The surface of a hemisphere is made of two parts

1. the curved surface
2. the circular base

The surface area of a hemisphere,

\(\large A=\pi r^2+\pi r^2\\ \large A=3\pi r^2\\ \large 1360=3\pi r^2\\ \large r^2=\frac{1360}{3\pi}\\ \large r=\sqrt{\frac{1360}{3\pi}}\\ \large r\approx12.0125...\\ \large r\approx12.0cm \)

c) Solve \(\large \frac{4}{3}\pi r^3=\frac{1}{3}\pi r^2h\\\)

a)

\(\large V=\frac{4}{3}\pi r^3\\ \large V=\frac{4}{3}\pi \times10^3\\ \large V=\frac{4000}{3}\pi \ cm^3\)

b)

\(\large V=\frac{1}{3}\pi r^2h\\ \large \frac{4000}{3}\pi=\frac{1}{3}\pi \times 10^2\times h\\ \large h=40\ cm\)

c)

\(\large \frac{4}{3}\pi r^3=\frac{1}{3}\pi r^2h\\ \large \frac{4}{\rlap{/}3}\rlap{/}\pi r^3=\frac{1}{\rlap{/}3}\rlap{/}\pi r^2h\\ \large 4r^3=r^2h\\ \large 4r=h\\ \large h=4r\)

The volume of a sphere is given by \(\large V=\frac{4}{3}\pi r^3\)

The Volume of the three spheres is

\(\large V=\frac{4}{3}\pi \times1^3+\frac{4}{3}\pi \times 6^3+\frac{4}{3}\pi \times 8^3\\ \large V=\frac{4}{3}\pi \times(1^3+6^3+8^3)\\ \large V=\frac{4}{3}\pi \times(729)\\ \)

The large sphere has the same volume as these three spheres

\(\large \frac{4}{3}\pi \times R^3=\frac{4}{3}\pi \times(729)\\ \large R^3=729\\ \large R=9\ cm \)

The volume of the cylindrical bar is

\(\large V=\frac{1}{3}\pi\times 6^2\times12\\ \large V=144\pi \)

The volume of one of the spheres is

\(\large V=\frac{4}{3}\pi\times 1.5^3\\ \large V=4.5\pi \)

Therefore, the number of spheres is \(\large \frac{144\pi}{4.5\pi}\)

\(\large 32\) spheres

There are three surfaces to find: the curved surface of the cone, the curved surface of the cylinder and the circular base.

To find the curved surface area of the cone, you need to find the slant height. Use Pythagoras' Theorem.

We need to find the slant height of the cone:

\(\large l^2=5^2+12^2\\ \large l^2=169\\ \large l=13\)

The surface area of the cone is

\(\large A_{cone}=\pi rl\\ \large A_{cone}=\pi\times5\times 13\\ \large A_{cone}=65\pi\)

The surface area of the cylinder is

\(\large A_{cylinder}=2\pi rh\\ \large A_{cylinder}=2\pi \times 5\times 15\\ \large A_{cylinder}=150\pi\)

The area of the base is

\(\large A_{base}=\pi r^2\\ \large A_{base}=\pi \times 5^2\\ \large A_{base}=25\pi \)

The total surface area is

\(\large =65\pi+150\pi+25\pi\\ \large=240\pi\)

Write an equation for the arc length and another equation for the sector area.

Solve the equations simultaneously.

a) The area of the shaded sector OAB is \(\large \frac{40\pi}{3}\) cm²

\(\large A=\frac{1}{2}r^2\theta\\ \large \frac{40\pi}{3}=\frac{1}{2}r^2\theta\\\)

The length of the minor arc AB is \(\large \frac{10\pi}{3}\) cm

\(\large l=r\theta\\ \large \frac{10\pi}{3}=r\theta\)

Let's solve these equations by substituting for \(\large r\theta\) into the area equation

\(\large \frac{1}{2}r^2\theta=\frac{40\pi}{3}\\\)

\(\large \frac{1}{2}r\times r\theta=\frac{40\pi}{3}\\\)

\(\large \frac{1}{2}r\times \frac{10\pi}{3}=\frac{40\pi}{3}\\\)

\(\large \frac{1}{2}r=4\)

r = 8 cm

We can now find the angle by substituting this value into the length equation

\(\large \frac{10\pi}{3}=8\times\theta\\ \large \theta=\frac{10\pi}{24}\\ \large \theta=\frac{5\pi}{12}\)

a) You can use the Sine Rule to find r.

b) You should divide the shaded area into two parts. Work out the green shaded area and blue shaded area separately.

a)

b) Divide the shaded area into two parts.

a) The area of the blue segment = area of sector - area of triangle

b) Use your graphical calculator to solve this equation.

a)

area of the blue segment = area of sector - area of triangle

\(\large A_{segment}=\frac{1}{2}r^2\theta-\frac{1}{2}r^2\sin\theta\)

The angle subtended by the the green sector is \(\large 2\pi-\theta\)

Area of the green sector is

\(\large A_{sector}=\frac{1}{2}r^2(2\pi-\theta)\)

The area of the green sector = three times the area of the blue segment

\(\large \frac{1}{2}r^2(2\pi-\theta)\) = 3 \(\large (\frac{1}{2}r^2\theta-\frac{1}{2}r^2\sin\theta)\)

\(\large \frac{1}{2}r^2(2\pi-\theta)=\frac{3}{2}r^2(\theta-\sin\theta)\)

\(\large 2\pi-\theta=3(\theta-\sin\theta)\\ \large 2\pi-\theta=3\theta-3\sin\theta\\ \large 3\sin\theta=4\theta-2\pi\\ \large \sin\theta=\frac{4\theta-2\pi}{3}\)

b) We can solve this equation using our graphical equation.You can plot the graphs of \(\large y=\sin\theta\\ \) and \(\large y=\frac{4\theta-2\pi}{3}\) and find the point of intersection

\(\large \theta \approx2.18\)

The diagram looks approaximately like the following

b) Area of red region = Area of kite OACB - Area of sector OAB

a) Triangle OAC is a right angled triangle, so we can use right-angled triangle trigonometry to work out the length AC (opposite to the angle \(\large\frac{\theta}{2}\)

\(\large \tan\frac{\theta}{2}=\frac{AC}{r}\)

\(\large AC=r\tan\frac{\theta}{2}\)

b)

a) You can use the Pythagorean identity \(\large \cos^2\theta+\sin^2\theta\equiv1\), to find \(\large \cos \hat A\)

b) Use the Cosine Rule, c² = a² + b² - 2ab cosC

This is a very difficult proof for SL students!

There is always more than one way to carry out this proof.

The easiest, is to start with the left hand side and consider that \(\large \tan\theta\equiv\frac{\sin\theta}{\cos\theta}\)