Question

Q25. A ladder AP of length 10m is kept against a wall AB such that it reaches a window at A, 6m high from the ground. The same ladder reaches another window at C when it is rested against the wall opposite to AB. If the walls are 14 m apart. Find the height of the window at C from the ground.

Answer 1

Explanation:

Given a ladder AP of length = 10m is kept against a wall AB such that it reaches a window at A, such that AB=6m .

Consider a triangle ABP

By Pythagoras theorem,

`(Hypotenuse)^2=(Perpendicular)^2+(Base)^2\\\\i.e.\ (10)^2=(6)^2+(BP)^2\\\Rightarrow100=36+(BP)^2\\\Rightarrow(BP)^2=64\\\Rightarrow\ BP=8\ m`

Now,The same ladder reaches another window at C when it is rested against the wall opposite to AB .

Let D be a point which represent the foot of the second building CD on the ground which is 14 m away from B.

∴ BD = BP+ PD

⇒ 14 = 8+PD

⇒PD =14-8 = 6m

Now consider another triangle CPD

Again by Pythagoras theorem,

`(Hypotenuse)^2=(Perpendicular)^2+(Base)^2\\\\i.e.\ (CP)^2=(CD)^2+(PD)^2\\\Rightarrow\ (10)^2=(CD)^2+(6)^2\\\\\Rightarrow100=(CD)^2+36\\\Rightarrow(CD)^2=64\\\Rightarrow\ CD=8\ m`

Therefore, the height of the window at C from the ground CD=8m