Question 61

Two poles of height 7 metre and 12 metre stand on a plane ground. If the distance between their feet is 12 metre, the distance between their top will be

Solution

Height of pole AB = 7 metre = DE and CD = 12 metre

BD = 12 metre = AE

Now, CE = CD-DE = 12-7 = 5 metre

In $$\triangle$$ ACE

AC = $$\sqrt{(AE)^2 + (CE)^2}$$

= $$\sqrt{12^2 + 5^2} = \sqrt{169}$$

= 13 metre

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