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

Last updated date: 18th Mar 2023

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Answer

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Hint: use the concept of direct linear variation

It has been given that a $5m{\text{ 60cm}}$high vertical pole is casting a shadow $3m{\text{ 20cm}}$long.

Now in first part we need to find the length of shadow cast by another pole which has a height of $10m{\text{ 50cm}}$

Let $x{\text{ m}}$be the length of the shadow that is cast by the pole of height$10m{\text{ 50cm}}$.

In the second part we have to tell the height of a pole which casts a shadow $5{\text{ m}}$long.

So let $y{\text{ m}}$be the height of the pole whose shadow is $5{\text{ m}}$long.

Now clearly it is a case of direct linear variation that is change in one entity directly changes another entity

So $\frac{{{\text{height of pol}}{{\text{e}}_1}}}{{length{\text{ of shado}}{{\text{w}}_1}}} = \dfrac{{{\text{height of pol}}{{\text{e}}_2}}}{{length{\text{ of shado}}{{\text{w}}_2}}}$

$ \Rightarrow \dfrac{{5.60}}{{3.20}} = \dfrac{{10.50}}{x}$ $\left[ {5m{\text{ 60cm = 5}}{\text{.60m}}} \right]$

Solving for x we get

$x = \dfrac{{10.50 \times 3.20}}{{5.60}} = 6$

Hence length of shadow cast by pole of height $10{\text{m 50cm}}$is 6m

Now again in second part is a case of direct linear variation that is change in one entity directly changes another entity we can write

$\dfrac{{{\text{height of pol}}{{\text{e}}_1}}}{{length{\text{ of shado}}{{\text{w}}_1}}} = \dfrac{{{\text{height of pol}}{{\text{e}}_3}}}{{length{\text{ of shado}}{{\text{w}}_3}}}$

$ \Rightarrow \dfrac{{5.60}}{{3.20}} = \dfrac{y}{5}$ $\left[ {5m{\text{ 60cm = 5}}{\text{.60m}}} \right]$

Solving for y we get

$y = \dfrac{{5.60 \times 5}}{{3.20}} = 8.75$

So height of pole which cast a shadow of length 5 m is 8.75m

Note- Whenever we come across such problem statements, the key concept is to think towards direct linear variation which states that change in one entity eventually directly changes another entity, this helps you reach the right solution.

It has been given that a $5m{\text{ 60cm}}$high vertical pole is casting a shadow $3m{\text{ 20cm}}$long.

Now in first part we need to find the length of shadow cast by another pole which has a height of $10m{\text{ 50cm}}$

Let $x{\text{ m}}$be the length of the shadow that is cast by the pole of height$10m{\text{ 50cm}}$.

In the second part we have to tell the height of a pole which casts a shadow $5{\text{ m}}$long.

So let $y{\text{ m}}$be the height of the pole whose shadow is $5{\text{ m}}$long.

Now clearly it is a case of direct linear variation that is change in one entity directly changes another entity

So $\frac{{{\text{height of pol}}{{\text{e}}_1}}}{{length{\text{ of shado}}{{\text{w}}_1}}} = \dfrac{{{\text{height of pol}}{{\text{e}}_2}}}{{length{\text{ of shado}}{{\text{w}}_2}}}$

$ \Rightarrow \dfrac{{5.60}}{{3.20}} = \dfrac{{10.50}}{x}$ $\left[ {5m{\text{ 60cm = 5}}{\text{.60m}}} \right]$

Solving for x we get

$x = \dfrac{{10.50 \times 3.20}}{{5.60}} = 6$

Hence length of shadow cast by pole of height $10{\text{m 50cm}}$is 6m

Now again in second part is a case of direct linear variation that is change in one entity directly changes another entity we can write

$\dfrac{{{\text{height of pol}}{{\text{e}}_1}}}{{length{\text{ of shado}}{{\text{w}}_1}}} = \dfrac{{{\text{height of pol}}{{\text{e}}_3}}}{{length{\text{ of shado}}{{\text{w}}_3}}}$

$ \Rightarrow \dfrac{{5.60}}{{3.20}} = \dfrac{y}{5}$ $\left[ {5m{\text{ 60cm = 5}}{\text{.60m}}} \right]$

Solving for y we get

$y = \dfrac{{5.60 \times 5}}{{3.20}} = 8.75$

So height of pole which cast a shadow of length 5 m is 8.75m

Note- Whenever we come across such problem statements, the key concept is to think towards direct linear variation which states that change in one entity eventually directly changes another entity, this helps you reach the right solution.

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