Two cables

On a flat plain, 2 columns are erected vertically upwards. One is 7 m high and the other 4 m. Cables are stretched between the top of one column and the foot of the other column. At what height will the cables cross? Assume that the cables do not sag.

Correct answer:

x =  2.5455 m

Step-by-step explanation:

$$\begin{gathered} a=7\text{ m } \\ b=4\text{ m } \\ a:d=x:d_2 \\ b:d=x:d_1 \\ {d}_1+d_2=d \\ b\mathrm{/}(d_1+d_2)\cdot d_1=x \\ a\mathrm{/}(d_1+d_2)=b\mathrm{/}(d_1+d_2)\cdot d_1\mathrm{/}{d}_2 \\ a:b=d_1:d_2 \\ a\mathrm{/}(d_1+d_2)=x\mathrm{/}{d}_2 \\ a\cdot d_2=x\cdot (d_1+d_2) \\ a\cdot d_2=x\cdot (d_2\cdot a\mathrm{/}b+d_2) \\ a=x\cdot (a\mathrm{/}b+1) \\ x={\displaystyle \frac{a}{1+a\mathrm{/}b}}={\displaystyle \frac{7}{1+7\mathrm{/}4}}={\displaystyle \frac{28}{11}}\doteq 2.5455\text{ m } \\ \text{ Verifying Solution: } \\ h=harmonic(a,b) \\ h={\displaystyle \frac{1}{1\mathrm{/}a+1\mathrm{/}b}}={\displaystyle \frac{1}{1\mathrm{/}7+1\mathrm{/}4}}={\displaystyle \frac{28}{11}}\doteq 2.5455\text{ m} \end{gathered}$$

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