5.1.4 Exercises

The son of the house is looking at the tree on the neighbouring property. He observes that the tree is completely covered by the hedge separating the two properties only if he stands close enough to the hedge. Now he is looking for the point at which he just cannot see the tree any more.
The boy is $1.40$ metres tall. If the boy stands $2.50$ metres away from the hedge, which is $2.40$ metres high, $1$ metre wide and clipped into a pointed shape at the top, the tree disappears from his sight.
What is the height of the tree if the middle of the trunk is $14.5$ metres away from the hedge?
Please carry out the calculation using variables and insert the values only at the end!
Result:
$\hspace{0.17em}\mathrm{m}$.
Hint

Take the width of the hedge into account!

Solution

The line segments are denoted as shown in the figure below.

Applying the second intercept theorem results in „$\frac{\text{full}}{\text{at front}}=\frac{\text{long}}{\text{short}}$“ :

${\displaystyle \frac{d_{\mathrm{KB}}}{d_{\mathrm{KH}}}=\frac{h_B-h_K}{h_H-h_K}\mathrm{ }\text{or}\mathrm{ }{h}_B=(h_H-h_K)·\frac{d_{\mathrm{KB}}}{d_{\mathrm{KH}}}+h_K\hspace{0.5em}.}$

The values are $d_{\mathrm{KH}}=2.5\hspace{0.17em}\mathrm{m}+\frac{1\hspace{0.17em}\mathrm{m}}{2}=3\hspace{0.17em}\mathrm{m}$ and $d_{\mathrm{KB}}=2.5\hspace{0.17em}\mathrm{m}+1\hspace{0.17em}\mathrm{m}+14.5\hspace{0.17em}\mathrm{m}=18\hspace{0.17em}\mathrm{m}$. Hence, it follows that

${\displaystyle h_B=(2.4\hspace{0.17em}\mathrm{m}-1.4\hspace{0.17em}\mathrm{m})·\frac{18\hspace{0.17em}\mathrm{m}}{3\hspace{0.17em}\mathrm{m}}+1.4\hspace{0.17em}\mathrm{m}=1\hspace{0.17em}\mathrm{m}·6+1.4\hspace{0.17em}\mathrm{m}=7.4\hspace{0.17em}\mathrm{m}\hspace{0.5em}.}$