L.C. MATHS

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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...

Circles

Prisms

Cylinders

Spheres

Cones

Trapezoidal Rule

## 2022 Paper 1 Question 8(b)

(b) If a solid is made up of faces with straight edges, then the following identity is often true:

\begin{align}C-E+F=2\end{align}

where:

• $$C$$ is the number of corners
• $$E$$ is the number of edges
• $$F$$ is the number of faces

(i) The value of $$E$$ for a cube is $$12$$. Write down the values of $$C$$ and $$F$$ for a cube, and
show that $$C-E+F=2$$ for these values.

(ii) Each of the faces of a different solid is in the shape of a triangle of area $$5\mbox{ cm}^2$$.
This solid has $$12$$ corners ($$C$$) and $$30$$ edges ($$E$$), and $$C-E+F=2$$ for this solid.
Work out the surface area of this solid, in $$\mbox{cm}^2$$.

(iii) The surface of a third solid is made up of $$h$$ hexagons and $$p$$ pentagons, where $$h,p\in\mathbb{N}$$. For this solid:

\begin{align}\frac{6h+5p}{3}-\frac{6h+5p}{2}+h+p=2\end{align}

Use this equation to find the number of pentagons in the surface of this solid (that is, the value of $$p$$).

(i) $$C=8$$ and $$F=6$$

(ii) $$100\mbox{ cm}^2$$

(iii) $$p=12$$

Solution

(i)

\begin{align}C=8 && F=6\end{align}

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

\begin{align}C-E+F&=8-12+6\\&=2\end{align}

as required.

(ii)

\begin{align}C-E+F=2\end{align}

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

\begin{align}12-30+F=2\end{align}

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

\begin{align}F&=2-12+30\\&=20\end{align}

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

\begin{align}\mbox{Surface area}&=20\times5\\&=100\mbox{ cm}^2\end{align}

(iii)

\begin{align}\frac{6h+5p}{3}-\frac{6h+5p}{2}+h+p=2\end{align}

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

\begin{align}2(6h+5p)-3(6h+5p)+6h+6p=12\end{align}

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

\begin{align}12h+10p-18h-15p+6h+6p=12\end{align}

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

\begin{align}p=12\end{align}

Video Walkthrough

## 2022 Paper 2 Question 8(a) &(c)

The top of a particular lighthouse is in the shape of a hemisphere on top of a cylinder.
The hemisphere and the cylinder both have a radius of $$3\mbox{ m}$$.

(a)

(i) Find the volume of the hemisphere. Give your answer in $$\mbox{m}^3$$ in terms of $$\pi$$.

(ii) The volume of the cylinder is $$36\pi\mbox{ m}^3$$.

Work out $$h$$, the height of the cylinder.

(i) $$18\pi\mbox{ m}^3$$

(ii) $$4\mbox{ m}$$

Solution

(i)

\begin{align}V&=\frac{2}{3}\pi r^3\\&=\frac{2}{3}\pi(3^3)\\&=18\pi\mbox{ m}^3\end{align}

(ii)

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

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

\begin{align}(3^2)h=36\end{align}

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

\begin{align}h&=\frac{36}{3^2}\\&=4\mbox{ m}\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) Assume that the Fastnet lighthouse can be seen from anywhere within a circle of radius $$50\mbox{ km}$$.

(i) Work out the area of the circle within which the Fastnet lighthouse can be seen.
Give your answer correct to the nearest $$\mbox{km}^2$$.

(ii) $$50\mbox{ km}=27\mbox{ nautical miles}$$.

Use this to work out how many $$\mbox{km}$$ are in $$1\mbox{ nautical mile}$$.

Give your answer correct to $$4$$ significant figures.

