Question
A $5\ m$ $60\ cm$ high vertical pole casts a shadow $3\ m 20\ cm$ long. Find at the same time the length of the shadow cast by another pole $10\ m$ $50\ cm$ high.
✓
Answer
Let the height of the vertical pole be $xm$ and the length of the shadow by $ym$.
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 $\frac{{{x_1}}}{{{y_1}}} = \frac{{{x_2}}}{{{y_2}}}$.
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
$\frac{{5.6}}{{3.2}} = \frac{{10.5}}{{{y_2}}}$
$\therefore $ $5.6y_2= 3.2$ $\times$ $10.5$
$\therefore $ ${y_2} = \frac{{3.2 \times 10.5}}{{5.6}}$
$\therefore $ $y_2= 6$
Hence, the length of the shadow is $6\ m$.
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 $\frac{{{x_1}}}{{{y_1}}} = \frac{{{x_2}}}{{{y_2}}}$.
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
$\frac{{5.6}}{{3.2}} = \frac{{10.5}}{{{y_2}}}$
$\therefore $ $5.6y_2= 3.2$ $\times$ $10.5$
$\therefore $ ${y_2} = \frac{{3.2 \times 10.5}}{{5.6}}$
$\therefore $ $y_2= 6$
Hence, the length of the shadow is $6\ m$.
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