L.C. MATHS

Course Content
Higher Level (by year)
0/17
Higher Level (by topic)
0/13
Ordinary Level (by year)
0/17
Ordinary Level (by topic)
0/12
Past Papers

## Length, Area & Volume

NOTE: Clicking an entry in the right column will take you directly to that question!

(BLUE = Paper 1, RED = Paper 2)

Since 2015, the following subtopic... Has explicitly appeared in...

Sectors of Circles

Other 2D Shapes

3D Objects

Trapezoidal Rule

## 2022 Paper 2 Question 7(a)-(d)

A company makes and sells candles of different shapes and sizes.

(a) A candle in the shape of a cylinder has a diameter of $$10\mbox{ cm}$$ and a volume of $$450\pi\mbox{ cm}^3$$.
Work out the height of this candle.

$$18\mbox{ cm}$$

Solution

\begin{align}\pi r^2h=450\pi\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\pi (5^2)h=450\pi\end{align}

\begin{align}\downarrow\end{align}

\begin{align}h&=\frac{450}{5^2}\\&=18\mbox{ cm}\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) A small candle in the shape of a cone has a volume of $$12\pi\mbox{ cm}^3$$.
A large candle, also in the shape of a cone, has a volume of $$150\pi\mbox{ cm}^3$$.
The height of the small candle is $$h$$, and the height of the large candle is $$2h$$.
The large candle has a radius that is $$k$$ times that of the small candle, where $$k\in\mathbb{R}$$.

Work out the value of $$k$$.

$$2.5$$

Solution

Small Candle

\begin{align}\frac{1}{3}\pi r^2 h=12\pi\end{align}

\begin{align}\downarrow\end{align}

\begin{align}r^2h=36\end{align}

$\,$

Large Candle

\begin{align}\frac{1}{3}\pi (kr)^2 (2h)=150\pi\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k^2(r^2h)=225\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k^2(36)=225\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k&=\sqrt{\frac{225}{36}}\\&=2.5\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(c) A third conical candle has its curved surface area covered in cloth.

The net of the cloth is shown in the diagram below (not to scale).
This net covers the cone perfectly, with no overlapping material.

In the diagram, $$|\angle BOA|=216^{\circ}$$, as shown. $$|OA|=8\mbox{ cm}$$.

Find the length of the arc from $$B$$ to $$A$$, in terms of $$\pi$$, and hence find the radius of the cone.

$$l=9.6\pi\mbox{ cm}$$ and $$r=4.8\mbox{ cm}$$

Solution

\begin{align}l&=2\pi r\left(\frac{\theta}{360^{\circ}}\right)\\&=2\pi (8)\left(\frac{216^{\circ}}{360^{\circ}}\right)\\&=9.6\pi\mbox{ cm}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2\pi r=9.6\pi\end{align}

\begin{align}\downarrow\end{align}

\begin{align}r&=\frac{9.6}{2}\\&=4.8\mbox{ cm}\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(d)

(i) A spherical ball of wax is used as a candle. The radius of the sphere is $$2.7\mbox{ cm}$$.
Find the volume of the sphere, correct to $$3$$ decimal places.

(ii) A horizontal slice is cut off this sphere so that the candle will balance on level surfaces.
The area of the circular base of this candle is $$5.4\mbox{ cm}^2$$.

Find the value of $$l$$, the vertical distance from the top of the candle to where the cut is made.
As shown in the diagram on the right, $$l>2.7\mbox{ cm}$$.

Give your answer in $$\mbox{cm}$$, correct to the nearest $$\mbox{mm}$$.