(i) $$7{,}854\mbox{ km}^2$$

(ii) $$1.852\mbox{ km}$$

Solution

(i)

\begin{align}A&=\pi r^2\\&=\pi(50^2)\\&\approx7{,}854\mbox{ km}^2\end{align}

(ii)

\begin{align}50\mbox{ km}=27\mbox{ nautical miles}\end{align}

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

\begin{align}\frac{50}{27}\mbox{ km}=1\mbox{ nautical mile}\end{align}

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

\begin{align}\approx1.852\mbox{ km}=1\mbox{ nautical mile}\end{align}

Video Walkthrough

## 2022 Paper 2 Question 9(a) & (b)

Seán has built a shed. The diagrams below show the dimensions of Seán’s shed.
The shed is in the shape of a prism. Its front face is in the shape of a triangle on top of a rectangle.
Its highest point is directly above the centre of its base.

(a) State which of the following statements is most likely to be true, and write down a possible height of Seán that would support your answer.

• The shed at the highest point is $$3$$ times as high as Seán.
• The shed at the highest point is $$5$$ times as high as Seán.
• The shed at the highest point is $$8$$ times as high as Seán.

$$1.7\mbox{ m}$$

Solution

The shed at the highest point is $$5$$ times as high as Seán.

\begin{align}h&=\frac{8.5}{5}\\&=1.7\mbox{ m}\end{align}

Video Walkthrough
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(b) Seán says that his shed has a capacity of over one million litres, where $$1\mbox{ m}^3=1{,}000\mbox{ litres}$$.
Work out the volume of Seán’s shed, to show that he is correct.

$$V=1{,}674{,}000\mbox{ litres}$$ which is over one million litres.

Solution

\begin{align}V&=\left[(7\times12)+\frac{1}{2}(12\times1.5)\right]\times 18\\&=1{,}674\mbox{ m}^3\\&=1{,}674{,}000\mbox{ litres}\end{align}

which is indeed over one million litres.

Video Walkthrough

## 2021 Paper 1 Question 8(a) - (c)

A square sheet of cardboard, of side $$10$$ units, is used to make an open box.
Squares of side $$x$$ units, where $$x\in\mathbb{R}$$, are removed from each corner of the cardboard and it is then folded along the dotted lines, as shown in the diagram below, in order to create the box.

(a) The length ($$l$$), breadth ($$b$$) and height ($$h$$) of the box are shown in the diagram above.
Write $$l$$, $$b$$, and $$h$$ in terms of $$x$$.

$$l=10-2x$$, $$b=10-2x$$ and $$h=x$$

Solution

\begin{align}l=10-2x\end{align}

\begin{align}b=10-2x\end{align}

\begin{align}h=x\end{align}

Video Walkthrough
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(b) Show that the volume of the box can be written as

\begin{align}V(x)=4x^3-40x^2+100x\end{align}

Solution

\begin{align}V&=lbh\\&=(10-2x)(10-2x)(x)\\&=(10-2x)(10x-2x^2)\\&=100x-20x^2-20x^2+4x^3\\&=4x^3-40x^2+100x\end{align}

as required.

Video Walkthrough
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(c) Explain why a box of height $$6$$ units cannot be made from the sheet of cardboard.

The length and breadth would be negative, which is not possible.

Solution

The length and breadth would be negative, which is not possible.

Video Walkthrough

## 2021 Paper 1 Question 10(b)

(b)

(i) John walks around a circular trail of radius $$0.5\mbox{ km}$$ at a steady speed of $$6\mbox{ km/h}$$.
How long will it take him to complete $$3$$ full circuits of the trail?

(ii) Mary decides to walk every day over a $$5$$ day period.
She walks a distance of $$3\mbox{ km}$$ on day one.
She increases the length of her walk by $$15\%$$ each day for the next four days.
Her average speed on day $$5$$ is $$4\mbox{ km/h}$$.
Find how long it will take her to complete her walk on day $$5$$.

(iii) One day, during John’s walk he meets Mary at point $$P$$ on the trail.
Mary is walking in the opposite direction at a steady speed of $$4\mbox{ km/h}$$.
John continues walking at $$6\mbox{ km/h}$$.
How far will he travel until he meets Mary again?

