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To divide a line segment \[\mathbf{AB}\]in the ratio\[\mathbf{5}\text{ }:\text{ }\mathbf{6}\], draw a ray \[\mathbf{AX}\]such that \[\angle \mathbf{BAX}\] is an acute angle, then draw a ray \[\mathbf{BY}\]parallel to \[\mathbf{AX}\]and the points \[{{\mathbf{A}}_{\mathbf{1}}},\text{ }{{\mathbf{A}}_{\mathbf{2}}},\text{ }{{\mathbf{A}}_{\mathbf{3}}},\text{ }\ldots \]and \[{{\mathbf{B}}_{\mathbf{1}}},\text{ }{{\mathbf{B}}_{\mathbf{2}}},\text{ }{{\mathbf{B}}_{\mathbf{3}}},\text{ }\ldots \]are located at equal distances on ray \[\mathbf{AX}\]and\[\mathbf{BY}\], respectively. Then the points joined are \[\left( \mathbf{A} \right)\text{ }{{\mathbf{A}}_{\mathbf{5}}}~\mathbf{and}\text{ }{{\mathbf{B}}_{\mathbf{6}}}~\] \[\left( \mathbf{B} \right)\text{ }{{\mathbf{A}}_{\mathbf{6}}}~\mathbf{and}\text{ }{{\mathbf{B}}_{\mathbf{5}}}~\] \[~\left( \mathbf{C} \right)\text{ }{{\mathbf{A}}_{\mathbf{4}}}~\mathbf{and}\text{ }{{\mathbf{B}}_{\mathbf{5}}}\] \[\left( \mathbf{D} \right)\text{ }{{\mathbf{A}}_{\mathbf{5}}}~\mathbf{and}\text{ }{{\mathbf{B}}_{\mathbf{4}}}\]

\[\left( \mathbf{A} \right)\text{ }{{\mathbf{A}}_{\mathbf{5}}}~\mathbf{and}\text{ }{{\mathbf{B}}_{\mathbf{6}}}\]

As per the inquiry,

A line portion \[AB\]in the proportion \[5:7\]

Along these lines,\[~A:B\text{ }=\text{ }5:7\]

 

Steps of development:

  1. Draw a beam \[AX,\]an intense point \[BAX.\]
  2. Draw a beam\[BY\text{ }||AX\], \[point\text{ }ABY\text{ }=\text{ }point\text{ }BAX.\]
  3. Presently, find the focuses \[A1,A2,A3,A4\text{ }and\text{ }A5\]on \[AX\]and\[~B1,B2,B3,B4,B5\text{ }and\text{ }B6\]

\[\left( Since\text{ }A:B\text{ }=\text{ }5:7 \right)\]

  1. Join\[{{A}_{5}}{{B}_{6}}.\]

 

\[{{A}_{5}}{{B}_{6}}.\]meet \[AB\]at a point\[~C\].

\[AC:BC=\text{ }5:6\]

NCERT Exemplar Class 10 Maths Chapter 10 Ex. 10.1 Question-3