(i) $$82.448\mbox{ cm}^3$$

(ii) $$5.1\mbox{ cm}$$

Solution

(i)

\begin{align}V&=\frac{4}{3}\pi r^3\\&=\frac{4}{3}\pi(2.7^3)\\&\approx82.448\mbox{ cm}^3\end{align}

(ii)

\begin{align}\pi r^2=5.4\end{align}

\begin{align}\downarrow\end{align}

\begin{align}r&=\sqrt{\frac{5.4}{\pi}}\\&=1.311…\end{align}

Let $$x$$ be vertical distance from the sphere’s centre to the horizontal slice.

\begin{align}x^2+(1.311…)^2=2.7^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\sqrt{2.7^2-(1.311…)^2}\\&=2.36…\end{align}

\begin{align}\downarrow\end{align}

\begin{align}l&=2.7+x\\&=2.7+2.36…\\&\approx 5.1\mbox{ cm}\end{align}

Video Walkthrough

## 2021 Paper 1 Question 3(a)

The diagram shows a cuboid with dimensions $$x$$, $$y$$ and $$z \mbox{ cm}$$.
The areas, in $$\mbox{cm}^2$$, of three of its faces are also shown.

(a) Find the volume of the cuboid in the form $$a\sqrt{b}\mbox{ cm}^3$$, where $$a,b\in\mathbb{N}$$.

$$8\sqrt{6}\mbox{ cm}^3$$

Solution

\begin{align}xy=4\sqrt{3}&&xz=2\sqrt{2}&&yz=8\sqrt{6}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(xy)(xz)(yz)=(2\sqrt{2})(8\sqrt{6})(4\sqrt{3})\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(xyz)^2=384\end{align}

\begin{align}\downarrow\end{align}

\begin{align}V^2=384\end{align}

\begin{align}\downarrow\end{align}

\begin{align}V&=\sqrt{384}\\&=\sqrt{64\times6}\\&=\sqrt{64}\times\sqrt{6}\\&=8\sqrt{6}\mbox{ cm}^3\end{align}

Video Walkthrough

## 2021 Paper 2 Question 5(a)

(a) Two identical right-circular solid cones meet along their bases
and fit exactly inside a sphere, as shown in the diagram.

(i) Prove that the volume of the remaining space inside the sphere is exactly half the total volume of the sphere.

(ii) The combined volume of the two cones is $$\dfrac{686}{3}\pi\mbox{ cm}^3$$.
Find the radius of one of the cones.

(ii) $$7\mbox{ cm}$$

Solution

(i)

The volume $$V$$ of the remaining space is:

\begin{align}V&=\frac{4}{3}\pi r^3-2\left[\frac{1}{3}\pi r^2(r)\right]\\&=\frac{4}{3}\pi r^3-\frac{2}{3}\pi r^3\\&=\frac{2}{3}\pi r^3\\&=\frac{1}{2}\left(\frac{4}{3}\pi r^3\right)\end{align}

i.e. it is half of the volume of the sphere, as required.

(ii)

\begin{align}2\left[\frac{1}{3}\pi r^2(r)\right]=\frac{686}{\pi}\pi\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2r^3=686\end{align}

\begin{align}\downarrow\end{align}

\begin{align}r&=\sqrt[3]{343}\\&=7\mbox{ cm}\end{align}

Video Walkthrough

## 2021 Paper 2 Question 7(a) & (e)

The diagram (Triangle $$ABC$$) shows the $$3$$ sections of a level triathlon course.
In order to complete the triathlon, each contestant must swim $$4\mbox{ km}$$ from $$C$$ to $$B$$, cycle from $$B$$ to $$A$$, and then run $$28\mbox{ km}$$ from $$A$$ to $$C$$.
Mary can cycle at an average speed of $$25\mbox{ km/hour}$$.
It takes her $$1$$ hour and $$12$$ minutes to cycle from $$B$$ to $$A$$.

(a) Show that the total length of the course is $$62\mbox{ km}$$.

Solution

\begin{align}L&=|CB|+|BA|+|AC|\\&=4+(25\times1.2)+28\\&=62\mbox{ km}\end{align}

as required.

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(e) Find the shortest distance from the point $$C$$ to the side $$AB$$.
Give your answer in $$\mbox{km}$$, correct to $$1$$ decimal place.

$$3.3\mbox{ km}$$

Solution

\begin{align}\mbox{Area}=\frac{1}{2}bh\end{align}

\begin{align}\downarrow\end{align}

\begin{align}50.1=\frac{1}{2}(30)h\end{align}

\begin{align}\downarrow\end{align}

\begin{align}h&=\frac{(2)(50.1)}{30}\\&\approx3.3\mbox{ km}\end{align}

Video Walkthrough

## 2020 Paper 2 Question 7(a)

(a) A company makes biodegradable paper cups in the shape of a right circular cone. Each cup has radius of $$3.3\mbox{ cm}$$ and a slant height of $$9\mbox{ cm}$$, as shown.

