construction)
`:. (B_(2)B_(5))/(BB_(5))=(AP)/(BP)` ( by B.P theorem)
`implies (3)/(5)=(AP)/(BP)` ( construction)
i.e., P divides AB externally in the ratio 3:5.


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