(i) $$94\mbox{ min}$$

(ii) $$79\mbox{ min}$$

(iii) $$1{,}885\mbox{ m}$$

Solution

(i)

\begin{align}t&=3\times\frac{2\pi r}{s}\\&=3\times\frac{2\pi(0.5)}{6}\\&=1.5707..\mbox{ hr}\\&\approx94\mbox{ min}\end{align}

(ii)

\begin{align}t&=\frac{d}{s}\\&=\frac{3(1.15^4)}{4}\\&=1.2117…\mbox{ hr}\\&\approx79\mbox{ min}\end{align}

(iii)

\begin{align}d&=\frac{6}{10}\times(2\pi r)\\&=\frac{6}{10}\times[2\pi(0.5)]\\&\approx1{,}885\mbox{ m}\end{align}

Video Walkthrough

## 2021 Paper 2 Question 2(b)

(b) The figure $$ABCDE$$ shown in the diagram consists of a large square $$ACDE$$ standing on the diagonal $$[AC]$$ of a smaller square $$ABCF$$.
The smaller square has a side length of $$2\mbox{cm}$$.
Find the area and perimeter of the figure $$ABCDE$$.

$$\mbox{Area}=10\mbox{ cm}^2$$ and $$\mbox{Perimeter}=12.49\mbox{ cm}$$

Solution

\begin{align}|AC|&=\sqrt{|AB|^2+|BC|^2}\\&=\sqrt{2^2+2^2}\\&\sqrt{8}\end{align}

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

\begin{align}\mbox{Area}&=(\sqrt{8})^2+\frac{1}{2}(2\times2)\\&=10\mbox{ cm}^2\end{align}

and

\begin{align}\mbox{Perimeter}&=\sqrt{8}+\sqrt{8}+\sqrt{8}+2+2\\&\approx12.49\mbox{ cm}\end{align}

Video Walkthrough

## 2021 Paper 2 Question 8(a) & (b)

The diagram shows the plan of a lake.
The line segment $$[PQ]$$ represents the distance from the pier, $$P$$, to the far side of the lake.
At equal intervals of $$10\mbox{ m}$$ along this line, perpendicular measures are made to the sides of the lake as shown.

(a)

(i) Use the Trapezoidal Rule to estimate the surface area of the lake.

(ii) If the lake is on average $$8\mbox{ m}$$ deep, estimate the volume of water in the lake.

(i) $$2{,}050\mbox{ m}^2$$

(ii) $$16{,}400\mbox{ m}^3$$

Solution

(i)

\begin{align}A&=\frac{10}{2}\left[0+0+2(110)\right]+\frac{10}{2}\left[10+0+2(90)\right]\\\&=2{,}050\mbox{ m}^2\end{align}

(ii)

\begin{align}V&=(2{,}050)(8)\\&=16{,}400\mbox{ m}^3\end{align}

Video Walkthrough
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(b)

(i) Calculate the volume , in $$\mbox{cm}^3$$, of a hemisphere of radius $$21\mbox {cm}$$.
Give your answer terms of $$\pi$$.

(ii) A buoy on the lake is in the shape of a hemisphere of radius $$21\mbox{ cm}$$ surmounted by a cone.
The volume of the cone is equal to the volume of the hemisphere.
Find $$h$$, the total height of the buoy.