(i) Show that the vertical height of the cup is $$8.37\mbox{ cm}$$, correct to $$2$$ decimal places.

(ii) Find the curved surface area of the cup correct to $$2$$ decimal places.

(iii) The diagram shows the net of the cup. Find, in degrees, the size of the angle $$\theta$$.

(ii) $$93.31\mbox{ cm}^2$$

(iii) $$132^{\circ}$$

Solution

(i)

\begin{align}9^2=h^2+3.3^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}h&=\sqrt{9^2-3.3^2}\\&\approx8.37\mbox{ cm}\end{align}

as required.

(ii)

\begin{align}A&=\pi rl\\&=\pi(3.3)(9)\\&\approx93.31\mbox{ cm}^2\end{align}

(iii)

\begin{align}2\pi(3.3)&=[2\pi(9)]\left(\frac{\theta}{360^{\circ}}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\theta&=\frac{(3.3)(360^{\circ})}{9}\\&=132^{\circ}\end{align}

Video Walkthrough

## 2019 Paper 1 Question 9(a) & (c)

Norman windows consist of a rectangle topped by a semi‐circle as shown above.
Let the height of the rectangle be $$y$$ metres and the radius of the semi‐circle be $$x$$ metres as shown. The perimeter of the window is $$P$$.

(a)

(i) Write $$P$$ in terms of $$x$$, $$y$$ and $$\pi$$.

(ii) In a particular Norman window the perimeter $$P=12$$ metres.

Show that $$y=\dfrac{12-(2+\pi)x}{2}$$ for $$0\leq x\leq \dfrac{12}{2+\pi}$$, where $$x\in\mathbb{R}$$.

(i) $$P=2x+2y+\pi x$$

Solution

(i)

\begin{align}P&=2x+2y+\frac{1}{2}(2\pi x)\\&=2x+2y+\pi x\end{align}

(ii)

\begin{align}2x+2y+\pi x=12\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2y=12-2x-\pi x\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y&=\frac{12-2x-\pi x}{2}\\&=\frac{12-(2+\pi)x}{2}\end{align}

as required.

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(c)

(i) The Norman window shown below has a perimeter of $$12$$ metres and $$y=\dfrac{12-(2+\pi)x}{2}$$.

Show that the function $$a(x)=\dfrac{24x-(\pi+4)x^2}{2}$$ represents the area of the
window, in terms of $$x$$ and $$\pi$$.

(ii) Find $$a'(x)$$.

(iii) Find the relationship between $$x$$ and $$y$$ when the area of the window in part (c)(i) is at its
maximum.

(ii) $$a'(x)=12-(\pi+4)x$$

(iii) The area is a maximum when the radius is equal to the height of the rectangle.

Solution

(i)

\begin{align}a(x)&=2xy+\frac{1}{2}(\pi x^2)\\&=2x\left(\frac{12-(2+\pi)x}{2}\right)+\frac{1}{2}(\pi x^2)\\&=\frac{24x-4x^2-2\pi x^2+\pi x^2}{2}\\&=\frac{24x-(\pi+4)x^2}{2}\end{align}

as required.

(ii)

\begin{align}a'(x)&=\frac{24-2(\pi+4)x}{2}\\&=12-(\pi+4)x\end{align}

(iii)

\begin{align}a'(x)=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}12-(\pi+4)x=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=\frac{12}{\pi+4}\end{align}

and

\begin{align}y&=\frac{12-(2+\pi)x}{2}\\&=\frac{12-(2+\pi)\left(\frac{12}{\pi+4}\right)}{2}\\&=\frac{12(\pi+4)-(2+\pi)(12)}{2(\pi+4)}\\&=\frac{6(\pi+4)-(2+\pi)(6)}{\pi+4}\\&=\frac{6\pi+24-12-6\pi}{\pi+4}\\&=\frac{12}{\pi+4}\\&=x\end{align}

Therefore, the area is a maximum when the radius is equal to the height of the rectangle.

