Foot of a 10 m long ladder leaning against a verticle wall is 6 m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.

Explanation

Let AB be a verticle wall and AC=10 m is a ladder. The top of the ladder reaches to A and distance of ladder from the base of the wall BC is 6 m .

In right angled `DeltaABC,AC^(2)=AB^(2)+BC^(2)` [by pythagoras theorem]
`rArr (10)^(2)=AB^(2)+(6)^(2)`
`rArr 100=AB^(2)+36`
`rArrAB^(2)=100-36=64`
`therefore AB=sqrt(64)=8cm`
Hence, the height of the point on the wall where the top of the ladder reaches is 8 cm.