• # question_answer 9)                 A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time (0 the length of the shadow cast by another pole 10 m 50 cm high (ii) the height of a pole which casts a shadow 5 m long.

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.