Ex.13.1 Q9 Direct and Inverse Proportions Solution - NCERT Maths Class 8

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Question

 A \(5\,\rm{m}\) \(60\,\rm {cm} \) high vertical pole casts a shadow \(3\;\rm{m}\) \(20 \;\rm{cm}\) long. Find at the same time

(i) Length of the shadow cast by another pole \(10 \;\rm{m}\) \(50 \;\rm{cm}\) high.

(ii) The height of the pole which casts a shadow of \(5\;\rm{m}\) long.

Text Solution

(i) Length of the shadow cast by another pole \(10\; \rm{m}\) \(50 \;\rm{cm}\) high.

Waht is Known?

\(5.6\, \rm{m}\) vertical pole casts a shadow of \(3.2\; \rm{m}\) long.

Waht is Unknown?

The length of a shadow cast by a pole \(10.5\; \rm{m}\) high.

Reasoning:

Two numbers \(x\) and \(y\) are said in direct proportion if

\[\begin{align}\frac{x}{y} = k,\quad x= y\,k\end{align}\]

Where \(k\) is a constant.

Steps:

\({{\bf{Height\; of \;the \;pole}}}\) \({{\bf{Length \;of\; the \;shadow}}}\)
\({5.6{\rm{m}}}\) \({3.2{\rm{m}}}\)
\({10.5{\rm{m}}}\) \(?\)

As the height of the pole increases the length of the shadow also increases. So, it is a direct proportion.

\[\begin{align}\frac{{{x_1}}}{{{y_1}}}&= \frac{{{x_2}}}{{{y_2}}}\\\frac{{5.6}}{{3.2}}&= \frac{{10.5}}{{{y_2}}}\\\,\,5.6\,\, \times &= 10.5 \times 3.2\\{y_2}&= \frac{{10.5 \times 3.2}}{{5.6}}\\{{\rm{y}}_2}& = 6\end{align}\]

If the height of the pole is \(10.5\; \rm{m}\), then length of the shadow is \(6\;\rm{m}\).

(ii) The height of the pole which casts a shadow of \(5\,\rm{m}\) long.

Waht is Known?

\(5.6 \,\rm{m}\) vertical pole casts a shadow of \(3.2\,\rm{m}\) long.

Waht is Unknown?

The height of the pole when the length of the shadow is \(5\,\rm{m}\) long.

Steps:

\[\begin{align}\frac{{{x_1}}}{{{y_1}}}&= \frac{{{x_2}}}{{{y_2}}}\\\frac{{5.6}}{{3.2}}&= \frac{{{x_2}}}{5}\\3.2x &= 5 \times 5.6\\{x_2}&= \frac{{5 \times 5.6}}{{3.2}}\\{x_2}&= 8.75\end{align}\]

If the height of the pole is \(5\,\rm{m}\), then length of the shadow is \(8.75\, \rm{m.}\)