Question

Two poles of height 7m'and 15m stands on a plane ground. If the distance
between their feet is 6m, then distance between their tops is ​

Answer 1

Let the he height of first tower =12m

height of second tower =5m

the distance between the two tower =12m

then, the length of first tower=(7+5)m the length of second tower. The right triangleABC can be constructed using the length of the two tower.

AB =Base of the two tower =12m

BC=height of the first tower w.r.t second tower AC=distance between the top of the two tower in the form of hypotenuse of ABC

using pythaguras theorem

AC^2=AB^2+BC^2

AC^2=5x5+12x12

AC =square root of (169)

AC =13 metres

Hence the distance between the two tower =13m

Answer 2

Then distance between their tops is ​10m.

Given:

Two poles AB= 7m and CD= 15m

Distance between both poles BD= 6 m

To Find:

Now, create a line segment AE parallel to line BD

So, CE=CD - ED = 15-7= 8m

In right angled △AEC,

$AC^{2} = AE^{2} +EC^{2}$

=> $AC^{2} = 6^{2}+ 8^{2}$

=> AC = $\sqrt{36+64}$ = $\sqrt{100}$

⇒AC=10 m

So, the distance between their tops  is 10 m.

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