Question

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

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

Given: Height of first pole = AB = 6 m

Height of second pole = CD = 11 m

Distance b/w feet of poles = AC = 12 m

To Find :-

Distance between the tops of the pole i.e., BD

Solution :-

Let we draw a line BE perpendicular to DC i.e. BE 1 DC Since AC is also perpendicular to DC as pole is vertical to ground,

So, BE = AC = 12 m

Similarly, AB = EC = 6 m

Now,

DE = DC-EC

DE= 11-6

DE = 5 m

Since 2 BED = 90* as BE 1 DC

A BED is right triangle

Using Pythagoras theorem in right angle triangle AEB

(Hypotenuse)^2 = (Height)^2 + (Base)^2

(BD)^2 = (DE)^2 + (BE)^2

(BD)^2 = (5)^2 + (12)^2

(BD)^2 = 25 + 144

(BD)^2 = 169

BD = 13 x 13

BD = (13)2

BD = 13

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