Question

I need complete working/steps for part (ii) only. The answer for part (ii) in the book is 1.72 m. Please tell me how to do it. It's the chapter of Pythagoras Theorem.

(ii) The upper end of the pole, P, slides down 0.9 m to a point on the wall, X. Calculate YR, the distance the lower end of the pole has slid away from its original position R. ​

Answer 1

1. 72m

Explanation:

length of the pole remains the same in both the parts (¡) and (¡¡), that is PR and YX.

Earlier,

Triangle PQR

PR² = RQ² + PQ²

PR² = 1.1² + 4.2²

PR =
`√(1.21 +17.64 )`

PR = 4. 34 m

Now,

In Triangle YXQ

YX² = XQ²+ YQ²

• Since, YX and PR are the lengths of the rod they'll be the same so YX = 4. 34m
• XQ = PQ - PX = 4.2 - 0.9 = 3.3
• YQ = YR + RQ

(4.34)² = (3.3)² + (YR + 1.1)²

18.85 - 10.89 = (YR + 1.1)²

7. 96 = (YR + 1.1)²

`√(7.96) = YR + 1.1`

2. 82 = YR + 1.1

YR = 2.82 - 1.1

YR = 1.72 m

All the calculations have been done with respect to the diagram provided. Spot the triangles then the line segments and using simple geometry you'll Crack it.