A ladder 25m long reaches a window of building 20m above the ground determine the distances of foot of the ladder from building
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Let AB be the height of ladder
AC be the height of the building
And BC be the distance
By Pythagoras theorem
AB^2=AC^2+BC^2
25^2=20^2+BC^2
625=400+BC^2
625-400=BC^2
225=BC^2
√225=BC
15=BC
Therefore the distance of the foot from the building is 15 m
AC be the height of the building
And BC be the distance
By Pythagoras theorem
AB^2=AC^2+BC^2
25^2=20^2+BC^2
625=400+BC^2
625-400=BC^2
225=BC^2
√225=BC
15=BC
Therefore the distance of the foot from the building is 15 m
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by Pythagoras Thorem,
AC*AC = AB*AB + BC*BC
(25)*(25) = (20)*(20) + BC*BC
625 = 400 + BC*BC
BC*BC = 625 - 400
BC*BC = 225
BC = 15m
AC*AC = AB*AB + BC*BC
(25)*(25) = (20)*(20) + BC*BC
625 = 400 + BC*BC
BC*BC = 625 - 400
BC*BC = 225
BC = 15m
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