Question

Two poles of 8m and 14m stand upright on a plane ground. If the distance between two tops is 10 m. Find the distance between their feet. ​

Answer 1

Given,

Heights of two poles = 8m and 14m

Distance between their top = 10m

To find,

The distance between their feet.

Solution,

If we imagine a vertical straight line which starts from the top of the first pole and perpendicularly ends on a certain point on the second pole, then the second pole is divided into two segments.

The Length of the upper segment of the second pole from that imaginary point will be

= (14-8) = 6m

Now, we can easily imagine a right angled triangle in this case, which has,

Base = Distance between two poles = Let, x m

Height = Upper segment of the second pole = 6m

Hypotenuse = 10m

According to the Pythagoras theorem,

(10)²=(x)²+(6)²

100 = x²+36

x² = 100-36

x² = 64

x = 8

Hence, the distance between their feet will be 8m.

Answer 2

SOLUTION

GIVEN

• Two poles of 8 m and 14 m stand upright on a plane ground

• The distance between two tops is 10 m

TO DETERMINE

The distance between their feet

EVALUATION

Let AB and CD be the poles

Then AB = 8 & CD = 14

Also the distance between two tops is 10 m

So BD = 10

Let E be a point such that AB = CE

From Figure

DE = CD - CE = 14 - 8 = 6

We have to find the distance between their feet i.e AC

Now BDE is a Right angled triangle

In context of the given problem Pythagoras theorem states that : In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides

Therefore

(Hypotenuse)² = ( Perpendicular)² + (Base)²

$\implies \sf{{BD}^{2} = {DE}^{2} + {BE}^{2} }$

$\implies \sf{{(10)}^{2} = {(6)}^{2} + {BE}^{2} }$

$\implies \sf{ {BE}^{2} = {(10)}^{2} - {(6)}^{2} }$

$\implies \sf{ {BE}^{2} = 100 - 36}$

$\implies \sf{ {BE}^{2} = 64}$

$\implies \sf{BE = 8}$

Now AC = BE = 8

FINAL ANSWER

The distance between their feet = 8 m

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