Q. 11

# Two poles of heig

Answer :

Given two poles of heights 6 m and 11 m stand vertically upright on a plane ground. Distance between their foot is 12 m. Let CD be the pole with height 6 m. AB is the pole with height 11m and DB = 12 m

Let us assume a point E on the pole AB which is 6m from the base of AB.

Hence AE = AB – 6 = 11 – 6 = 5m

We know that the Pythagoras theorem state that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Now, in right triangle AEC,

AC2 = AE2 + EC2

Since CDEB forms a rectangle and opposite sides of rectangle are equal,

AC2 = 52 + 122

= 25 + 144

= 169

AC = 13

The distance between their tops is 13 m.

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