(i) $$6{,}174\pi\mbox{ cm}^3$$

(ii) $$63\mbox{ cm}$$

Solution

(i)

\begin{align}V&=\frac{2}{3}\pi r^3\\&=\frac{2}{3}\pi(21^3)\\&=6{,}174\pi\mbox{ cm}^3\end{align}

(ii)

\begin{align}\frac{1}{3}\pi r^2H=6{,}174\pi\end{align}

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

\begin{align}\frac{1}{3} (21^2)H=6{,}174\end{align}

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

\begin{align}H&=\frac{(3)(6{,}174)}{(21^2)}\\&=42\mbox{ cm}\end{align}

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

\begin{align}h&=H+r\\&=42+21\\&=63\mbox{ cm}\end{align}

Video Walkthrough

## 2021 Paper 2 Question 9(b)

(b) An aircraft flies from airport $$A$$ to airport $$B$$, and then on to airport $$C$$, at the same altitude.
The pilot records the flight summary on the given diagram.

(i) Find the distance from airport $$B$$ to airport $$C$$. Give your answer correct to the nearest $$\mbox{km}$$.

(ii) When the plane was directly over airport $$C$$, the pilot was instructed to “circle” until a runway was available. She therefore flew $$10\mbox{ km}$$ away from $$C$$ before turning and flying along a circular arc of $$70^{\circ}$$ and then returning to the airport.
This path was all flown at the same altitude.
Find the total distance travelled.
Give your answer, in $$\mbox{km}$$, correct to two decimal places.

(i) $$324\mbox{ km}$$

(ii) $$32.22\mbox{ km}$$

Solution

(i)

\begin{align}\frac{|BC|}{\sin47^{\circ}}=\frac{260}{\sin36^{\circ}}\end{align}

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

\begin{align}|BC|&=\frac{260(\sin47^{\circ})}{\sin36^{\circ}}\\&\approx324\mbox{ km}\end{align}

(ii)

\begin{align}d&=10+10+2\pi(10)\left(\frac{70^{\circ}}{360^{\circ}}\right)\\&\approx32.22\mbox{ km}\end{align}

Video Walkthrough

## 2020 Paper 2 Question 7(c)

A vertical mobile phone mast, $$[DC]$$, of height $$h\mbox{ m}$$, is secured with two cables: $$[AC]$$ of length $$x\mbox{ m}$$,
and $$[BC]$$ of length $$y\mbox{ m}$$, as shown in the diagram.
The angle of elevation to the top of the mast from $$A$$ is $$30^{\circ}$$ and from $$B$$ is $$45^{\circ}$$.

(c)

(i) The mast can provide a strong signal for an area in the shape of a regular hexagon of side $$8\mbox{ km}$$, as shown in the diagram.
Find the area of the hexagon.
Give your answer in $$\mbox{km}^2$$, correct to $$2$$ decimal places.

(ii) A circle which touches all vertices of the hexagon can show areas where the signal is weak. One of these areas is shaded in the diagram.
Give your answer in $$\mbox{km}^2$$, correct to $$1$$ decimal place.

(i) $$166.28\mbox{ km}^2$$

(ii) $$5.8\mbox{ km}^2$$

Solution

(i)

\begin{align}A&=6\times\left[\frac{1}{2}(8)(8)(\sin60^{\circ})\right]\\&\approx166.28\mbox{ km}^2\end{align}

(ii)

\begin{align}A&=\frac{\pi(8^2)-166.28}{6}\\&\approx5.8\mbox{ km}^2\end{align}

Video Walkthrough

## 2020 Paper 2 Question 9

A rectangular sheet of aluminium is used to make a cylindrical can of radius $$r\mbox{ cm}$$ and height
$$10\mbox{ cm}$$, as shown below. The aluminium does not overlap in the finished can.

(a)

(i) Show that $$r$$, the radius of the cylinder, is $$3\mbox{ cm}$$.

(ii) Find the distance $$y$$. Give your answer correct to the nearest centimetre.

(iii) Find the area, in $$\mbox{cm}^2$$, of the waste aluminium after the top, bottom and side of the cylindrical can have been removed from the rectangular sheet.
Give your answer correct to the nearest $$\mbox{cm}^2$$.

