Chapter 15, Problem 34CR

To determine

To Find: Thedistance between the ground and the point at which small ball touches the wall.

Answer to Problem 34CR

  $21cm$

Explanation of Solution

Given:

  

  $\begin{array}{l}\text{Radius of Small ball}=r=4cm\\ \text{Radius of Big ball}=R=9cm\end{array}$

Formula Used:

  $\text{Pythagoras Theorem}:h^2=b^2+p^2$

Calculation:

  $\begin{array}{l}Let,\\ \text{The perpendicular distance between the centre of small ball and the large ball be}x\\ \text{And }y\text{ be the base }=R-r=9-4=5\\ \text{Radius of Small ball}=r=4cm\\ \text{Radius of Big ball}=R=9cm\\ {13}^2=y^2+x^2\\ \text{⇒}169=5^2+x^2\\ \text{⇒}{x}^2=169-25\\ \text{⇒}x=\sqrt{144}=12cm\\ \text{Total distance between small ball and the ground is }=x+R=12+9=21cm\end{array}$

Conclusion:

Thus, the required height is $21cm$ .