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Draw a line segment of length 7 cm. 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 `AX"||"BY` such that `angleAand angleB` are acute angles. 3. Divide AX and BY in 3 and 5 parts equally by compass and mark `A_(1),A_(2),A_(3),B_(1),B_(2),B_(3),B_(4)and B_(5)` respectively. 4. Join `A_(3)B_(5)` which intersect AB at P and divides `AP: PB = 3:5.` Hence P is the REQUIRED point on AB which divide it in `3:5.` Verification (Justicicatio) : `In DeltaAA_(3)P and DeltaBB_(5)P` `""AX"||"BY""("by construction")` `""angleA=angleB""("alt.angles")` `""angleA_(3)PA=angleB_(5)PB""("vertically opp. angles")` `therefore""DeltaA A_(3)P~DeltaBB_(5)P""("by AA criterion of SIMILARITY")` `implies""(A A_(3))/(BB_(5))=(AP)/(BP)""("LET ecah equal part = c cm" thereforeA A_(1)=A_(1)A_(2)=B_(1)B_(2)...=x)` `implies""(3x)/(5x)=(AP)/(BP) implies""AP:BP=3:5` 
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