Question 7

Two towers 10 meters apart, are 4 m and 6 m high respectively. What will be the height of pointof intersection of lines joining the top of each tower to the bottom of opposite tower?

Solution
Screenshot_20220628_141604

Let AB and CD be the towers of length 4m and 6m respectively.

Let the length of BF be x therefore, the length of FC will be 10-x.

Triangle BEF is similar to triangle BDC

So, $$\frac{EF}{DC}=\frac{BF}{BC}=\frac{BE}{BD}$$

or, $$\frac{EF}{6}=\frac{x}{10}=\frac{BE}{BD}$$

Therefore, $$EF=\frac{3x}{5}$$

Similarly, triangle EFC is similar to triangle ABC

So, $$\frac{EF}{AB}=\frac{CF}{BC}=\frac{EC}{AC}$$

or, $$\frac{EF}{4}=\frac{10-x}{10}=\frac{EC}{AC}$$

or, $$EF=\ \frac{2\left(10-x\right)}{5}$$

Therefore, $$\frac{2\left(10-x\right)}{5}=\frac{3x}{5}$$

or, $$x=4$$

EF= $$\frac{3x}{5}=2.4\ m$$

Video Solution

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