# 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.

Video Solution
Direct And Inverse Proportions
Ex 13.1 | Question 9

## 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:

 Height of the pole 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}}\\ 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 $$8.75\, \rm{m.}$$

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