Video Walkthrough

## 2019 Paper 2 Question 7

(a) A cattle feeding trough of uniform cross section and $$2.5\mbox{ m}$$ in length, is shown in Figure 1.
The front of the trough (segment $$ABC$$) is shown in Figure 2.
The front of the trough is a segment of a circle of radius $$0\mbox{ cm}$$.
The height of the trough, $$|DB|$$, is $$30\mbox{ cm}$$.
$$|OA|=|OC|=|OB|=90\mbox{ cm}$$. $$[OB]\perp[AC]$$.

(i) Find $$|AD|$$. Give your answer in the form $$a\sqrt{B}\mbox{ cm}$$, where $$a,b\in\mathbb{Z}$$.

(ii) Find $$|\angle DOA|$$. Give your answer in radians, correct to $$2$$ decimal places.

(iii) Find the area of the segment $$ABC$$. Give your answer in $$\mbox{m}^2$$ correct to $$2$$ decimal places.

(iv) Find the volume of the trough. Give your answer in $$\mbox{m}^3$$, correct to $$2$$ decimal places.

(i) $$30\sqrt{5}\mbox{ cm}$$

(ii) $$0.84\mbox{ rad}$$

(iii) $$0.28\mbox{ m}^2$$

(iv) $$0.70\mbox{ m}^3$$

Solution

(i)

(ii)

(iii)

\begin{align}A&=\frac{1}{2}r^2\theta-\frac{1}{2}bh\\&=\frac{1}{2}(90^2)(2\times0.84)-\frac{1}{2}(2\times30\sqrt{5})(90-30)\\&\approx0.28\mbox{ m}^2\end{align}

(iv)

\begin{align}V&=Al\\&=(0.28)(2.5)\\&=0.70\mbox{ m}^3\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) A sand timer for games is shown in the diagram.
Each half of the timer consists of a hemisphere, a cylinder of height $$3.5\mbox{ cm}$$ and a cone of height $$1.5\mbox{ cm}$$.
All of the parts have a radius of $$1.25\mbox{ cm}$$.

(i) The upper half of the timer is full of sand.
Find the volume of sand in the upper half of the timer.
Give your answer in $$\mbox{cm}^3$$ correct to $$2$$ decimal places.

(ii) Sand flows from the top half of the timer into the bottom part.
As it flows the top surfaces in both parts remain level.
At a certain time, $$98\%$$ of the sand has flowed into the bottom half of the timer.
Find $$h$$, the height of the remaining sand (in the conical part of the top of the timer).
Give your answer in $$\mbox{cm}$$, correct to $$2$$ decimal places.

(i) $$23.73\mbox{ cm}^3$$

(ii) $$0.87\mbox{ cm}$$

Solution

(i)

\begin{align}V&=\frac{2}{3}\pi(1.25)^3+\pi(1.25^2)(3.5)+\frac{1}{3}\pi(1.25^2)(1.5)\\&\approx23.73\mbox{ cm}^3\end{align}

(ii)

\begin{align}\frac{r}{h}=\frac{1.25}{1.5}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}r=\frac{5h}{6}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{3}\pi \left(\frac{5h}{6}\right)^2(h)=(0.02)(23.73)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}h&=\sqrt[3]\frac{(36)(3)(0.02)(23.73)}{25\pi}\\&\approx0.87\mbox{ cm}\end{align}

Video Walkthrough

## 2018 Paper 1 Question 8(b)

The graph of the symmetric function $$f(x)=\dfrac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$$ is shown below.

(b) The co-ordinates of $$B$$ are $$\left(-1,\frac{1}{\sqrt{2\pi e}}\right)$$. Find the area of the shaded rectangle in the diagram above. Give your answer correct to $$3$$ decimal places.

$$0.484\mbox{ units}^2$$

Solution

\begin{align}A&=l\times w\\&=\left(\frac{1}{\sqrt{2\pi e}}\right)\times(1+1)\\&\approx0.484\mbox{ units}^2\end{align}

Video Walkthrough

## 2018 Paper 2 Question 7

A section of a garden railing is shown below. This section consists of nine cylindrical bars, labelled $$A$$ to $$I$$, with a solid sphere attached to the centre of the top of each bar.
The volume of each sphere from $$B$$ to $$E$$ is $$1.75$$ times the volume of the previous sphere.

