Question

A ${\text{5m 60cm}}$ high vertical pole casts a shadow ${\text{3m 20cm}}$ long. Find at the same time
\[\left( 1 \right)\]The length of the shadow cast by another pole${\text{10m 50cm}}$ high
\[\left( 2 \right)\]The height of a pole which casts a shadow $5m$ long.

Answer 1

Hint: First we convert all the given data in cm.
Here we use the concept of linear variation and direct proportion method.
Simplify the equations.
Finally, we get the required answer.

Complete step-by-step answer:
First we have to convert all terms in $cm$
$ \Rightarrow {\text{1 m = 100 cm}}$
That is we can write it as,
$5m60cm = 5m + 60cm$
We have to multiply \[100\] to convert into \[m\]
 =$5 \times 100cm + 60cm$
On multiply the terms we get,
$ = 500 + 60$
Let us add the terms we get,
$ = 560cm$
Then we take,
$3m20cm = 3m + 20cm$
We have to multiply \[100\] to convert into \[m\]
$ = 3 \times 100cm + 20cm$
On multiply the terms we get,
$ = 300 + 20$
Let us add the terms we get,
$ = 320$
Also we take,
$10m50cm = 10m + 50cm$
We have to multiply \[100\] to convert into
$ = 10 \times 100cm + 50cm$
On multiply the terms we get,
$ = 1000 + 50$
We have to add the terms we get,
$ = 1050cm$
Now, our question becomes like that,
A $560cm$ high vertical pole casts a shadow $320cm$ long.
We need to find length if shadow cast by pole of height $1050cm$
Let \[y\]be the length of shadow cast by pole of height $1050cm$
 Height of pole (in cm) =$560, 1050$
Length of shadow (in cm) = $320,y$
Here clearly we have seen that the height of the pole increases, so length of shadow also increases.
We have to use it in direct proportion.
That is we can write it as, $\dfrac{{560}}{{320}} = \dfrac{{1050}}{y}$
We just cancel the zero in the first fraction we get,
$\dfrac{{56}}{{32}} = \dfrac{{1050}}{y}$
Let us take a cross multiplication we get,
$56 \times y = 1050 \times 32$
We take a variable as in LHS and remaining take as in RHS we get,
$y = \dfrac{{1050 \times 32}}{{56}}$
Now, we cancel the terms of the same multiple
$y = \dfrac{{1050 \times 16}}{{28}}$
Again, cancelled
$y = \dfrac{{1050 \times 4}}{7}$
We just divided the fraction by $7$
$y = 150 \times 4$
When multiplying both the terms
$y = 600cm$
Converting the $cm$ to $m$
We have to divided 100 we get,
$y = 6m$
Length of shadow is $6m$
\[\left( 2 \right)\]Now, we need to find height of casting a shadow of 5m i.e.$500cm$
Let \[x\]be the height of pole which casts a shadow of $500cm$
Height of pole (in $cm$) =$560$ and $x$
Length of shadow (in $cm$) =$320$ and $500$
Now, height of the pole and length of shadow are in direct proportion
$\dfrac{{560}}{{320}} = \dfrac{x}{{500}}$
 We just cancel the zero in the first fraction
$\dfrac{{56}}{{32}} = \dfrac{x}{{500}}$
We just cross multiplying the both fractions
$56 \times 500 = x \times 32$
Multiplying both the terms,
$28000 = 32x$
Separate the variable term
$\dfrac{{28000}}{{32}} = x$
Get the value of variable
$x = 875$
$x = 875cm$
Converting the $cm$ to $m$
$x = 8m75cm$
Height of the pole is $8m75cm$

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