(ii) $$19\mbox{ cm}$$

(iii) $$57\mbox{ cm}^2$$

Solution

(i)

\begin{align}r&=\frac{16-10}{2}\\&=3\mbox{ cm}\end{align}

as required.

(ii)

\begin{align}y&=2\pi r\\&=2\pi(3)\\&\approx19\mbox{ cm}\end{align}

(iii)

\begin{align}A&=(16)(19)-(10)(19)-2\pi(3^2)\\&\approx57\mbox{ cm}^2\end{align}

Video Walkthrough
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(b)

(i) Find the volume of a spherical ice cube of radius $$1.5\mbox{ cm}$$.
Give your answer in terms of $$\pi$$.

(ii) Three of the spherical ice cubes of radius $$1.5\mbox {cm}$$ are added to a cylinder of internal radius $$3.5zmbox{ cm}$$ which is partially filled with water.
All of the ice cubes are completely submerged in the water and the water does not overflow.
Find the rise, $$h\mbox{ cm}$$, in the water level. Give your answer correct to $$1$$ decimal place.

(i) $$4.5\pi\mbox{ cm}^3$$

(ii) $$1.1\mbox{ cm}$$

Solution

(i)

\begin{align}V&=\frac{4}{3}\pi r^3\\&=\frac{4}{3}\pi(1.5^3)\\&=4.5\pi\mbox{ cm}^3\end{align}

(ii)

\begin{align}3\times 4.5\pi=\pi(3.5^2)h\end{align}

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

\begin{align}h&=\frac{3\times4.5}{3.5^2}\\&\approx1.1\mbox{ cm}\end{align}

Video Walkthrough

## 2019 Paper 1 Question 5(a)

Harry draws a scale diagram of the portion of his garden that is covered in lawn.
His diagram is shown below.
Each box on the grid is $$1\mbox{ cm}\times 0.5\mbox{ cm}$$.
Each $$\mbox{cm}$$ on Harry’s diagram represents $$3\mbox{ m}$$.
In order to estimate the area of the lawn Harry divides the diagram into eight sections.

(a) Use the trapezoidal rule to estimate the area of the lawn using the scale: $$1\mbox{ cm}=3\mbox{ cm}$$.

$$324\mbox{ m}^2$$

Solution

\begin{align}A&=\frac{3}{2}[0+0+2(15+18+15+12+15+18+15)]\\&=324\mbox{ m}^2\end{align}

Video Walkthrough

## 2019 Paper 2 Question 5

(a) The crescent, shown in the shaded part of the diagram, was created by removing a disc of radius $$2.5\mbox{ cm}$$ from a disc of radius $$3\mbox{ cm}$$.
Find the area and the perimeter of the crescent.
Give each answer correct to two decimal places.

$$\mbox{Area}=8.64\mbox{ cm}^2$$ and $$\mbox{Perimeter}=34.56\mbox{ cm}$$

Solution

\begin{align}\mbox{Area}&=\pi(3^2)-\pi(2.5^2)\\&\approx8.64\mbox{ cm}^2\end{align}

and

\begin{align}\mbox{Perimeter}&=2\pi(3)+2\pi(2.5)\\&\approx34.56\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) An empty inverted cone of vertical height $$12\mbox{ cm}$$ and radius $$7\mbox{ cm}$$ is filled with water from a pipe.
The water flows from the pipe at a steady rate of $$0.5$$ litres per minute.
Find the time it takes to fill the cone.

$$74\mbox{ s}$$

Solution

\begin{align}V&=\frac{1}{3}\pi r^2h\\&=\frac{1}{3}\pi(7^2)(12)\\&=196\pi\mbox{ cm}^3\end{align}

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

\begin{align}t&=\frac{196\pi}{0.5\times1{,}000}\\&=1.2315..\mbox{ min}\\&\approx74\mbox{ s}\end{align}

Video Walkthrough

## 2019 Paper 2 Question 8(a)

(a) A solid sphere of radius $$3\mbox{ cm}$$ is placed inside a cylinder and then water is poured into the cylinder until it is full, as shown in the diagram.

