To divide a line segment \[\mathbf{AB}\]in the ratio\[\mathbf{4}:\mathbf{7}\], a ray \[\mathbf{AX}\]is drawn first such that \[\mathbf{BAX}\]is an acute angle and then points \[{{\mathbf{A}}_{\mathbf{1}}},\text{ }{{\mathbf{A}}_{\mathbf{2}}},\text{ }{{\mathbf{A}}_{\mathbf{3}}},\ldots .\]are located at equal distances on the ray \[\mathbf{AX}\]and the point \[\mathbf{B}\]is joined to \[\left( \mathbf{A} \right)\text{ }{{\mathbf{A}}_{\mathbf{12}}}~\left( \mathbf{B} \right)\text{ }{{\mathbf{A}}_{\mathbf{11}}}~\left( \mathbf{C} \right)\text{ }{{\mathbf{A}}_{\mathbf{10}}}~\left( \mathbf{D} \right)\text{ }{{\mathbf{A}}_{\mathbf{9}}}\]
SOLUTION:-
\[\left( \mathbf{B} \right)\text{ }{{\mathbf{A}}_{\mathbf{11}}}\]
As per the inquiry,
A line section\[~AB\] in the proportion \[4:7\]
Thus, \[A:B\text{ }=\text{ }4:7\]
Presently,
Draw a beam \[AX\]making an acute angle \[BAX\]
Least number of focuses situated at equivalent distances on the beam,
Hatchet \[=\text{ }A+B\text{ }=\text{ }4+7=\text{ }11\]
\[{{A}_{1}},\text{ }{{A}_{2}},\text{ }{{A}_{3}},\text{ }\ldots \ldots \ldots .\]
are situated at equivalent distances on the beam \[AX.\]
Point \[B\]is joined to the last point is
\[~{{A}_{11}}.\]
