Question

Two poles of height 50m and 10m stand upright on a plane ground. If the distance between their tops is 41m, find the distance between their feet
Please give Step by Step answer

Answer 1

Answer:

12

Step-by-step explanation:

Let AB and CD are the poles of heights 15m and 10m such thatAC is 13m.

Now,

CE=BD=10m

CD=BE=10m

∴AE=AB−BE=15−10=5m

Now, in triangle AEC

⇒(13) ^2 =5^2+(EC)^2

⇒(EC)^2 =169−25

⇒EC= √144

⇒EC=12

∴BD=12m

Therefore 12m is the distance between the fett of the poles.

Answer 2

Answer:

Hey there !!

➔ Let the two poles AB and CD are stand upright on the ground of 15m and 10m respectively.

➔ AB is the distance between the tops of two poles is of 13m.

➔ And, DB is the distance between their feet.

➤Now,

=> AE = AB - BE

=> AE = 15 - 10.[BE = CD]

=> AE = 5m.

In ∆AEC,

➔ By Pythagoras theorem,

=> ( AC )² = ( AE)² + ( CE )².

=> (13)² = (5)² + (CE)².

=> 169 = 25 + (CE)².

=> (CE)² = 169 - 25.

=> CE = √144.

=> CE = 12m.

And, CE = BD.

So, BD = 12m. ( Distance between their feet ).

✔✔Hence, it is solved✔✔