A ladder 13 m long is placed on the ground in such a way that it touches the
top of a vertical wall 12 m high. Find the distance of the foot of the ladder
from the bottom of the wall.

3 m

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4 m

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5 m

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7 m

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Solution

The correct option is C 5 m
According to the Pythagoras theorem , $(p e r p e n d i c u l a r)^{2} + (b a s e)^{2} = (h y p o t e n u s e)^{2}$
So, $(b a s e)^{2} = (h y p o t e n u s e)^{2} - (p e r p e n d i c u l a r)^{2} = (13)^{2} - (12)^{2}$ $= 169 - 144 = 25$
So, base= 5m.

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A ladder 13 m long is placed on the ground in such a way that it touches the top of a vertical wall 12 m high. Find the distance of the foot of the ladder from the bottom of the wall.

The correct option is C 5 mAccording to the Pythagoras theorem , (perpendicular)2+(base)2=(hypotenuse)2 So, (base)2=(hypotenuse)2−(perpendicular)2=(13)2−(12)2 =169−144=25 So, base= 5m.

A ladder 13 m long is placed on the ground in such a way that it touches the
top of a vertical wall 12 m high. Find the distance of the foot of the ladder
from the bottom of the wall.

The correct option is C 5 m
According to the Pythagoras theorem , $(p e r p e n d i c u l a r)^{2} + (b a s e)^{2} = (h y p o t e n u s e)^{2}$
So, $(b a s e)^{2} = (h y p o t e n u s e)^{2} - (p e r p e n d i c u l a r)^{2} = (13)^{2} - (12)^{2}$ $= 169 - 144 = 25$
So, base= 5m.