(a) The radius of sphere $$A$$ is $$3\mbox{ cm}$$. Find the sum of the volumes of the five spheres $$A$$, $$B$$, $$C$$, $$D$$ and $$E$$.
Give your answer correct to the nearest $$\mbox{cm}^3$$.

$$2324\mbox{ cm}^3$$

Solution

\begin{align}V_A&=\frac{4}{3}\pi r^3\\&=\frac{4}{3}\pi(3^3)\\&=36\pi\end{align}

\begin{align}\downarrow\end{align}

\begin{align}S_5&=\frac{36\pi(1-1.75^5)}{1-1.75}\\&\approx2324\mbox{ cm}^3\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b)

(i) The surface area of sphere $$E$$ can be taken to be $$503\mbox{ cm}^2$$.
The height of the railing at $$E$$ (i.e. the sum of the heights of bar $$E$$ and sphere $$E$$) is $$1.2$$ metres.
Find the height of bar $$E$$, in $$\mbox{cm}$$, correct to $$1$$ decimal place.

(ii) The radius of each bar is $$1$$ cm. The volume of bar $$A$$ is $$71.3\pi\mbox{ cm}^3$$.
The heights of the bars $$A$$, $$B$$, $$C$$, $$D$$ and $$E$$ form an arithmetic sequence.
Find, in $$\mbox{cm}$$, the height of each bar.

(i) $$107.3\mbox{ cm}$$

(ii) $$71.3\mbox{ cm}, 80.3\mbox{ cm}, 89.3\mbox{ cm}, 98.3\mbox{ cm}, 107.3\mbox{ cm}$$

Solution

(i)

\begin{align}4\pi r^2=503\end{align}

\begin{align}\downarrow\end{align}

\begin{align}r&=\sqrt{\frac{503}{4\pi}}\\&=6.33…\end{align}

\begin{align}\downarrow\end{align}

\begin{align}h&=120-r\\&=120-2(6.33…)\\&\approx107.3\mbox{ cm}\end{align}

(ii)

\begin{align}V_A=\pi r^2 h_A\end{align}

\begin{align}\downarrow\end{align}

\begin{align}h_A&=\frac{V_A}{\pi r^2}\\&=\frac{71.3\pi}{\pi(1^2)}\\&=71.3\mbox{ cm}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}d&=\frac{107.3-71.3}{4}=9\end{align}

\begin{align}\downarrow\end{align}

\begin{align}71.3\mbox{ cm}&&80.3\mbox{ cm}&&89.3\mbox{ cm}&&98.3\mbox{ cm}&&107.3\mbox{ cm}\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(c) There is a wall on each side of the section of railing, as shown in the diagram below which is
not to drawn to scale. The distance from wall to wall is $$1.5\mbox{ m}$$. The distance from the wall to
bar $$A$$ is $$20\mbox{ cm}$$ and similarly from the other wall to bar $$I$$ is $$20\mbox{ cm}$$.
The radius of each bar is $$1\mbox{ cm}$$. The gap between each bar is identical.
Find the size of this gap.

$$11.5\mbox{ cm}$$

Solution

\begin{align}\mbox{Sum of gaps}&=150-20-20-9(2\times1)\\&=92\mbox{ cm}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\mbox{Gap}&=\frac{92}{8}\\&=11.5\mbox{ cm}\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(d) The sphere on bar $$A$$ and the sphere on bar $$B$$ are to be joined by a straight rod as shown in the diagram below which is not to drawn to scale.
Find the length of the shortest rod that will join sphere $$A$$ to sphere $$B$$.
Give your answer in $$\mbox{cm}$$, correct to $$1$$ decimal place.

