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To divide a line segment \[\mathbf{AB}\]in the ratio\[\mathbf{5}:\mathbf{7}\], first a ray \[\mathbf{AX}\]is drawn so that \[\mathbf{BAX}\]is an acute angle and then at equal distances points are marked on the ray \[\mathbf{AX}\]such that the minimum number of these points is (A) \[\mathbf{8}\](B) \[\mathbf{10}\](C)\[~\mathbf{11}\] (D) \[\mathbf{12}\]
To divide a line segment \[\mathbf{AB}\]in the ratio\[\mathbf{5}:\mathbf{7}\], first a ray \[\mathbf{AX}\]is drawn so that \[\mathbf{BAX}\]is an acute angle and then at equal distances points are marked on the ray \[\mathbf{AX}\]such that the minimum number of these points is (A) \[\mathbf{8}\](B) \[\mathbf{10}\](C)\[~\mathbf{11}\] (D) \[\mathbf{12}\]
CBSE | Class 10 | Exercise 10.1 | Maths | NCERT Exemplar
To divide a line segment \[\mathbf{AB}\]in the ratio\[\mathbf{5}:\mathbf{7}\], first a ray \[\mathbf{AX}\]is drawn so that \[\mathbf{BAX}\]is an acute angle and then at equal distances points are marked on the ray \[\mathbf{AX}\]such that the minimum number of these points is (A) \[\mathbf{8}\](B) \[\mathbf{10}\](C)\[~\mathbf{11}\] (D) \[\mathbf{12}\]

SOLUTION:-

\[\left( D \right)\text{ }12\]

As indicated by the inquiry,

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

In this way, \[A:B\text{ }=\text{ }5:7\]

Presently,

Draw a beam \[AX\]making an intense point\[\angle BAX\],

Imprint \[A+B\]focuses at equivalent distance.

In this way, we have \[A=5\text{ }and\text{ }B=7\]

Thus, least number of these focuses \[=\text{ }A+B\text{ }=\text{ }5+7\text{ }=12\]

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