Question

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

Answer 1

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 I 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 =5m

Since z BED = 90° as BE 1 DC

A BED is right triangle

Using Pythagoras theorem in right angle triangle AEB

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

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

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

(BD) = 25 + 144

(BD)2 = 169

BD = V169

BD = V13 x 13

BD = /(13)2

BD = 13

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