$$10.0\mbox{ cm}$$

Solution

\begin{align}V_B&=1.75V_A\\&=1.75(36\pi)\\&=63\pi\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{4}{3}\pi r_B^3=63\pi\end{align}

\begin{align}\downarrow\end{align}

\begin{align}r_B&=\sqrt[3]{\frac{3(63)}{4}}\\&=3.62…\mbox{ cm}\end{align}

The base $$b$$ of the triangle shown has a length of

\begin{align}b&=11.5+(2\times1)\\&=13.5\mbox{ cm}\end{align}

The height $$h$$ of the triangle shown is instead

\begin{align}h&=(9-r_A)+r_B\\&=(9-3)+3.62\\&=9.62\mbox{ cm}\end{align}

The hypotenuse $$H$$ of the triangle is therefore

\begin{align}H&=\sqrt{b^2+h^2}\\&=\sqrt{13.5^2+9.62^2}\\&16.576…\mbox{ cm}\end{align}

Therefore, the rod length $$L$$ is

\begin{align}L&=H-r_A-r_B\\&=16.576…-3-3.62\\&\approx10.0\mbox{ cm}\end{align}

Video Walkthrough

## 2017 Paper 2 Question 5(d)

$$ABCD$$ is a rectangle.

$$F\in[AB]$$, $$G\in[BC]$$, $$[FD]\cap[AG]={E}$$, and $$FD \perp AG$$.

$$|AE|=12\mbox{ cm}$$, $$|EG|=27\mbox{ cm}$$, $$|FE|=5\mbox{ cm}$$, $$|AD|=31.2\mbox{ cm}$$ and $$|AB|=36\mbox{ cm}$$.

(d) Find the area of the quadrilateral $$GCDE$$.

$$680.4\mbox{ cm}^2$$

Solution

\begin{align}A_{GCDE}&=A_{ABCD}-A_{ABG}-A_{AFD}+A_{AFE}\\&=(31.2)(36)-\frac{1}{2}(36)(15)-\frac{1}{2}(31.2)(13)+\frac{1}{2}(5)(12)\\&=680.4\mbox{ cm}^2\end{align}

Video Walkthrough

## 2017 Paper 2 Question 7

Two solid cones, each of radius $$R\mbox{ cm}$$ and height $$R\mbox{ cm}$$ are welded together at their vertices and placed in the smallest possible hollow cylinder, as shown in Figure 1 below.

(a) Show that the capacity (volume) of the empty space in the cylinder is equal to the capacity of an empty sphere of radius $$R\mbox{ cm}$$ (Figure 2).

Solution

\begin{align}V&=\pi r^2h_1-2\left(\frac{1}{3}\pi r^2h_2\right)\\&=\pi R^2(2R)-2\left(\frac{1}{3}\pi R^2(R)\right)\\&=2\pi R^3-\frac{2}{3}\pi R^3\\&=\frac{4}{3}\pi R^3\end{align}

as required.

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) In the remainder of this question, $$R=12\mbox{ cm}$$. Water is poured into both the cylinder and the
sphere to a depth of 6 cm as shown below (Figure 3 and Figure 4 respectively).

(i) Find $$|AB|$$, the radius of the circular surface of the water in the sphere (Figure 4).
Give your answer in the form ܽ$$a\sqrt{b}\mbox{ cm}$$, where $$a,b\in\mathbb{N}$$.

(ii) Find $$|CD|$$, the radius of the cone at water level, as shown in Figure 3.

(iii) Verify that the area of the surface of the water in the sphere is equal to the area of the surface of the water in the cylinder.

(i) $$6\sqrt{3}\mbox{ cm}$$

(ii) $$6\mbox{ cm}$$

Solution

(i)

\begin{align}|AB|&=\sqrt{12^2-(12-6)^2}\\&=\sqrt{108}\\&=6\sqrt{3}\mbox{ cm}\end{align}

(ii)

\begin{align}\frac{r}{12}=\frac{6}{12}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}r=6\mbox{ cm}\end{align}

(iii)

\begin{align}A_C&=\pi(12^2)-\pi(6^2)\\&=108\pi\end{align}

and

\begin{align}A_S&=\pi(6\sqrt{3})^2\\&=108\pi\end{align}

as required.

