Answer:
Let the height of the vertical pole be x m and the length of the shadow be y m. As the height of the vertical pole increases, the length of the shadow also increases in the same ratio. It is a case of direct proportion. We make use of the relation of the type \[\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}\]. (i) Here \[{{x}_{1}}=5\,m\,\,60\,cm\,=5.60\,m\] \[{{y}_{1}}=3\,\,m\,20\,cm\,\,=3.20\,m\] \[{{x}_{2}}=10\,m\,50\,cm\,=10.50\,m\] Therefore, \[\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{2}}}{{{y}_{2}}}\] gives \[\frac{5.6}{3.2}\,=\frac{10.5}{{{y}_{2}}}\] \[\Rightarrow \] \[5.6\,{{y}_{2}}=3.2\times 10.5\] \[\Rightarrow \] \[{{y}_{2}}=\frac{3.2\times 10.5}{5.6}\] \[\Rightarrow \] \[{{y}_{2}}=6\] Hence, the length of the shadow is 6 m. (ii) Here \[{{x}_{1}}=5\,m\,\,60\,\,cm=560\,\,cm\] \[{{y}_{1}}=\,3\text{ }m\text{ }20\text{ }cm=320\text{ }cm\] \[{{y}_{3}}=5\text{ }m=500\text{ }cm\] Therefore, \[\frac{{{x}_{1}}}{{{y}_{1}}}=\frac{{{x}_{3}}}{{{y}_{3}}}\] gives \[\frac{560}{320}\,=\frac{{{x}_{3}}}{500}\] \[\Rightarrow \] \[320\,{{x}_{3}}=560\times 500\] \[\Rightarrow \] \[{{x}_{3}}=\frac{560\times 500}{320}\] \[\Rightarrow \] \[{{x}_{3}}=875\] Hence, the height of the pole is 8 m 75 cm.
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