Question

A ladder 10m long is made to rest against a wall. Its lower end touches. the ground at a distance of 6m from the wall. At what height above the ground the upper end of the ladder rests against the wall?​

Answer 1

by pythagoras theorem,

100-36=x²

64=x²

x=8m

at 8m height the upper end of ladder is placed....

HOPE IT'S HELPFUL....

Answer 2

Answer:

As we know that Wall will be perpendicular to the ground so the ladder will make a right angle triangle with ground and wall. So, right angle triangle will be formed as shown in attachment.

Now,

ln a right angled △ABC

${Hypotenuse}^{2} = {Perpendicular}^{2} + {Base}^{2} \: \: \: \:(PGT) \\ \implies {10}^{2} = {6}^{2} + {h}^{2} \\ \implies \: h = \sqrt{ {10}^{2} - {6}^{2} } \\ \implies \: h = \sqrt{100 - 36} \\ \implies \: h = \sqrt{64} \: \\ \implies \: \orange{\bold{\boxed{\boxed{h = 8}}}}$ So ,at 8 metre height above the ground upper and of the ladder rest against the wall.