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Draw a line segment of length (7.6;cm) and divide it in the ratio (5:8.)Measure the two parts.

Question

Draw a line segment of length \(7.6\;cm\) and divide it in the ratio \(5:8.\)Measure the two parts.

Answer

see the solution below
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Solution

Steps of construction·
1.Draw a line segment\(AB=7.6\;cm\)
2.Draw a ray \(AX,\)making an acute \(\angle BAX\).
3. Along\(AX,\). mark\(5+8=13\)points \(A_{1},A_{2},A_{3},A_{4},\cdots ,A_{12},A_{13}\)such that
\(AA_{1}=A_{1}A_{2}=A_{2}A_{3}=A_{3}A_{4}=A_{4}A_{5}=A_{5}A_{6}=A_{6}A_{7}\)\(=A_{7}A_{8}=A_{8}A_{9}=A_{9}A_{10}=A_{10}A_{11}=A_{11}A_{12}=A_{12}A_{13}\)
4. Join\(A_{13}B.\)
5. From\(A_{5}\)draw\(A_{5}O|| A_{13}B\) meeting \(AB\) at \(O\). [By making an angle equal to \(\angle AA_{13}B]\)
Then, \(O\) is the point on AB which divides it in the ratio \(5:8\).
So,\(AO:OB=5:8\)

Justification
Let \(AA_{1}=A_{1}A_{2}=A_{2}A_{3}=A_{3}A_{4}=A_{4}A_{5}\)
\(=A_{5}A_{6}=A_{5}A_{7}=\cdots =A_{12}A_{13}=x\)
\(\Delta ABA_{13},\)we have
\(A_{5}O|| A_{13}B\)
\(\therefore \frac {AO}{OB}=\frac {AA_{5}}{A_{5}A_{13}}=\frac {5x}{8x}=\frac {5}{8}\)      (By basic proportionality theorem)
Hence,\(AO:OB=5:8\)
On measuring, we find that\(AO=2.9\;cm\)
and \(OB=4.7\;cm\)
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