# Daily Maths Revision – Week 9 Walkthrough

If you’re struggling with this week’s questions, this walkthrough should help you get the skills you need.

This week, we’re exploring:

### Question 1: Similar Solids

Two shapes are similar if one is an enlargement of the other. The key to solving problems involving similar shapes is finding the scale factor.

The two prisms below are similar. The surface area of the smaller prism is 243cm2. Find the surface area of the larger prism.

First, we compare the two lengths we’re given. By dividing the larger length by the smaller length, we can find the scale factor for length.

Scale factor for length = $\frac{8}{6}$
= $\frac{4}{3}$

So, if we multiply any of the lengths on the smaller shape by this scale factor, we will get the corresponding length on the larger shape.

The area has been enlarged in two dimensions, so we need to square the scale factor for length to find the scale factor for area.

Scale factor for area = $\left( \frac{4}{3} \right)^2$
= $\frac{16}{9}$

Surface area of the larger shape = 243 × $\frac{16}{9}$
= 432cm2

Using the same shapes, find the volume of the smaller prism given that the volume of the larger prism is 320cm3.

The volume has been enlarged in three dimensions, so we need to cube the scale factor for length to find the scale factor for volume.

Scale factor for volume = $\left( \frac{4}{3} \right)^3$
= $\frac{64}{27}$

Volume of the smaller shape = 320 ÷ $\frac{64}{27}$
= 135cm3

This is the connection between the scale factors for length, area (including surface area) and volume:

Scale factor for length = $k$
Scale factor for area = $k^2$
Scale factor for volume = $k^3$

### Question 2: Pythagoras’ Theorem in 3D

When using Pythagoras’ theorem to solve problems in 3D, the most difficult part is visualising the problem. It can help to draw out the 2D triangles you are working with, as well as labelling the 3D shape. Colour coding can also help.

Find the length of AH in the cuboid below.

We will start by finding the length AD. We will use triangle ABD as shown below:

This right-angled triangle has shorter sides of 21cm and 12cm and we are looking for the hypotenuse.

AD2 = AB2 + BD2
AD2 = 212 + 122
AD2 = 585
AD = $\sqrt{585}$

Now, we can use this length with triangle ADH to find the length of AH.

This right-angled triangle has shorter sides of $\sqrt{585}$ cm and 16cm and we are looking for the hypotenuse.

AH2 = AD2 + DH2
AH2 = ($\sqrt{585}$)2 + 162
AH2 = 841
AH = 29cm

The length AH is 29cm.

### Question 3: Capture Recapture

Capture recapture is a statistics related question that involves equivalent fractions.

Essentially, a sample of a population is taken and marked somehow. This sample is then released into the population and, after some time, a second sample is taken. The size of the population is estimated by the number of marked items in the second sample.

It works on a few assumptions:

1. The size of the population is fixed – it can’t increase or decrease significantly between the two samples.
2. The population is contained. For example, this method wouldn’t work with wild birds as they are likely to migrate in the time between the two samples being taken.
3. The first sample has enough time to integrate with the population before the second sample is taken.

Cole catches 10 rabbits from a warren. He tags them and releases them back into the warren. The next day, Cole catches 20 rabbits from the warren. 4 of them are tagged. Estimate the number of rabbits in the warren.

Let’s call the whole population $x$. 10 of the population are tagged. We assume the same proportion of tagged rabbits are in the second sample. We can write these proportions as two fractions and set them equal.

\begin{aligned} && \frac{4}{20} &= \frac{10}{x} && \end{aligned}

Now, we need to solve this equation for $x$.

\begin{aligned} && \frac{4}{20} &= \frac{10}{x} && \\&\times x &&& \times x \\&& \frac{4}{20} \times x &= 10 && \\&\times 20 &&& \times 20 \\ && 4x &= 200 && \\&\div 4 &&& \div x \\ && x&=50 \end{aligned}

There are approximately 50 rabbits in the warren.

### Question 4: Changing the Subject of a Formula

To change the subject of a formula, we need to make sure we keep both sides of the formula balanced. Just like when you are solving equations, whatever you do to one side, you must do to the other.

Let’s make $x$ the subject of the formula below.

\begin{aligned} && \boldsymbol{x} &= \boldsymbol{\frac{a(m-x)}{2b-y}} && \end{aligned}

At first glance, it may look like $x$ is already the subject of this formula. However, there is a second occurrence of $x$ on the right-hand side of the formula. For $x$ to be the subject, it needs to be in the form $x$ = … with no other $x$ terms on the other side of the formula.

We start by multiplying by the denominator of the fraction.

\begin{aligned} && x &= \frac{a(m-x)}{2b-y} && \\&\times (2b-y) &&& \times (2b-y) \\ && x(2b-y) &= a(m-x) && \\&\text{expand} &&& \text{expand} \\ && 2bx -xy &= am-ax && \\&+ax &&& +ax \\ && 2bx -xy +ax&= am && \end{aligned}

Next, we need to factorise the $x$ out of the expression on the left hand side.

\begin{aligned} 2bx-xy+ax&=am\\x(2b-y+a)&=am \end{aligned}

Finally, we divide by the expression in the backets:

\begin{aligned} \boldsymbol{x} &= \boldsymbol{\frac{am}{2b-y+a}} \end{aligned}

### Question 5: Expanding Double Brackets

There are a lot of different methods to expand double brackets. We will look at the grid method as this is the simplest method.

Expand and simplify:

$\huge{ \boldsymbol{(2x+1)(3x-7)}}$

We start by drawing a grid. Write one of the brackets across the top and the other down the side – it doesn’t matter which way round.

Now, we simply multiply the term in each row by each column. Make sure you are careful with the negatives.

\begin{aligned} (2x+1)(3x-7)&=6x^2+3x-14x-7\\&=\boldsymbol{6x^2-11x-7} \end{aligned}

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