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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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