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InterviewSolution
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Draw a line segment of length 7cm. Find a point P on it which divides it in the ratio 3:5. |
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Answer» Solution :Steps of construction 1. Draw a line segment AB=7 cm. 2. Draw a ray AX , making an acute `angleBAX`. 3. Along AX, MARK 3+5=8 points `A_(1),A_(2),A_(3),A_(4),A_(5),A_(6),A_(6),A_(7),A_(8)` such that `A A_(1)=A_(1)A_(2)=A_(2)A_(3)=A_(3)A_(4)=A_(4)A_(5)=A_(5)A_(6)=A_(6)A_(7)=A_(7)A_(8)` 4. Join `A_(8)B`. From `A_(3)`, draw `A_(3)C||A_(8)` B meeting AB at C. [ by making an angle equal to `angleBA_(8)A " at " A_(3)`] Then, C is the point on AB which divides it in the RATIO 3:5. Thus, AC:CB=3:5  JUSTIFICATION Let `A A_(1)=A_(1)A_(2)=A_(2)A_(3)=A_(3)A_(4)= ....=A_(7)A_(8)=x` In `DeltaABA_(8)` , we have `A_(3)C||A_(8)B` `:. (AC)/(CB)=(A A_(3))/(A_(3)A_(8))=(3x)/(5x)=(3)/(5)` Hence, `AC:CB=3:5` |
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