8th Class Mathematics Direct and Inverse Proportions

  • 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.

    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.

More Questions

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

adversite



LIMITED OFFER HURRY UP! OFFER AVAILABLE ON ALL MATERIAL TILL TODAY ONLY!

You need to login to perform this action.
You will be redirected in 3 sec spinner

Free
Videos