(i) Find the volume of the sphere, in terms of $$\pi$$.

(ii) The sphere is now removed. The internal radius of the cylinder is $$5\mbox{ cm}$$.
Find the drop, in $$\mbox{cm}$$, in the height of the water.

(iii) The cylinder has a height of $$18\mbox{ cm}$$. The curved surface of the cylinder is cut from a rectangular piece of metal measuring $$35\mbox{ cm}$$ by $$20\mbox{ cm}$$, as shown. Find how much metal will be left over when the curved surface of the cylinder is cut out.
Give your answer correct to $$1$$ decimal place.

(i) $$36\pi\mbox{ cm}^3$$

(ii) $$\dfrac{36}{25}\mbox{ cm}$$

(iii) $$34.5\mbox{ cm}^2$$

Solution

(i)

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

(ii)

\begin{align}36\pi=\pi(5^2)h\end{align}

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

\begin{align}h&=\frac{36}{5^2}\\&=\frac{36}{25}\mbox{ cm}\end{align}

(iii)

\begin{align}35\times20-2\pi(5)(18)\approx134.5\mbox{ cm}^2\end{align}

Video Walkthrough

## 2019 Paper 2 Question 9(d)

(d)

(i) Consider the triangle $$RST$$.
Use the Cosine Rule to find an expression for $$\cos \theta$$,
where $$\theta$$ is the measure of the angle $$TRS$$.
Hence show that $$\theta=106^{\circ}$$, correct to the nearest degree.

(ii) John sails directly from $$S$$ to $$T$$. Mary sails from $$S$$ to $$T$$ along the minor arc $$ST$$.
Find the difference between the distance that John sails and the distance that Mary sails. Give your answer correct to the nearest $$\mbox{ km}$$.

(iii) The sea in this region is estimated to have an average of $$1$$ ship per $$25$$ square kilometres at any time.
Use this estimate to find the number of ships in the sector $$RST$$.

(i) $$\cos\theta=-\dfrac{7}{25}$$

(ii) $$25\mbox{ km}$$

(iii) $$370$$

Solution

(i)

\begin{align}|TS|&=2|QS|\\&=2(80)\\&=160\mbox{ km}\end{align}

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

\begin{align}160^2=100^2+100^2-2(100)(100)\cos\theta\end{align}

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

\begin{align}\cos\theta&=\frac{160^2-100^2-100^2}{2(100)(100)}\\&=-\frac{7}{25}\end{align}

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

\begin{align}\theta&=\cos^{-1}\left(\frac{7}{25}\right)\\&\approx106^{\circ}\end{align}

as required.

(ii)

\begin{align}2\pi(100)\times\frac{106^{\circ}}{360^{\circ}}-160\approx25\mbox{ km}\end{align}

(iii)

\begin{align}\mbox{Area}&=\pi(100)^2\times\frac{106^{\circ}}{360^{\circ}}\\&=9{,}250.245…\mbox{ km}^2\end{align}

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

\begin{align}N&=\frac{9{,}250.245…}{25}\\&\approx370\end{align}

Video Walkthrough

## 2018 Paper 2 Question 5

(a) The square $$ABCD$$ has sides of length $$7\mbox{ cm}$$.
The vertices of the square $$PQRS$$ lie on the perimeter of $$ABCD$$, as shown in the diagram, with $$|AQ|=5\mbox{ cm}$$.
Find the area of the square $$PQRS$$.

$$29\mbox{ cm}^2$$

Solution

\begin{align}|PQ|&=\sqrt{|AP|^2+|AQ|^2}\\&=\sqrt{2^2+5^2}\\&=\sqrt{29}\end{align}

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

\begin{align}A&=|PQ|^2\\&=(\sqrt{29})^2\\&=29\mbox{ cm}^2\end{align}

Video Walkthrough