Question

Two poles of heights 6 m and 11 m
stand on a plane ground. If the distance
between their tops feet is 12m, find
the distance of their tops​

Answer 1

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Two poles AB=11 m and CD=6m

Distance between both poles BD=12 m

Then, create a line segment CE parallel to line BD

So, AE=AB−CD=11−5=6 m

In right angled △AEC,

$= (AC)^{2} =(AE)^{2} +(AE)^{2} =(12)^{2} + (5)^{2}$

=144+25=169

⇒AC=13 m

So, the distance between their tops is 13 m.

$\blue{i \: hope \: it \: helpfull \: for \: you}$

Answer 2

Answer:

Two poles of height 6m and 11m stands on a plane ground, where distance between their feet is 12m, the distance between the tops of the two poles is 13 m.

Step-by-step explanation:

Let, BD be the distance between the two poles

Assume, BD = x

As △BED is a right-angled triangle right angled at E, therefore 'x' is the hypotenuse.

Now apply the Pythagoras theorem on △BED ,

Hypotenuse2 = Perpendicular2 + Base2

=> BD2 = ED2 + BE2

=> x2 = 122 + 52

=> x2 = 144 + 25

=> x2 = 169

=> x = 13 m

Therefore, the distance between the top of the two poles is 13 m

I hope your will understand