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(c) The mathematician Cavalieri discovered that, at the same depth, the volume of water in the available space in the cylinder is equal to the volume of water in the sphere.
Use this discovery to find the volume of water in the sphere when the depth is $$6\mbox{ cm}$$.
Give your answer in terms of $$\pi$$.

$$360\pi\mbox{ cm}^3$$

Solution

\begin{align}V&=\pi(12^2)(6)-\left[\frac{1}{3}\pi(12^2)(12)-\frac{1}{3}\pi(6^2)(6)\right]\\&=360\pi\mbox{ cm}^3\end{align}

Video Walkthrough

## 2015 Paper 1 Question 3(a)

Let $$f(x)=-x^2+12x-27$$, $$x\in\mathbb{R}$$.

(a)

(i) Complete Table $$1$$ below.

$$x$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$

$$f(x)$$

$$0$$

$$5$$

$$8$$

(ii) Use Table $$1$$ and the trapezoidal rule to find the approximate area of the region bounded by the graph of $$f$$ and the $$x$$-axis.

(i)

$$x$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$

$$f(x)$$

$$0$$

$$5$$

$$8$$

$$9$$

$$8$$

$$5$$

$$0$$

(ii) $$35\mbox{ square units}$$

Solution

(i)

$$x$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$

$$f(x)$$

$$0$$

$$5$$

$$8$$

$$9$$

$$8$$

$$5$$

$$0$$

(ii)

\begin{align}A&=\frac{h}{2}[y_1+y_n+2(y_2+y_3+…+y_{n-1})]\\&=\frac{1}{2}[0+0+2(5+8+9+8+5)]\\&=35\mbox{ square units}\end{align}

Video Walkthrough

## 2015 Paper 2 Question 7(a) & (b)

A flat machine part consists of two circular ends attached to a plate, as shown (diagram not to scale).
The sides of the plate, $$HK$$ and $$PQ$$, are tangential to each circle.
The larger circle has centre $$A$$ and radius $$4r\mbox{ cm}$$.
The smaller circle has centre $$B$$ and radius $$r\mbox{ cm}$$.
The length of $$[HK]$$ is $$8r\mbox{ cm}$$ and $$|AB|=20\sqrt{73}\mbox{ cm}$$.

(a) Find $$r$$, the radius of the smaller circle. (Hint: Draw $$BT\parallel KH$$, $$T\in AH$$.)

$$20\mbox{ cm}$$

Solution

\begin{align}|AT|^2+|BT|^2=|AB|^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(3r)^2+(8r)^2=(20\sqrt{73})^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}9r^2+64r^2=29{,}200\end{align}

\begin{align}\downarrow\end{align}

\begin{align}73r^2=29{,}200\end{align}

\begin{align}\downarrow\end{align}

\begin{align}r&=\sqrt{\frac{29{,}200}{73}}\\&=20\mbox{ cm}\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) Find the area of the quadrilateral $$ABKH$$.

$$8{,}000\mbox{ cm}^2$$

Solution

\begin{align}A&=|BT|\times|BK|+\frac{1}{2}|AT|\times|BT|\\&=(8r)(r)+\frac{1}{2}(3r)(8r)\\&=8r^2+12r^2\\&=20r^2\\&=20(20^2)\\&=8{,}000\mbox{ cm}^2\end{align}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(c)

(i) Find $$|\angle HAP|$$, in degrees, correct to one decimal place.

(ii) Find the area of the machine part, correct to the nearest $$2\mbox{ cm}$$.

(i) $$138.9^{\circ}$$

(ii) $$28{,}833\mbox{ cm}^2$$

Solution

(i)

\begin{align}|\angle HAP|&=2\times|\angle HAB|\\&=2\tan^{-1}\left(\frac{8r}{3r}\right)\\&=2\tan^{-1}\left(\frac{8}{3}\right)\\&\approx138.9^{\circ}\end{align}

(ii)

\begin{align}A&=2(8000)+\pi (80^2)\left(\frac{360^{\circ}-138.9^{\circ}}{360^{\circ}}\right)+\pi (20^2)\left(\frac{138.9^{\circ}}{360^{\circ}}\right)\\&\approx28{,}833\mbox{ cm}^2\end{align}

Video Walkthrough