(a)OR(b)

In Figure $\displaystyle 4$, two circles with centres at O and O' of radii $\displaystyle 2$ r and r respectively, touch each other internally at A. A chord AB of the bigger circle meets the smaller circle at C. Show that C bisects AB.

Figure $\displaystyle 4$

\begin{tabular}[t]{|l|l|l|} \hline Q13. \[\begin{aligned} & O P \perp B C \Rightarrow(12-x)^{2}=x^{2}+8^{2} \\ & \Rightarrow x=\frac{10}{3} \mathrm{~cm} \\ & \therefore B C=\frac{10}{3} \mathrm{~cm} \end{aligned} \] (a) Sides of pool are \(\displaystyle 7-2 x\) and \(\displaystyle 12-2 x\) \[\begin{aligned} & \Rightarrow(7-2 x)(12-2 x)=36 \\ & \Rightarrow 4 x^{2}-38 x+48=0 \end{aligned} \] (b) \[\begin{gathered} \Rightarrow 2 x^{2}-19 x+24=0 \\ \Rightarrow(x-8)(2 x-3)=0 \\ x \neq 8 m \therefore x=\frac{3}{2} m \end{gathered} \] ∴ width of sidewalk around the pool is \(\displaystyle \frac{3}{2} \mathrm{~m}\) (a) \(\displaystyle 2 \pi r=44\) \[\Rightarrow r=7 \] Therefore, \(\displaystyle l=\sqrt{24^{2}+7^{2}}=25 \mathrm{~cm}\) Paper required for $\displaystyle 4$ caps \(\displaystyle =4 \times \pi r l\) \[\begin{aligned} & =4 \times \frac{22}{7} \times 7 \times 25 \\ & =2200 \mathrm{~cm}^{2} \text { or } 700 \pi \mathrm{~cm}^{2} \end{aligned} \] (b) Volume of cake \(\displaystyle =\frac{22}{7} \mathrm{X} 12 \times 12 \times 14\) \(\displaystyle =6336 \mathrm{~cm}^{3}\) \(\displaystyle 6336 \mathrm{~cm}^{3}\) is nearest to \(\displaystyle 6500 \mathrm{~cm}^{3}\) Now, \(\displaystyle 650 \mathrm{~cm}^{3}=100 \mathrm{~g}\) \[\Rightarrow 6500 \mathrm{~cm}^{3}=1 \mathrm{~kg} \] \(\displaystyle \therefore 1 \mathrm{~kg}\) cake should be ordered.