(a) The total surface area of two spheres are in the ratio 9 : 49. If the radius of the smaller sphere is 12 cm, find, correct to the nearest \(cm^{3}\), th...
Question 1 Report
(a) The total surface area of two spheres are in the ratio 9 : 49. If the radius of the smaller sphere is 12 cm, find, correct to the nearest \(cm^{3}\), the volume of the bigger sphere.
(b) A cyclist starts from a point X and rides 3 km due West to a point Y. At Y, he changes direction and rides 5 km North- West to a point Z.
(i) How far is he from the starting point, correct to the nearest km? ; (ii) Find the bearing of Z from X, to the nearest degree.
Answer Details
(a) Surface areas of similar spheres are in the ratio of the squares of their radii:
\[\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{9}{49}\Rightarrow \frac{r_{1}}{r_{2}}=\frac{3}{7}\]The smaller radius \(r_{1}=12\) cm corresponds to the ratio part 3, so one part \(=\dfrac{12}{3}=4\) cm, and the bigger radius \(r_{2}=7\times4=28\) cm.
\[V=\frac{4}{3}\pi r_{2}^{3}=\frac{4}{3}(3.142)(28)^{3}=\frac{4}{3}(3.142)(21952)\approx 91964\text{ cm}^{3}\]Volume of bigger sphere \(\approx 91{,}964\text{ cm}^{3}\).
(b) Take \(X\) as origin, West as the negative \(x\)-direction. \(Y=(-3,0)\). North-West is the bearing \(315^{\circ}\), i.e. direction \((-\sin45^{\circ},\cos45^{\circ})\), so
\[Z=(-3,0)+5(-\sin45^{\circ},\cos45^{\circ})=(-3-3.536,\ 3.536)=(-6.536,\ 3.536)\](i) Distance from \(X\):
\[XZ=\sqrt{6.536^{2}+3.536^{2}}=\sqrt{42.72+12.50}=\sqrt{55.22}\approx 7.4\ \text{km}\approx 7\text{ km}\](ii) \(Z\) lies to the North and West of \(X\). The angle west of due North is \(\tan^{-1}\dfrac{6.536}{3.536}=\tan^{-1}1.848=61.6^{\circ}\).
\[\text{Bearing of }Z\text{ from }X=360^{\circ}-61.6^{\circ}\approx 298^{\circ}\]Bearing \(\approx 298^{\circ}\).
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